Operators 2: Hermitian Operators and Commutators

Chem 3240 · Lecture 3.5b

Davit Potoyan

Where we are

Last time: on a grid every operator is a matrix, and \hat{H}\psi = E\psi becomes H\mathbf{v} = E\mathbf{v}
The eigenvalues are the values a measurement can return
Postulate 2: every observable is a linear Hermitian operator
Today: which matrices may stand for an observable? And what if two of them do not commute?

Recap: eigenvectors of a matrix

A\mathbf{v} = \lambda\mathbf{v}\ \text{ with }\ \mathbf{v} \neq 0 \qquad\Longleftrightarrow\qquad \det(A - \lambda I) = 0

A = \left(\begin{smallmatrix} 2 & 1 \\ 1 & 2 \end{smallmatrix}\right): \lambda = 3 along (1, 1) and \lambda = 1 along (1, -1). They add up to the trace and multiply to the determinant

Hermitian: the matrix version of a real number

numbers	matrices
conjugate z^*: flip the sign of i	adjoint A^\dagger: swap rows and columns, then conjugate
real: z^* = z	Hermitian: A^\dagger = A

Hermitian operators are the “real numbers” of quantum mechanics: their eigenvalues are real

Adjoint and Hermitian operators

The adjoint \hat{A}^\dagger: move \hat{A} from the ket to the bra

\langle \phi \vert \hat{A}\psi \rangle = \langle \hat{A}^\dagger\phi \vert \psi \rangle

Hermitian, \hat{A}^\dagger = \hat{A}: the operator moves across unchanged

\begin{aligned} &\text{integral:}\qquad & \int \phi^*\,\big(\hat{A}\psi\big)\,dx &= \int \big(\hat{A}\phi\big)^*\,\psi\,dx \\ &\text{Dirac:} & \langle \phi \vert \hat{A}\psi \rangle &= \langle \hat{A}\phi \vert \psi \rangle \\ &\text{matrix:} & A_{jk} &= A_{kj}^* \end{aligned}

Example: which matrices are Hermitian?

A = \begin{pmatrix} 2 & 1-i \\ 1+i & 3 \end{pmatrix} \quad B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \quad C = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} \quad D = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}

A: A_{12} = 1-i = A_{21}^*, real diagonal: Hermitian
B: B_{12} = 2 \neq B_{21}^* = 3: not Hermitian
C: the diagonal must be real, and i \neq i^* = -i: not Hermitian
D: D_{12} = -i = D_{21}^*: Hermitian, the pattern of \hat{p} on the grid

Why observables must be Hermitian

A measurement needs three things from its operator:

Real eigenvalues: measured values are real numbers
Orthogonal eigenfunctions: different outcomes exclude each other
A complete basis: every state is a sum of eigenfunctions (next lecture)

\hat{A}\phi_n = a_n\phi_n \quad\Longrightarrow\quad \begin{aligned} &a_n = a_n^* \\ &\langle \phi_m \vert \phi_n \rangle = \delta_{mn} \\ &\psi = \textstyle\sum_n c_n\,\phi_n \end{aligned}

Example: real eigenvalues from the quadratic formula

Show that every Hermitian 2\times 2 matrix \begin{pmatrix} a & b \\ b^* & d \end{pmatrix}, with a and d real, has real eigenvalues.

(a - \lambda)(d - \lambda) - b\,b^* = 0 \qquad\Longrightarrow\qquad \lambda = \frac{a + d}{2} \pm \sqrt{\Big(\frac{a - d}{2}\Big)^2 + \lvert b\rvert^2}

Under the root sits a sum of squares, never negative, so \lambda is real. Matrix A above: \lambda = \tfrac{5}{2} \pm \tfrac{3}{2} = 4 and 1.

The coupling \lvert b\rvert pushes the two levels apart: bonding and antibonding orbitals

Why the eigenvalues are real, in two lines

Let \hat{A}\phi = a\phi with \langle \phi \vert \phi \rangle = 1 and \hat{A} Hermitian. Show that a is real.

\langle \phi \vert \hat{A}\phi \rangle = \langle \phi \vert a\phi \rangle = a, \qquad\qquad \langle \hat{A}\phi \vert \phi \rangle = \langle a\phi \vert \phi \rangle = a^*

Hermitian says the two are equal, so a = a^*.

The step that matters: a constant leaves a ket as it is, but leaves a bra conjugated

\langle a\phi \vert \psi \rangle = \int (a\phi)^*\,\psi\,dx = a^*\!\int \phi^*\,\psi\,dx = a^*\,\langle \phi \vert \psi \rangle

Hermitian eigenvectors are perpendicular

Symmetric: (1, 1)\cdot(1, -1) = 0, orthogonal eigenvectors
Not symmetric: real eigenvalues, but the eigenvectors overlap. A rotation has none at all

Live: perpendicular or not?

What goes wrong without Hermiticity

Add a non-Hermitian part: the eigenvalues leave the real axis and the eigenvectors start to overlap
Real eigenvalues are not enough: \left(\begin{smallmatrix} 1 & 1 \\ 0 & 2 \end{smallmatrix}\right) has 1 and 2, but overlapping eigenvectors

Example: the derivative on two points

On two grid points \dfrac{d}{dx} \to D_1 = \dfrac{1}{2\Delta x}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. Find the eigenvalues of D_1 and of P = -i\hbar D_1.

\begin{aligned} \det(D_1 - \lambda I) &= \lambda^2 + \frac{1}{4\Delta x^2} = 0 &&\Rightarrow\quad \lambda = \pm\frac{i}{2\Delta x} \\ P = -i\hbar D_1 &= \frac{\hbar}{2\Delta x}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} &&\Rightarrow\quad