Chapter 9: Eigenvalue Problems and Spectral Theory — How Measurement Works Mathematically

Opening: The Mathematical Heart of Measurement

"The physical question 'What will I measure?' translates exactly into the mathematical question 'What are the eigenvalues?'"

In Chapter 6, you learned the measurement postulate: when you measure an observable $\hat{A}$, the only possible outcomes are its eigenvalues, and the system collapses into the corresponding eigenstate. In Chapter 8, you learned Dirac notation — the language that makes this postulate computationally powerful. This chapter is where those two threads meet.

The eigenvalue problem — finding the values $a$ and states $|a\rangle$ such that $\hat{A}|a\rangle = a|a\rangle$ — is the single most important mathematical problem in quantum mechanics. Every measurement you will ever compute, every energy level you will ever predict, every transition probability you will ever calculate begins with solving an eigenvalue equation. Perturbation theory (Chapters 17--18) is about approximately solving eigenvalue problems. Variational methods (Chapter 19) are about bounding eigenvalues from above. Scattering theory (Chapter 22) involves the continuous eigenvalues of the Hamiltonian. Quantum computing (Chapters 25, 40) encodes information in eigenstates.

This chapter develops the eigenvalue problem systematically, starting from the discrete case you saw in Chapter 8 (spin-1/2, harmonic oscillator) and extending to the continuous case (position and momentum eigenstates). Along the way, we will need the Dirac delta function — a mathematical object that is "not a function" in the classical sense, but is absolutely indispensable. We will state the spectral theorem, the master result that guarantees every observable can be decomposed into its eigenvalues and projection operators. We will derive the Fourier transform as a natural change of basis. And we will close by explaining — at a conceptual level — why the proper mathematical framework for all of this is not the Hilbert space alone, but a larger structure called the rigged Hilbert space.

By the end of this chapter, you will understand the mathematical machinery that makes quantum measurement precise.

9.1 The Eigenvalue Equation for Observables

The fundamental equation

Let $\hat{A}$ be a Hermitian operator representing an observable (energy, position, momentum, spin component, angular momentum — any measurable quantity). The eigenvalue equation is:

$$\hat{A}|a\rangle = a|a\rangle$$

Here: - $|a\rangle$ is an eigenstate (or eigenvector, or eigenket) of $\hat{A}$. - $a$ is the corresponding eigenvalue — a real number (guaranteed by the Hermiticity of $\hat{A}$, as we proved in Chapter 8). - The label $a$ on the ket $|a\rangle$ is the eigenvalue itself. This is standard Dirac convention: we label eigenstates by their eigenvalues.

The eigenvalue equation says: when $\hat{A}$ acts on $|a\rangle$, the result is just a scalar multiple of $|a\rangle$. The state $|a\rangle$ is "unchanged in direction" by $\hat{A}$ — it is merely scaled. This is the defining property of an eigenvector.

💡 Key Insight — The eigenvalue equation connects mathematics to physics through the measurement postulate. If you measure $\hat{A}$ on a system in state $|a\rangle$, the outcome is $a$ with certainty (probability 1). The eigenvalues are the possible measurement results. The eigenstates are the states of definite value for that observable.

Why Hermiticity matters

In Chapter 8, we proved two consequences of Hermiticity ($\hat{A} = \hat{A}^\dagger$):

1. Eigenvalues are real. If $\hat{A}|a\rangle = a|a\rangle$, then $a \in \mathbb{R}$. This is physically necessary — measurement outcomes must be real numbers.

2. Eigenstates corresponding to distinct eigenvalues are orthogonal. If $\hat{A}|a\rangle = a|a\rangle$ and $\hat{A}|a'\rangle = a'|a'\rangle$ with $a \neq a'$, then $\langle a|a'\rangle = 0$.

Let us re-derive property (2) as a warm-up, since the technique is fundamental.

Proof of orthogonality. Start with $\hat{A}|a'\rangle = a'|a'\rangle$. Take the inner product with $\langle a|$:

$$\langle a|\hat{A}|a'\rangle = a'\langle a|a'\rangle$$

Now use Hermiticity. Since $\hat{A} = \hat{A}^\dagger$, we have $\langle a|\hat{A} = a\langle a|$ (the eigenvalue equation for the bra). Therefore:

$$\langle a|\hat{A}|a'\rangle = a\langle a|a'\rangle$$

Equating the two expressions:

$$a\langle a|a'\rangle = a'\langle a|a'\rangle \implies (a - a')\langle a|a'\rangle = 0$$

Since $a \neq a'$, we must have $\langle a|a'\rangle = 0$. $\square$

🔄 Spaced Review (Chapter 6) — In Chapter 6, we proved these same properties using wave mechanics: $\int \psi_a^*(x) \hat{A} \psi_{a'}(x) \, dx$. The Dirac notation proof is cleaner because it works for any Hilbert space, not just $L^2(\mathbb{R})$. The physics is identical; the notation is more general.

Degeneracy

When two or more linearly independent eigenstates share the same eigenvalue, the eigenvalue is said to be degenerate. If $\hat{A}|a, r\rangle = a|a, r\rangle$ for $r = 1, 2, \ldots, g_a$, then $a$ has degeneracy $g_a$, and the eigenstates $\{|a, 1\rangle, |a, 2\rangle, \ldots, |a, g_a\rangle\}$ span a $g_a$-dimensional subspace called the eigenspace of $a$.

Within a degenerate eigenspace, the eigenstates are not automatically orthogonal — any linear combination of degenerate eigenstates is again an eigenstate with the same eigenvalue. However, we can always choose an orthonormal set via the Gram-Schmidt procedure. We assume this has been done: within each eigenspace, the eigenstates are orthonormal.

Example: Hydrogen atom. The energy eigenvalue $E_n = -13.6\,\text{eV}/n^2$ (introduced in Chapter 2 and solved in Chapter 5) has degeneracy $g_n = n^2$ (ignoring spin). For $n = 2$, there are four degenerate states: $|2,0,0\rangle$, $|2,1,-1\rangle$, $|2,1,0\rangle$, $|2,1,1\rangle$, all with the same energy. The additional labels $l$ and $m$ (from the angular momentum operators $\hat{L}^2$ and $\hat{L}_z$) break the degeneracy by distinguishing states within the eigenspace. This is the first hint that we often need a complete set of commuting observables (CSCO) to fully label eigenstates — a concept we will develop in Chapter 10.

🔗 Connection — Degenerate perturbation theory (Chapter 18) is specifically about what happens when a perturbation lifts the degeneracy of an eigenvalue. The hydrogen fine structure is the paradigmatic example: the $n = 2$ states, all degenerate in the Coulomb potential, split into distinct energy levels when spin-orbit coupling is included.

9.2 Discrete Spectra: Bound States Revisited in Dirac Notation

The discrete eigenvalue problem

An operator $\hat{A}$ has a discrete spectrum if its eigenvalues form a countable set $\{a_1, a_2, a_3, \ldots\}$ (finite or countably infinite). The eigenstates satisfy:

$$\hat{A}|a_n\rangle = a_n|a_n\rangle, \qquad n = 1, 2, 3, \ldots$$

with the orthonormality condition:

$$\langle a_m|a_n\rangle = \delta_{mn}$$

and the completeness relation:

$$\sum_n |a_n\rangle\langle a_n| = \hat{I}$$

These three equations — the eigenvalue equation, orthonormality, and completeness — are the triad that defines a complete, orthonormal eigenbasis for $\hat{A}$.

Spin-1/2: The paradigm of the discrete spectrum

The simplest non-trivial example is the spin-1/2 system, our running example from Chapter 8. Let us work through the eigenvalue problem for all three spin operators systematically.

$\hat{S}_z$ eigenstates (the standard basis):

$$\hat{S}_z|\uparrow\rangle = +\frac{\hbar}{2}|\uparrow\rangle, \qquad \hat{S}_z|\downarrow\rangle = -\frac{\hbar}{2}|\downarrow\rangle$$

The eigenvalues are $\pm\hbar/2$. In the matrix representation (using the $\hat{S}_z$ eigenbasis):

$$S_z = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \frac{\hbar}{2}\sigma_z$$

This is already diagonal — as it must be, since we are working in its own eigenbasis.

$\hat{S}_x$ eigenstates:

$$S_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \frac{\hbar}{2}\sigma_x$$

The eigenvalue equation $S_x|\chi\rangle = \lambda|\chi\rangle$ gives the characteristic equation:

$$\det(S_x - \lambda I) = 0 \implies \det\begin{pmatrix} -\lambda & \hbar/2 \\ \hbar/2 & -\lambda \end{pmatrix} = \lambda^2 - \frac{\hbar^2}{4} = 0$$

So $\lambda = \pm\hbar/2$ — the same eigenvalues as $\hat{S}_z$. The eigenstates are:

$$|+\rangle_x = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle)$$

$$|-\rangle_x = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}}(|\uparrow\rangle - |\downarrow\rangle)$$

You can verify: $S_x|+\rangle_x = +(\hbar/2)|+\rangle_x$ and $S_x|-\rangle_x = -(\hbar/2)|-\rangle_x$.

✅ Checkpoint — Verify by direct matrix multiplication that $S_x|+\rangle_x = +(\hbar/2)|+\rangle_x$. If this does not work out, review matrix multiplication in Section 8.4 before continuing.

$\hat{S}_y$ eigenstates:

$$S_y = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \frac{\hbar}{2}\sigma_y$$

By the same procedure, the eigenvalues are again $\pm\hbar/2$, and the eigenstates are:

$$|+\rangle_y = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix} = \frac{1}{\sqrt{2}}(|\uparrow\rangle + i|\downarrow\rangle)$$

$$|-\rangle_y = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix} = \frac{1}{\sqrt{2}}(|\uparrow\rangle - i|\downarrow\rangle)$$

💡 Key Insight — All three spin operators have the same eigenvalues ($\pm\hbar/2$) but different eigenstates. The $\hat{S}_x$ eigenstates are superpositions of the $\hat{S}_z$ eigenstates with real coefficients; the $\hat{S}_y$ eigenstates involve imaginary coefficients. This is why a state can be "spin-up along $z$" while being a perfect 50/50 superposition of "spin-up" and "spin-down" along $x$: $|\uparrow\rangle = \frac{1}{\sqrt{2}}(|+\rangle_x + |-\rangle_x)$.

The quantum harmonic oscillator: a countably infinite discrete spectrum

The QHO (Chapter 4) provides the paradigmatic example of a discrete spectrum with infinitely many eigenvalues:

$$\hat{H}|n\rangle = E_n|n\rangle = \left(n + \frac{1}{2}\right)\hbar\omega|n\rangle, \qquad n = 0, 1, 2, \ldots$$

In Dirac notation, the ladder operators from Chapter 8 give the elegant algebraic relations:

$$\hat{a}|n\rangle = \sqrt{n}|n-1\rangle, \qquad \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$$

$$\hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right) = \hbar\omega\left(\hat{N} + \frac{1}{2}\right)$$

where $\hat{N} = \hat{a}^\dagger\hat{a}$ is the number operator with eigenvalues $n = 0, 1, 2, \ldots$

The eigenstates satisfy $\langle m|n\rangle = \delta_{mn}$ and $\sum_{n=0}^{\infty}|n\rangle\langle n| = \hat{I}$. Any state in the QHO Hilbert space can be expanded:

$$|\psi\rangle = \sum_{n=0}^{\infty} c_n|n\rangle, \qquad c_n = \langle n|\psi\rangle, \qquad \sum_{n=0}^{\infty}|c_n|^2 = 1$$

🔄 Spaced Review (Chapter 8) — In Chapter 8, we introduced ladder operators and showed that $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$ by demanding that $\langle n+1|n+1\rangle = 1$. Verify this by computing $\langle n|\hat{a}\hat{a}^\dagger|n\rangle$ using the commutation relation $[\hat{a}, \hat{a}^\dagger] = 1$.

Expanding a general state in the eigenbasis

Given the eigenvalue equation $\hat{A}|a_n\rangle = a_n|a_n\rangle$ and completeness $\sum_n |a_n\rangle\langle a_n| = \hat{I}$, any state $|\psi\rangle$ can be expanded:

$$|\psi\rangle = \hat{I}|\psi\rangle = \sum_n |a_n\rangle\langle a_n|\psi\rangle = \sum_n c_n|a_n\rangle$$

where $c_n = \langle a_n|\psi\rangle$ is the probability amplitude for measuring $a_n$. The probability of obtaining $a_n$ is:

$$P(a_n) = |c_n|^2 = |\langle a_n|\psi\rangle|^2$$

The expectation value of $\hat{A}$ in state $|\psi\rangle$ is:

$$\langle\hat{A}\rangle = \langle\psi|\hat{A}|\psi\rangle = \sum_n a_n|c_n|^2 = \sum_n a_n P(a_n)$$

This is the weighted average of eigenvalues, weighted by probabilities — exactly the formula for the expected value of a random variable in probability theory.

Worked example. A spin-1/2 particle is prepared in the state $|\psi\rangle = \frac{1}{\sqrt{3}}|\uparrow\rangle + \sqrt{\frac{2}{3}}|\downarrow\rangle$. What are the possible outcomes of measuring $S_z$, with what probabilities?

Solution. The eigenvalues of $\hat{S}_z$ are $\pm\hbar/2$. The expansion coefficients are $c_\uparrow = \langle\uparrow|\psi\rangle = 1/\sqrt{3}$ and $c_\downarrow = \langle\downarrow|\psi\rangle = \sqrt{2/3}$.

$$P(+\hbar/2) = |c_\uparrow|^2 = \frac{1}{3}, \qquad P(-\hbar/2) = |c_\downarrow|^2 = \frac{2}{3}$$

$$\langle\hat{S}_z\rangle = \frac{1}{3}\cdot\frac{\hbar}{2} + \frac{2}{3}\cdot\left(-\frac{\hbar}{2}\right) = -\frac{\hbar}{6}$$

Now suppose we measure $S_x$ instead. We need to expand $|\psi\rangle$ in the $\hat{S}_x$ eigenbasis:

$$c_{+x} = {}_{x}\langle+|\psi\rangle = \frac{1}{\sqrt{2}}(\langle\uparrow| + \langle\downarrow|)\left(\frac{1}{\sqrt{3}}|\uparrow\rangle + \sqrt{\frac{2}{3}}|\downarrow\rangle\right) = \frac{1}{\sqrt{6}} + \frac{\sqrt{2}}{\sqrt{6}} = \frac{1 + \sqrt{2}}{\sqrt{6}}$$

$$P(+\hbar/2)_x = |c_{+x}|^2 = \frac{(1 + \sqrt{2})^2}{6} = \frac{3 + 2\sqrt{2}}{6} \approx 0.971$$

This illustrates a profound point: the same state $|\psi\rangle$ gives very different probabilities depending on which observable we choose to measure. The state is not uncertain in some vague sense — it encodes precise probabilities for every possible measurement, and these probabilities depend on the eigenbasis of the measured observable.

The variance and uncertainty

The variance of the measurement of $\hat{A}$ in state $|\psi\rangle$ is:

$$(\Delta A)^2 = \langle\hat{A}^2\rangle - \langle\hat{A}\rangle^2 = \sum_n a_n^2 |c_n|^2 - \left(\sum_n a_n|c_n|^2\right)^2$$

The uncertainty (standard deviation) is $\Delta A = \sqrt{(\Delta A)^2}$.

For a state that is an eigenstate of $\hat{A}$, say $|\psi\rangle = |a_k\rangle$, we have $c_n = \delta_{nk}$, so $\langle\hat{A}\rangle = a_k$ and $\langle\hat{A}^2\rangle = a_k^2$, giving $\Delta A = 0$. An eigenstate has zero uncertainty in the corresponding observable — the measurement always yields the same result. This is consistent with the measurement postulate: if you measure $\hat{A}$ on an eigenstate $|a_k\rangle$, you get $a_k$ with probability 1.

Worked example: Uncertainty of $\hat{S}_x$ in $|\uparrow\rangle$. A spin-1/2 particle is in the state $|\uparrow\rangle$ (eigenstate of $\hat{S}_z$). What is the uncertainty in a measurement of $\hat{S}_x$?

Since $|\uparrow\rangle = \frac{1}{\sqrt{2}}(|+\rangle_x + |-\rangle_x)$, we have $P(+\hbar/2)_x = P(-\hbar/2)_x = 1/2$.

$$\langle\hat{S}_x\rangle = \frac{1}{2}\cdot\frac{\hbar}{2} + \frac{1}{2}\cdot\left(-\frac{\hbar}{2}\right) = 0$$

$$\langle\hat{S}_x^2\rangle = \frac{1}{2}\cdot\left(\frac{\hbar}{2}\right)^2 + \frac{1}{2}\cdot\left(-\frac{\hbar}{2}\right)^2 = \frac{\hbar^2}{4}$$

$$\Delta S_x = \sqrt{\frac{\hbar^2}{4} - 0} = \frac{\hbar}{2}$$

The spin-up state $|\uparrow\rangle$ has maximum uncertainty in $S_x$: each measurement gives $+\hbar/2$ or $-\hbar/2$ with equal probability, and the standard deviation equals the magnitude of each eigenvalue. Compare this with $\Delta S_z = 0$ for the same state. This is the uncertainty principle for spin: $\Delta S_x \cdot \Delta S_y \geq \frac{\hbar}{2}|\langle\hat{S}_z\rangle| = \frac{\hbar^2}{4}$.

💡 Key Insight — The eigenvalue problem connects three seemingly different questions: (1) "What values can I measure?" (eigenvalues), (2) "What is the probability of each outcome?" (expansion coefficients squared), and (3) "How uncertain is the outcome?" (variance from the expansion coefficients). All three are answered by the same calculation: expand $|\psi\rangle$ in the eigenbasis of the observable being measured.

Simultaneous eigenstates and commuting observables

A natural question arises: can a state be an eigenstate of two different observables simultaneously? In general, no — but when two operators commute, the answer is yes.

Theorem. If $[\hat{A}, \hat{B}] = 0$, then $\hat{A}$ and $\hat{B}$ share a complete set of simultaneous eigenstates. That is, there exist states $|a, b\rangle$ such that $\hat{A}|a, b\rangle = a|a, b\rangle$ and $\hat{B}|a, b\rangle = b|a, b\rangle$.

Proof sketch. Let $\hat{A}|a\rangle = a|a\rangle$. Then $\hat{A}(\hat{B}|a\rangle) = \hat{B}(\hat{A}|a\rangle) = a(\hat{B}|a\rangle)$, so $\hat{B}|a\rangle$ is also an eigenstate of $\hat{A}$ with eigenvalue $a$. If $a$ is non-degenerate, $\hat{B}|a\rangle$ must be proportional to $|a\rangle$, which means $|a\rangle$ is also an eigenstate of $\hat{B}$. If $a$ is degenerate, $\hat{B}$ maps the eigenspace of $a$ to itself, and within this eigenspace we can diagonalize $\hat{B}$ to find simultaneous eigenstates. $\square$

Example: $\hat{L}^2$ and $\hat{L}_z$ commute (you will prove this in Chapter 12). Their simultaneous eigenstates are the spherical harmonics $|l, m\rangle$, with $\hat{L}^2|l, m\rangle = \hbar^2 l(l+1)|l, m\rangle$ and $\hat{L}_z|l, m\rangle = m\hbar|l, m\rangle$.

Counter-example: $\hat{S}_x$ and $\hat{S}_z$ do not commute ($[\hat{S}_x, \hat{S}_z] = -i\hbar\hat{S}_y \neq 0$). There is no state that is simultaneously an eigenstate of both. This is why a particle in $|\uparrow\rangle$ (eigenstate of $\hat{S}_z$) has $\Delta S_x = \hbar/2 \neq 0$ — measuring $S_x$ on such a state gives a genuinely random outcome.

🔗 Connection — The concept of commuting observables leads directly to the complete set of commuting observables (CSCO), developed in Chapter 10. A CSCO is a maximal set of mutually commuting Hermitian operators whose simultaneous eigenstates are uniquely labeled (no residual degeneracy). For the hydrogen atom, $\{\hat{H}, \hat{L}^2, \hat{L}_z, \hat{S}_z\}$ is a CSCO. Every state is uniquely labeled by the quantum numbers $|n, l, m, m_s\rangle$.

9.3 Continuous Spectra: Position and Momentum Eigenstates

The essential difference

Not all observables have discrete spectra. The position operator $\hat{x}$ and the momentum operator $\hat{p}$ have continuous spectra: their eigenvalues range over all real numbers.

The eigenvalue equations are:

$$\hat{x}|x\rangle = x|x\rangle, \qquad x \in \mathbb{R}$$

$$\hat{p}|p\rangle = p|p\rangle, \qquad p \in \mathbb{R}$$

These equations look just like the discrete case, but the continuous nature of the spectrum changes everything:

The crucial difference is the appearance of the Dirac delta function $\delta(x - x')$ in place of the Kronecker delta $\delta_{mn}$. This is not a mere notational change — it signals that position eigenstates $|x\rangle$ are not normalizable. They are not elements of the Hilbert space $L^2(\mathbb{R})$. They are, in a precise sense, "idealized" states that no physical system can occupy. Yet they are indispensable mathematical tools.

⚠️ Common Misconception — Students often think that $|x\rangle$ represents a particle localized at position $x$. In a sense it does, but $|x\rangle$ is not a physically realizable state — it has infinite norm and infinite uncertainty in momentum. A physical particle localized "near $x$" is a narrow wave packet, not a position eigenstate. The position eigenstate is the mathematical limit of an infinitely narrow packet. We use it as a computational tool, not as a physical state.

Position eigenstates and the wave function

The completeness relation for position eigenstates is:

$$\int_{-\infty}^{\infty} |x\rangle\langle x| \, dx = \hat{I}$$

Apply this to any ket $|\psi\rangle$:

$$|\psi\rangle = \int_{-\infty}^{\infty} |x\rangle\langle x|\psi\rangle \, dx = \int_{-\infty}^{\infty} \psi(x)|x\rangle \, dx$$

where we have defined:

$$\psi(x) \equiv \langle x|\psi\rangle$$

This is the wave function — the object you have been working with since Chapter 2. The Dirac notation reveals what it really is: the expansion coefficient of $|\psi\rangle$ in the position eigenbasis. The wave function $\psi(x)$ is to $|x\rangle$ what $c_n$ is to $|a_n\rangle$.

The normalization condition $\langle\psi|\psi\rangle = 1$ becomes:

$$\langle\psi|\psi\rangle = \int \langle\psi|x\rangle\langle x|\psi\rangle \, dx = \int \psi^*(x)\psi(x) \, dx = \int |\psi(x)|^2 \, dx = 1$$

This is exactly the normalization condition from Chapter 2, derived here by inserting a complete set of position eigenstates.

Momentum eigenstates

Similarly, the completeness relation for momentum eigenstates is:

$$\int_{-\infty}^{\infty} |p\rangle\langle p| \, dp = \hat{I}$$

and the momentum-space wave function is:

$$\phi(p) \equiv \langle p|\psi\rangle$$

The overlap between position and momentum eigenstates is one of the most important formulas in quantum mechanics:

$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$

This was introduced in Chapter 8 and derived from the requirement that $\hat{p}|p\rangle = p|p\rangle$ in the position representation (i.e., $-i\hbar\frac{d}{dx}\langle x|p\rangle = p\langle x|p\rangle$). Its complex conjugate gives:

$$\langle p|x\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{-ipx/\hbar}$$

🔄 Spaced Review (Chapter 7) — In Chapter 7, we used the momentum-space wave function to analyze wave packet propagation. A Gaussian wave packet has $\phi(p) = (2\pi\sigma_p^2)^{-1/4}\exp(-(p - p_0)^2/4\sigma_p^2)$, peaked at $p_0$ with width $\sigma_p$. The position-space wave packet is the Fourier transform. Now we see this is simply a change of basis — from $|p\rangle$ to $|x\rangle$ — mediated by the overlap $\langle x|p\rangle$.

The position-representation action of operators

Using the position basis, any operator equation can be translated into a differential equation. The position operator acts as:

$$\langle x|\hat{x}|\psi\rangle = x\langle x|\psi\rangle = x\psi(x)$$

The momentum operator acts as:

$$\langle x|\hat{p}|\psi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle = -i\hbar\frac{d\psi}{dx}$$

These are the position-representation forms you have been using since Chapter 2. The Dirac notation makes clear that they are derived quantities — consequences of the abstract operator equations and the choice of position basis.

The eigenvalue equation $\hat{H}|\psi\rangle = E|\psi\rangle$, projected onto $\langle x|$, becomes:

$$\langle x|\hat{H}|\psi\rangle = E\langle x|\psi\rangle \implies \left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\right]\psi(x) = E\psi(x)$$

This is the time-independent Schrodinger equation from Chapter 3. The abstract eigenvalue problem and the differential equation are two faces of the same coin.

9.4 The Dirac Delta Function: Not a Function, but Essential

Why we need it

The orthonormality condition for position eigenstates,

$$\langle x|x'\rangle = \delta(x - x')$$

requires the Dirac delta function (or more properly, the Dirac delta distribution). This object is central to the mathematical structure of continuous-spectrum quantum mechanics. You have likely encountered it before — in Chapter 3 (normalization of free-particle states), Chapter 5 (3D completeness relations), and Chapter 8 (position-momentum overlap). Here we develop it systematically.

Definition and key properties

The Dirac delta $\delta(x)$ is defined not by its value at each point, but by what it does under an integral:

$$\int_{-\infty}^{\infty} f(x)\delta(x - a) \, dx = f(a)$$

for any continuous function $f(x)$. This is the sifting property — the delta function "sifts out" the value of $f$ at $x = a$.

Informally, $\delta(x - a)$ is "zero everywhere except at $x = a$, where it is infinite, in such a way that it integrates to 1." More precisely, $\delta$ is not a function at all — it is a distribution (or generalized function) that maps test functions to numbers via integration.

The key properties of the delta function are:

1. Normalization: $$\int_{-\infty}^{\infty} \delta(x) \, dx = 1$$

2. Sifting property: $$\int_{-\infty}^{\infty} f(x)\delta(x - a) \, dx = f(a)$$

3. Symmetry: $$\delta(-x) = \delta(x)$$

4. Scaling: $$\delta(ax) = \frac{1}{|a|}\delta(x), \qquad a \neq 0$$

5. Composition with a function: $$\delta(g(x)) = \sum_i \frac{\delta(x - x_i)}{|g'(x_i)|}$$

where the sum runs over all simple zeros $x_i$ of $g(x)$.

6. Derivative (distributional): $$\int_{-\infty}^{\infty} f(x)\delta'(x - a) \, dx = -f'(a)$$

🔴 Warning — Never evaluate $\delta(0)$. The expression $\delta(0) = \infty$ is sometimes written for heuristic purposes, but the delta function only has meaning inside an integral. Writing $\delta(0)$ as if it were a number will lead to nonsense. The same applies to $\delta^2(x)$, which is not well-defined.

Representations of the delta function

The delta function can be represented as the limit of various sequences of ordinary functions:

Gaussian representation: $$\delta(x) = \lim_{\sigma \to 0} \frac{1}{\sigma\sqrt{2\pi}} e^{-x^2/2\sigma^2}$$

Lorentzian representation: $$\delta(x) = \lim_{\epsilon \to 0^+} \frac{1}{\pi} \frac{\epsilon}{x^2 + \epsilon^2}$$

Sinc representation: $$\delta(x) = \lim_{L \to \infty} \frac{\sin(Lx)}{\pi x} = \lim_{L \to \infty} \frac{1}{2\pi}\int_{-L}^{L} e^{ikx} \, dk$$

Fourier integral representation (the most important for quantum mechanics): $$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx} \, dk$$

or equivalently, with the quantum mechanical convention:

$$\delta(x - x') = \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} e^{ip(x - x')/\hbar} \, dp$$

💡 Key Insight — The Fourier representation $\delta(x) = \frac{1}{2\pi}\int e^{ikx} \, dk$ is the mathematical backbone of the position-momentum relationship. It encodes the completeness of momentum eigenstates when viewed from the position representation. We will use it extensively in Section 9.6.

The delta function in Dirac notation

The orthonormality of position eigenstates gives:

$$\langle x|x'\rangle = \delta(x - x')$$

Using the completeness relation for momentum eigenstates:

$$\langle x|x'\rangle = \int \langle x|p\rangle\langle p|x'\rangle \, dp = \int \frac{1}{2\pi\hbar} e^{ip(x - x')/\hbar} \, dp = \delta(x - x')$$

This is consistent — the Fourier representation of the delta function emerges naturally from inserting a complete set of momentum eigenstates. This is not a coincidence. It is the mathematical expression of the fact that the position and momentum bases are related by a Fourier transform.

Similarly:

$$\langle p|p'\rangle = \int \langle p|x\rangle\langle x|p'\rangle \, dx = \frac{1}{2\pi\hbar}\int e^{i(p' - p)x/\hbar} \, dx = \delta(p - p')$$

🧪 Thought Experiment — Imagine preparing a particle in a "position eigenstate" $|x_0\rangle$. Its momentum-space wave function would be $\phi(p) = \langle p|x_0\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{-ipx_0/\hbar}$, a plane wave of constant amplitude. The probability density $|\phi(p)|^2 = 1/(2\pi\hbar)$ is uniform — every momentum is equally likely. This is the ultimate expression of the uncertainty principle: perfect knowledge of position implies total ignorance of momentum. In practice, this state is unphysical ($\int |\phi(p)|^2 dp = \infty$), but it is the correct mathematical limit.

Worked example: Verifying the Fourier representation

Let us verify the Fourier integral representation by showing that it satisfies the sifting property. We need to show:

$$\int_{-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik(x - a)} dk\right] f(x) \, dx = f(a)$$

Write $f(x)$ in terms of its Fourier transform $\tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int f(x)e^{-ikx}dx$, so $f(x) = \frac{1}{\sqrt{2\pi}}\int \tilde{f}(k)e^{ikx}dk$. Then:

$$\int \delta(x-a) f(x) \, dx = \int \frac{1}{2\pi}\int e^{ik(x-a)}dk \, f(x) \, dx = \frac{1}{2\pi}\int e^{-ika}\left[\int f(x) e^{ikx} dx\right]dk$$

The bracketed integral is $\sqrt{2\pi}\tilde{f}(-k)$, and the remaining integral reconstructs $f(a)$ by the Fourier inversion theorem. The details involve interchanging the order of integration, which is justified in distribution theory even though the integrals do not converge absolutely.

The derivative of the delta function

The distributional derivative $\delta'(x)$ is defined by integration by parts:

$$\int_{-\infty}^{\infty} f(x)\delta'(x - a) \, dx = -f'(a)$$

The idea: integrate $\int f \delta' dx$ by parts, treating $\delta$ as an ordinary function (boundary terms vanish for test functions that decrease rapidly):

$$\int f\delta' dx = [f\delta]_{-\infty}^{\infty} - \int f'\delta \, dx = 0 - f'(a) = -f'(a)$$

Higher derivatives follow the same pattern:

$$\int f(x)\delta^{(n)}(x - a) \, dx = (-1)^n f^{(n)}(a)$$

The derivative of the delta function appears in quantum mechanics when computing matrix elements of the momentum operator:

$$\langle x|\hat{p}|x'\rangle = -i\hbar\frac{\partial}{\partial x}\delta(x - x') = -i\hbar\delta'(x - x')$$

This formula shows that the momentum operator, in the position-position matrix representation, is proportional to the derivative of the delta function — it is a "nearest-neighbor" operator in continuous space.

Three-dimensional delta function

In three dimensions, the delta function generalizes to:

$$\delta^3(\mathbf{r} - \mathbf{r}') = \delta(x - x')\delta(y - y')\delta(z - z')$$

with the sifting property:

$$\int f(\mathbf{r})\delta^3(\mathbf{r} - \mathbf{r}') \, d^3r = f(\mathbf{r}')$$

This appears in the normalization of position eigenstates in three dimensions: $\langle\mathbf{r}|\mathbf{r}'\rangle = \delta^3(\mathbf{r} - \mathbf{r}')$.

In spherical coordinates, the three-dimensional delta function takes the form:

$$\delta^3(\mathbf{r} - \mathbf{r}') = \frac{1}{r^2}\delta(r - r')\delta(\cos\theta - \cos\theta')\delta(\phi - \phi')$$

The factor $1/r^2$ compensates for the Jacobian $r^2\sin\theta$ in the volume element $d^3r = r^2\sin\theta \, dr \, d\theta \, d\phi$, ensuring that $\int \delta^3(\mathbf{r} - \mathbf{r}') d^3r = 1$.

✅ Checkpoint — Before proceeding, verify that you can evaluate these integrals: 1. $\int_0^\infty x^3 \delta(x - 2) \, dx = ?$ 2. $\int_{-\infty}^{\infty} e^{3x}\delta(2x + 4) \, dx = ?$ 3. $\int_{-\infty}^{\infty} \cos(x)\delta'(x) \, dx = ?$

(Answers: 8, $\frac{1}{2}e^{-6}$, 0.) If any of these are unclear, re-read the identities in this section.

9.5 The Spectral Theorem and Spectral Decomposition

Statement of the spectral theorem

The spectral theorem is the master result of this chapter. It says:

Spectral Theorem (Finite-Dimensional). Every Hermitian operator $\hat{A}$ on a finite-dimensional Hilbert space $\mathcal{H}$ can be written as:

$$\hat{A} = \sum_n a_n |a_n\rangle\langle a_n|$$

where $\{a_n\}$ are the eigenvalues and $\{|a_n\rangle\}$ are the corresponding orthonormal eigenstates. If eigenvalue $a_n$ is degenerate with degeneracy $g_n$, the projector $|a_n\rangle\langle a_n|$ is replaced by the projection operator onto the eigenspace: $\hat{P}_{a_n} = \sum_{r=1}^{g_n}|a_n, r\rangle\langle a_n, r|$.

For operators with a continuous spectrum, the sum becomes an integral:

$$\hat{A} = \int a \, |a\rangle\langle a| \, da$$

And for operators with both discrete and continuous parts (a mixed spectrum), we have:

$$\hat{A} = \sum_n a_n |a_n\rangle\langle a_n| + \int a \, |a\rangle\langle a| \, da$$

🔵 Historical Note — The spectral theorem for self-adjoint operators on Hilbert spaces was proved by John von Neumann in 1929, building on work by David Hilbert and Erwin Schmidt. Von Neumann recognized that quantum mechanics required a rigorous mathematical framework, and the spectral theorem provided it. His 1932 book Mathematische Grundlagen der Quantenmechanik remains a landmark. The finite-dimensional version is a consequence of the fact that every Hermitian matrix can be diagonalized by a unitary transformation — a result from linear algebra.

What the spectral theorem means physically

The spectral decomposition

$$\hat{A} = \sum_n a_n \hat{P}_n, \qquad \hat{P}_n = |a_n\rangle\langle a_n|$$

decomposes the observable $\hat{A}$ into a sum of projection operators, each weighted by the corresponding eigenvalue. This directly encodes the measurement postulate:

1. The possible outcomes of measuring $\hat{A}$ are the eigenvalues $\{a_n\}$.
2. The probability of outcome $a_n$ is $P(a_n) = \langle\psi|\hat{P}_n|\psi\rangle = |\langle a_n|\psi\rangle|^2$.
3. After measurement, if outcome $a_n$ is obtained, the state collapses to $\hat{P}_n|\psi\rangle / \|\hat{P}_n|\psi\rangle\|$.
4. The expectation value is $\langle\hat{A}\rangle = \sum_n a_n P(a_n) = \sum_n a_n\langle\psi|\hat{P}_n|\psi\rangle$.

The spectral theorem guarantees that this scheme is mathematically consistent for any Hermitian operator on any Hilbert space.

Computing functions of operators

One of the most powerful consequences of the spectral decomposition is that it allows us to define functions of operators. If $f$ is any function defined on the eigenvalues of $\hat{A}$, then:

$$f(\hat{A}) = \sum_n f(a_n)|a_n\rangle\langle a_n|$$

This definition is natural: $f(\hat{A})$ acts on an eigenstate of $\hat{A}$ by replacing the eigenvalue $a$ with $f(a)$:

$$f(\hat{A})|a_n\rangle = f(a_n)|a_n\rangle$$

Example: Time evolution operator. The Hamiltonian $\hat{H}$ has spectral decomposition $\hat{H} = \sum_n E_n|E_n\rangle\langle E_n|$. The time-evolution operator is:

$$\hat{U}(t) = e^{-i\hat{H}t/\hbar} = \sum_n e^{-iE_n t/\hbar}|E_n\rangle\langle E_n|$$

This is the result from Chapter 7, now derived directly from the spectral theorem. Acting on a general state:

$$|\psi(t)\rangle = \hat{U}(t)|\psi(0)\rangle = \sum_n e^{-iE_n t/\hbar}\langle E_n|\psi(0)\rangle|E_n\rangle = \sum_n c_n e^{-iE_n t/\hbar}|E_n\rangle$$

Each energy eigenstate picks up a phase factor $e^{-iE_n t/\hbar}$ — precisely the result we derived in Chapter 7 by solving the time-dependent Schrodinger equation.

🔗 Connection — This formula is the mathematical foundation for everything in Part IV (approximation methods). Perturbation theory (Chapters 17--18) finds approximate eigenvalues $E_n$ and eigenstates $|E_n\rangle$ for Hamiltonians that are too complicated to solve exactly. The variational method (Chapter 19) gives upper bounds on the ground state energy $E_0$. In every case, the spectral theorem tells us how to use the eigenvalues and eigenstates once we have them.

Example: Inverse operator. If $\hat{A}$ has no zero eigenvalue, then:

$$\hat{A}^{-1} = \sum_n \frac{1}{a_n}|a_n\rangle\langle a_n|$$

Example: Square root of an operator. If all eigenvalues of $\hat{A}$ are non-negative:

$$\hat{A}^{1/2} = \sum_n \sqrt{a_n}|a_n\rangle\langle a_n|$$

Worked example: Spectral decomposition of $\hat{S}_z$

The spin-1/2 operator $\hat{S}_z$ has eigenvalues $+\hbar/2$ and $-\hbar/2$ with eigenstates $|\uparrow\rangle$ and $|\downarrow\rangle$. Its spectral decomposition is:

$$\hat{S}_z = \frac{\hbar}{2}|\uparrow\rangle\langle\uparrow| + \left(-\frac{\hbar}{2}\right)|\downarrow\rangle\langle\downarrow| = \frac{\hbar}{2}\left(|\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|\right)$$

Let us verify this. In the matrix representation:

$$|\uparrow\rangle\langle\uparrow| = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad |\downarrow\rangle\langle\downarrow| = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$

$$\hat{S}_z = \frac{\hbar}{2}\left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right] = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \checkmark$$

Now compute $e^{-i\hat{S}_z\theta/\hbar}$ (rotation about the $z$-axis by angle $\theta$):

$$e^{-i\hat{S}_z\theta/\hbar} = e^{-i\theta/2}|\uparrow\rangle\langle\uparrow| + e^{i\theta/2}|\downarrow\rangle\langle\downarrow| = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{pmatrix}$$

For $\theta = 2\pi$: $e^{-i\hat{S}_z \cdot 2\pi/\hbar} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\hat{I}$. A spin-1/2 particle picks up a minus sign under a $2\pi$ rotation — it must go through $4\pi$ to return to its original state. This remarkable fact, a direct consequence of the half-integer eigenvalues, has been experimentally verified using neutron interferometry.

📊 By the Numbers — The neutron $4\pi$ rotation was verified by Werner et al. (1975) using a silicon crystal interferometer. The neutron beam was split, one path passed through a magnetic field that produced a controllable phase shift, and the interference pattern was measured as a function of the rotation angle. The $4\pi$ periodicity — predicted by the spectral decomposition above — was confirmed to high precision.

Spectral decomposition for continuous spectra

For the position operator:

$$\hat{x} = \int_{-\infty}^{\infty} x \, |x\rangle\langle x| \, dx$$

This means: $\hat{x}$ is a "weighted sum" of projection operators $|x\rangle\langle x|$, each weighted by the position value $x$. The expectation value is:

$$\langle\hat{x}\rangle = \int x \, |\psi(x)|^2 \, dx$$

Similarly, for the momentum operator:

$$\hat{p} = \int_{-\infty}^{\infty} p \, |p\rangle\langle p| \, dp$$

with expectation value:

$$\langle\hat{p}\rangle = \int p \, |\phi(p)|^2 \, dp$$

The spectral decomposition for the Hamiltonian of a system with both bound and scattering states (mixed spectrum) takes the form:

$$\hat{H} = \sum_{n}^{\text{bound}} E_n|E_n\rangle\langle E_n| + \int_0^{\infty} E \, |E\rangle\langle E| \, dE$$

where the sum runs over the discrete bound-state energies and the integral runs over the continuous scattering-state energies. The completeness relation is correspondingly:

$$\sum_{n}^{\text{bound}} |E_n\rangle\langle E_n| + \int_0^{\infty} |E\rangle\langle E| \, dE = \hat{I}$$

⚖️ Interpretation — The spectral theorem is sometimes called the "diagonalization" of the operator. In the eigenbasis, any Hermitian operator is "diagonal" — it acts simply by multiplication. This is why physicists seek the eigenbasis: it is the representation in which the operator (and hence the corresponding measurement) is simplest.

9.6 Fourier Transforms as Change of Basis

From position to momentum

The Fourier transform — which you may have encountered in mathematics or signal processing courses — arises naturally in quantum mechanics as a change of basis from position to momentum.

Starting from the momentum-space wave function:

$$\phi(p) = \langle p|\psi\rangle$$

Insert the completeness relation for position:

$$\phi(p) = \int \langle p|x\rangle\langle x|\psi\rangle \, dx = \int \frac{1}{\sqrt{2\pi\hbar}} e^{-ipx/\hbar} \psi(x) \, dx$$

This is the Fourier transform from position to momentum representation:

$$\boxed{\phi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \psi(x) \, e^{-ipx/\hbar} \, dx}$$

From momentum to position

The inverse is obtained by inserting the completeness relation for momentum:

$$\psi(x) = \langle x|\psi\rangle = \int \langle x|p\rangle\langle p|\psi\rangle \, dp = \int \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} \phi(p) \, dp$$

$$\boxed{\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \phi(p) \, e^{ipx/\hbar} \, dp}$$

💡 Key Insight — The Fourier transform is not an arbitrary mathematical trick. It is the natural consequence of two completeness relations and the overlap $\langle x|p\rangle = (2\pi\hbar)^{-1/2}e^{ipx/\hbar}$. In Dirac notation, there is nothing to memorize — you simply insert complete sets of states and use the known overlap. The "sign conventions" that plague Fourier transform tables (where does the minus sign go? what normalization?) are automatically determined by the physics: $\langle x|p\rangle$ has the plus sign, $\langle p|x\rangle$ has the minus sign.

Parseval's theorem: Conservation of probability

The total probability must be the same whether computed in position or momentum space:

$$\int |\psi(x)|^2 \, dx = \langle\psi|\psi\rangle = \int |\phi(p)|^2 \, dp$$

This is Parseval's theorem (or Plancherel's theorem). In Dirac notation, the proof is immediate:

$$\int |\psi(x)|^2 \, dx = \int \langle\psi|x\rangle\langle x|\psi\rangle \, dx = \langle\psi|\left(\int |x\rangle\langle x| \, dx\right)|\psi\rangle = \langle\psi|\hat{I}|\psi\rangle = \langle\psi|\psi\rangle$$

The same argument with momentum gives $\int|\phi(p)|^2 dp = \langle\psi|\psi\rangle$. Since both equal $\langle\psi|\psi\rangle$, they equal each other. This is a beautiful example of how the completeness relation guarantees physical consistency.

Worked example: Fourier transform of a Gaussian

Consider the Gaussian wave function:

$$\psi(x) = \left(\frac{1}{2\pi\sigma_x^2}\right)^{1/4} e^{-x^2/4\sigma_x^2}$$

(normalized, centered at $x = 0$, with position uncertainty $\Delta x = \sigma_x$). The Fourier transform is:

$$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\left(\frac{1}{2\pi\sigma_x^2}\right)^{1/4} \int_{-\infty}^{\infty} e^{-x^2/4\sigma_x^2} e^{-ipx/\hbar} \, dx$$

This is a Gaussian integral (complete the square in the exponent):

$$\phi(p) = \left(\frac{2\sigma_x^2}{\pi\hbar^2}\right)^{1/4} e^{-\sigma_x^2 p^2/\hbar^2}$$

The momentum-space wave function is also a Gaussian, with width $\sigma_p = \hbar/(2\sigma_x)$. Note:

$$\Delta x \cdot \Delta p = \sigma_x \cdot \frac{\hbar}{2\sigma_x} = \frac{\hbar}{2}$$

The Gaussian saturates the uncertainty bound $\Delta x \Delta p \geq \hbar/2$ — it is the minimum-uncertainty state. This is a deep property: the Gaussian is the unique state that achieves equality in the Heisenberg uncertainty relation.

🔗 Connection — In Chapter 27, we will see that coherent states of the quantum harmonic oscillator are Gaussians that maintain this minimum uncertainty as they evolve in time. They are the "most classical" quantum states, and their Fourier transform properties are central to quantum optics.

Change of basis for operators

In the momentum representation, the position operator becomes a derivative:

$$\langle p|\hat{x}|\psi\rangle = i\hbar\frac{\partial}{\partial p}\phi(p)$$

This is the "dual" of $\langle x|\hat{p}|\psi\rangle = -i\hbar\frac{d\psi}{dx}$. The derivation uses integration by parts:

$$\langle p|\hat{x}|\psi\rangle = \int \langle p|x\rangle x \langle x|\psi\rangle \, dx = \frac{1}{\sqrt{2\pi\hbar}}\int x \, \psi(x) \, e^{-ipx/\hbar} \, dx$$

$$= \frac{1}{\sqrt{2\pi\hbar}}\int \psi(x) \left(i\hbar\frac{\partial}{\partial p}\right) e^{-ipx/\hbar} \, dx = i\hbar\frac{\partial}{\partial p}\phi(p)$$

The position and momentum representations are "mirror images" of each other:

This symmetry is the Fourier transform in action.

Convolution theorem

The Fourier transform converts multiplication in one representation into convolution in the other. If $\psi(x)$ has Fourier transform $\phi(p)$, and $V(x)$ has Fourier transform $\tilde{V}(p)$, then:

$$\mathcal{F}[V(x)\psi(x)] = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \tilde{V}(p - p')\phi(p') \, dp'$$

This is the convolution of $\tilde{V}$ and $\phi$. It explains why the potential energy — a multiplicative operator in position space — becomes an integral operator in momentum space: it mixes different momentum components.

The convolution theorem is the mathematical reason why: - A local potential in position space (e.g., $V(x) = \frac{1}{2}m\omega^2 x^2$) becomes non-local in momentum space. - The kinetic energy — a differential operator in position space — is local (multiplicative) in momentum space: $\hat{T} = p^2/(2m)$.

🔄 Spaced Review (Chapter 6) — In Chapter 6, we noted that the commutation relation $[\hat{x}, \hat{p}] = i\hbar$ prevents $\hat{x}$ and $\hat{p}$ from being simultaneously diagonal. The Fourier transform makes this concrete: in the position basis, $\hat{x}$ is diagonal and $\hat{p}$ is a derivative; in the momentum basis, $\hat{p}$ is diagonal and $\hat{x}$ is a derivative. The non-commutativity is reflected in the impossibility of finding a representation where both operators are multiplicative.

9.7 Momentum Space Quantum Mechanics

The Schrodinger equation in momentum space

The time-independent Schrodinger equation $\hat{H}|\psi\rangle = E|\psi\rangle$, projected onto $\langle p|$, becomes:

$$\frac{p^2}{2m}\phi(p) + \int \tilde{V}(p - p')\phi(p') \, dp' = E\phi(p)$$

where $\tilde{V}(q)$ is the Fourier transform of the potential:

$$\tilde{V}(q) = \frac{1}{2\pi\hbar}\int V(x) \, e^{-iqx/\hbar} \, dx$$

The kinetic energy is a simple multiplication in momentum space (it is diagonal in the momentum basis), but the potential energy becomes an integral operator (a convolution). This is the complementary situation to position space, where the potential is diagonal but the kinetic energy involves a second derivative.

💡 Key Insight — The choice of representation is a strategic decision. If the kinetic energy dominates (free particle, scattering problems), the momentum representation simplifies the problem. If the potential is simple (harmonic oscillator, finite well), the position representation is typically easier. Expert quantum mechanicists choose the representation that makes the problem most tractable.

When momentum space is useful

Momentum-space quantum mechanics is particularly valuable in several contexts:

1. The free particle. With $V = 0$, the Schrodinger equation in momentum space is trivially $\frac{p^2}{2m}\phi(p) = E\phi(p)$, with solutions $\phi(p) = \delta(p \pm \sqrt{2mE})$.

2. Scattering theory (Chapter 22). Cross sections are naturally expressed in terms of momentum transfer $\Delta p = p_f - p_i$, and the Born approximation uses the Fourier transform of the potential directly.

3. Solid-state physics (Chapter 26). Band structure is naturally formulated in momentum space (Brillouin zones, Bloch wavevectors).

4. Quantum field theory (Chapter 37). Fields are expanded in momentum modes, and interactions are naturally described in momentum space (Feynman diagrams).

Example: The delta-function potential in momentum space

Consider the attractive delta-function potential $V(x) = -\alpha\delta(x)$ (with $\alpha > 0$), which we solved in position space in Chapter 3. Let us solve it in momentum space to illustrate the technique.

The Fourier transform of the potential is:

$$\tilde{V}(q) = \frac{1}{2\pi\hbar}\int (-\alpha)\delta(x) \, e^{-iqx/\hbar} \, dx = -\frac{\alpha}{2\pi\hbar}$$

This is constant — the delta-function potential is "flat" in momentum space.

The Schrodinger equation becomes:

$$\frac{p^2}{2m}\phi(p) - \frac{\alpha}{2\pi\hbar}\int \phi(p') \, dp' = E\phi(p)$$

Define $C \equiv \int \phi(p') \, dp'$ (a constant). Then:

$$\phi(p) = \frac{\alpha C}{2\pi\hbar} \cdot \frac{1}{\frac{p^2}{2m} - E}$$

For a bound state, $E < 0$. Write $E = -|E|$:

$$\phi(p) = \frac{\alpha C}{2\pi\hbar} \cdot \frac{1}{\frac{p^2}{2m} + |E|}$$

Self-consistency: integrate both sides over $p$ (both sides must give $C$):

$$C = \frac{\alpha C}{2\pi\hbar}\int_{-\infty}^{\infty} \frac{dp}{\frac{p^2}{2m} + |E|}$$

Cancel $C$ (assuming $C \neq 0$ for a non-trivial solution):

$$1 = \frac{\alpha}{2\pi\hbar}\int_{-\infty}^{\infty}\frac{dp}{\frac{p^2}{2m} + |E|} = \frac{\alpha}{2\pi\hbar} \cdot \frac{2m}{\sqrt{2m|E|}} \cdot \pi = \frac{m\alpha}{\hbar\sqrt{2m|E|}}$$

Solving for $|E|$:

$$\sqrt{2m|E|} = \frac{m\alpha}{\hbar} \implies 2m|E| = \frac{m^2\alpha^2}{\hbar^2} \implies |E| = \frac{m\alpha^2}{2\hbar^2}$$

$$\boxed{E = -\frac{m\alpha^2}{2\hbar^2}}$$

This matches the result from Chapter 3, obtained there by solving a differential equation with matching conditions. The momentum-space approach avoids boundary conditions entirely — it trades them for an integral equation, which in this case is simpler.

The bound-state wave function in momentum space is a Lorentzian:

$$\phi(p) = \frac{N}{p^2 + \kappa^2}, \qquad \kappa = \frac{m\alpha}{\hbar}$$

where $N$ is a normalization constant. The Fourier transform back to position space gives $\psi(x) \propto e^{-\kappa|x|/\hbar}$, the exponentially decaying bound state from Chapter 3.

✅ Checkpoint — The delta-function potential has exactly one bound state. Can you explain from the momentum-space formulation why there cannot be more? (Hint: the integral equation has a unique self-consistency condition.)

Uncertainty principle from the Fourier transform

The position-momentum uncertainty principle $\Delta x \cdot \Delta p \geq \hbar/2$ can be derived directly from properties of the Fourier transform, without reference to commutators or the generalized uncertainty principle.

The key mathematical result is the Fourier uncertainty principle: for any function $f(x)$ and its Fourier transform $\tilde{f}(k)$, the product of their widths (defined as root-mean-square widths about their means) satisfies:

$$\Delta x \cdot \Delta k \geq \frac{1}{2}$$

with equality if and only if $f$ is a Gaussian. Since $p = \hbar k$, this translates directly to:

$$\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$$

This is a theorem of Fourier analysis, not a postulate of quantum mechanics. The physical content of quantum mechanics is not the uncertainty principle itself, but the identification of $\psi(x)$ and $\phi(p)$ as a Fourier transform pair — which follows from $\langle x|p\rangle = (2\pi\hbar)^{-1/2}e^{ipx/\hbar}$.

⚖️ Interpretation — The uncertainty principle is sometimes described as "measurement disturbs the system" (the Heisenberg microscope picture). While disturbance does occur, the uncertainty principle is more fundamental than any particular measurement scheme. It is a mathematical property of wave-like systems: a function and its Fourier transform cannot both be sharply peaked. In quantum mechanics, this mathematical property has the physical consequence that position and momentum cannot both be precisely defined for any quantum state.

Energy-time "uncertainty relation"

A common source of confusion is the energy-time "uncertainty relation" $\Delta E \cdot \Delta t \gtrsim \hbar/2$. Unlike position-momentum, this is not a straightforward consequence of a commutation relation, because time is a parameter in non-relativistic quantum mechanics, not an operator. However, the spectral theorem provides the correct interpretation.

If a state $|\psi\rangle$ has energy uncertainty $\Delta E$, then the shortest time $\Delta t$ over which the state changes appreciably is:

$$\Delta t \sim \frac{\hbar}{\Delta E}$$

More precisely, the "survival probability" $|\langle\psi(0)|\psi(t)\rangle|^2$ first deviates significantly from 1 after a time of order $\hbar/\Delta E$. For a state expanded as $|\psi\rangle = \sum_n c_n|E_n\rangle$, the time evolution $|\psi(t)\rangle = \sum_n c_n e^{-iE_n t/\hbar}|E_n\rangle$ involves phases that dephase on a timescale set by the spread of energies. States with sharply defined energy (small $\Delta E$) evolve slowly; states with large energy spread evolve quickly. This is the energy-time uncertainty relation, derived from the spectral decomposition.

9.8 Rigged Hilbert Space: The Right Mathematical Setting

The problem with the Hilbert space

Throughout this chapter, we have treated position eigenstates $|x\rangle$ and momentum eigenstates $|p\rangle$ as if they were legitimate members of the Hilbert space. But they are not.

The Hilbert space $\mathcal{H} = L^2(\mathbb{R})$ consists of square-integrable functions — functions for which $\int|\psi(x)|^2 dx < \infty$. Position eigenstates have "wave functions" $\langle x'|x\rangle = \delta(x' - x)$, which satisfy $\int|\delta(x' - x)|^2 dx' = \delta(0) = \infty$. They are not square-integrable. Similarly, momentum eigenstates have $\langle x|p\rangle = (2\pi\hbar)^{-1/2}e^{ipx/\hbar}$, with $\int|e^{ipx/\hbar}|^2 dx = \infty$. They are not in $\mathcal{H}$ either.

Yet we use them constantly. The completeness relations $\int|x\rangle\langle x| dx = \hat{I}$ and $\int|p\rangle\langle p| dp = \hat{I}$ are essential tools. The spectral theorem for operators with continuous spectra requires these generalized eigenstates. Are we being mathematically inconsistent?

The answer: rigged Hilbert spaces

The resolution is a mathematical framework called the rigged Hilbert space (also known as the Gelfand triple, after the mathematician Israel Gelfand). The idea is to enlarge the Hilbert space to accommodate objects like $|x\rangle$ and $\delta(x)$, while simultaneously restricting the class of "well-behaved" states.

A rigged Hilbert space is a triple:

$$\Phi \subset \mathcal{H} \subset \Phi'$$

where:

• $\mathcal{H}$ is the ordinary Hilbert space ($L^2(\mathbb{R})$ for a particle on a line).
• $\Phi$ is a dense subspace of $\mathcal{H}$ consisting of "very well-behaved" functions (smooth, rapidly decreasing — the Schwartz space $\mathcal{S}(\mathbb{R})$ is the canonical choice). These are the physically preparable states.
• $\Phi'$ is the dual space of $\Phi$ — the space of continuous linear functionals on $\Phi$. This is larger than $\mathcal{H}$ and includes distributions like $\delta(x - a)$ and plane waves $e^{ipx/\hbar}$.

The position "eigenstate" $|x\rangle$ and the momentum "eigenstate" $|p\rangle$ live in $\Phi'$, not in $\mathcal{H}$. They are generalized eigenstates — elements of the larger space that satisfy the eigenvalue equation in a distributional sense.

💡 Key Insight — Think of the rigged Hilbert space as three nested boxes:

Inner box ($\Phi$): The states we can actually prepare in a lab — smooth, localized, rapidly decreasing wave functions. Every physical state lives here.

Middle box ($\mathcal{H}$): All square-integrable functions. This is the standard Hilbert space, closed under limits. It contains some mathematical states (discontinuous functions, for instance) that are not physically preparable but are useful limits.

Outer box ($\Phi'$): Everything in $\mathcal{H}$, plus distributions like $\delta(x)$ and plane waves. These are the generalized eigenstates of operators with continuous spectra. They are essential mathematical tools, but they are not physical states.

Why this matters

The rigged Hilbert space resolves several puzzles:

1. Dirac's formalism is rigorous. The bra-ket notation, completeness relations, and spectral decompositions for continuous spectra are all well-defined within the rigged Hilbert space framework. Dirac's physical intuition, which led him to treat $|x\rangle$ as an eigenstate long before the mathematics was worked out, was correct.

2. The spectral theorem extends to continuous spectra. In the ordinary Hilbert space, only operators with purely discrete spectra have eigenvectors. In the rigged Hilbert space, operators with continuous spectra have generalized eigenvectors in $\Phi'$.

3. Eigenfunction expansions are justified. The expansion $|\psi\rangle = \int \psi(x)|x\rangle dx$ is well-defined when $|\psi\rangle \in \Phi$ and $|x\rangle \in \Phi'$, because the pairing between $\Phi$ and $\Phi'$ is well-defined even when the inner product on $\mathcal{H}$ is not.

4. The delta function orthogonality $\langle x|x'\rangle = \delta(x - x')$ makes sense. It means that the functional $|x\rangle \in \Phi'$, when applied to the functional $|x'\rangle$, gives the distribution $\delta(x - x')$. This is precise in distribution theory, even though $\delta(0) = \infty$ as a "number."

🔵 Historical Note — The rigged Hilbert space framework was developed by Gelfand and Vilenkin (1964) and independently by Maurin (1968). It was brought into the physics literature by Bohm and Gadella in the 1960s–1980s, who showed that it provides the correct mathematical setting for Dirac's formalism. Most physics textbooks (including Griffiths, Sakurai, and Shankar) use Dirac's formalism without mentioning rigged Hilbert spaces, which speaks to the power of Dirac's physical intuition — he got the right answers decades before the rigorous framework existed.

A concrete example: The free-particle energy eigenstates

Consider the free particle with Hamiltonian $\hat{H} = \hat{p}^2/(2m)$. The energy eigenstates are momentum eigenstates $|p\rangle$ with energy $E = p^2/(2m)$. These states are plane waves in position space:

$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}$$

This function is not in $L^2(\mathbb{R})$: it does not decay at infinity, so $\int|\langle x|p\rangle|^2 dx = \infty$. Yet the completeness relation $\int|p\rangle\langle p|dp = \hat{I}$ is essential for expanding any physical state as a wave packet:

$$|\psi\rangle = \int \phi(p)|p\rangle dp$$

where $\phi(p) = \langle p|\psi\rangle \in L^2(\mathbb{R})$ is the square-integrable momentum-space wave function. The state $|\psi\rangle$ belongs to $\Phi$ (if it is smooth and rapidly decreasing) or at least to $\mathcal{H}$, even though the "basis vectors" $|p\rangle$ belong to the larger space $\Phi'$.

This is exactly analogous to expanding a vector in terms of "basis vectors" that are infinitely long — the expansion coefficients are finite even though the basis vectors are not. The rigged Hilbert space makes this analogy precise.

The nuclear spectral theorem

The mathematical payoff of the rigged Hilbert space is the nuclear spectral theorem (Gelfand-Maurin theorem), which states:

For any self-adjoint operator $\hat{A}$ on $\mathcal{H}$, there exists a complete system of generalized eigenvectors in $\Phi'$ satisfying $\hat{A}|a\rangle = a|a\rangle$, and any $|\psi\rangle \in \Phi$ can be expanded in terms of these generalized eigenvectors.

This theorem justifies the spectral decomposition for operators with continuous spectra — something that the ordinary Hilbert space theory alone cannot do. It is the mathematical theorem behind every calculation in this chapter involving position or momentum eigenstates.

Practical implications

For most calculations in this textbook, you do not need to think explicitly about rigged Hilbert spaces. The rules of Dirac notation — completeness relations, eigenstate expansions, delta function orthogonality — will give you correct answers. But knowing that the framework exists serves two purposes:

1. Confidence: When you use $|x\rangle$ and $\delta(x - x')$, you are not doing anything mathematically illegitimate. The rigged Hilbert space provides the rigorous foundation.

2. Understanding edge cases: Occasionally, you will encounter situations where naive application of Hilbert space rules gives paradoxes (e.g., the "normalization" of plane waves, the self-adjointness of the momentum operator on a half-line). The rigged Hilbert space framework explains why these subtleties arise and how to resolve them.

⚠️ Common Misconception — "The Dirac delta function is not rigorous, so quantum mechanics is not rigorous." This is false. Distribution theory (developed by Laurent Schwartz in 1950, earning him the Fields Medal) and the rigged Hilbert space framework (developed by Gelfand and collaborators) place the entire formalism on solid mathematical foundations. The rigor exists; it is just not always displayed in physics textbooks, because the notation is designed so that you get the right answer without it.

9.9 Summary and Project Checkpoint

What you have learned

This chapter developed the eigenvalue problem — the mathematical heart of quantum measurement — in full generality. Here is the map of what we covered:

Section 9.1 established the eigenvalue equation $\hat{A}|a\rangle = a|a\rangle$ and its fundamental consequences: real eigenvalues, orthogonal eigenstates, and the role of degeneracy.

Section 9.2 worked through discrete spectra in detail, using spin-1/2 and the QHO as anchor examples. You computed eigenvalues, eigenstates, probabilities, and expectation values entirely in Dirac notation.

Section 9.3 extended the framework to continuous spectra (position and momentum), identifying the key difference: Kronecker deltas become Dirac deltas, sums become integrals, normalization changes from $\langle a_m|a_n\rangle = \delta_{mn}$ to $\langle x|x'\rangle = \delta(x - x')$.

Section 9.4 developed the Dirac delta function systematically — its defining property (sifting), its representations (Gaussian, Lorentzian, Fourier integral), and its use in quantum mechanics.

Section 9.5 stated and applied the spectral theorem: $\hat{A} = \sum_n a_n|a_n\rangle\langle a_n|$ for discrete spectra, with the integral generalization for continuous spectra. You saw how this enables functions of operators, especially the time-evolution operator.

Section 9.6 derived the Fourier transform as a change of basis between position and momentum representations, using nothing more than completeness relations and the overlap $\langle x|p\rangle$.

Section 9.7 applied the machinery to do quantum mechanics entirely in momentum space, solving the delta-function potential as an illustrative example.

Section 9.8 explained the rigged Hilbert space — the mathematical structure that makes Dirac's formalism rigorous — at a conceptual level.

The five essential results

1. Eigenvalue equation: $\hat{A}|a\rangle = a|a\rangle$ — the mathematical formulation of "what values can be measured?"

2. Spectral decomposition: $\hat{A} = \sum_n a_n|a_n\rangle\langle a_n|$ — every observable is a sum of its eigenvalues times projection operators.

3. Probability rule: $P(a_n) = |\langle a_n|\psi\rangle|^2$ — the Born rule, derived directly from the expansion in the eigenbasis.

4. Fourier transform as change of basis: $\phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\int \psi(x) e^{-ipx/\hbar} dx$ — position and momentum representations are related by the overlap $\langle x|p\rangle$.

5. Delta function orthonormality: $\langle x|x'\rangle = \delta(x - x')$ — the continuous-spectrum analogue of $\langle m|n\rangle = \delta_{mn}$, justified by the rigged Hilbert space.

🔄 Spaced Review — Before proceeding to Chapter 10 (Symmetry and Conservation Laws), verify that you can: - Solve $2 \times 2$ eigenvalue problems completely (eigenvalues, eigenstates, spectral decomposition) - Write the spectral decomposition of $\hat{S}_x$, $\hat{S}_y$, $\hat{S}_z$, and compute $e^{-i\hat{S}_n\theta/\hbar}$ - Derive the Fourier transform from Dirac notation using completeness - Apply the sifting property of $\delta(x - a)$ under an integral - Explain in one sentence why $|x\rangle$ is not in the Hilbert space but is still mathematically legitimate

Project Checkpoint: Quantum Simulation Toolkit v0.9

In the code accompanying this chapter, you will add three modules to your growing quantum toolkit:

1. eigensolve() — A general eigenvalue solver that takes a Hermitian matrix and returns eigenvalues, eigenstates, and the spectral decomposition.

2. spectral_decompose() — Constructs the spectral decomposition $\hat{A} = \sum a_n|a_n\rangle\langle a_n|$ and verifies reconstruction.

3. fourier_transform() — Numerical Fourier transform between position and momentum representations using FFT.

These modules will be used in Chapter 10 (checking that symmetry operators commute with the Hamiltonian by comparing eigenspaces), Chapter 11 (eigenvalues of tensor product operators), and Chapters 17--19 (approximate eigenvalue methods).

🔗 Connection — The eigenvalue solver is the single most-used function in the entire toolkit. Perturbation theory (Ch 17), degenerate perturbation theory (Ch 18), and the variational method (Ch 19) all require computing eigenvalues and eigenstates of modified Hamiltonians. The Fourier transform module will be essential for scattering theory (Ch 22) and quantum optics (Ch 27).

Key Terms Introduced in This Chapter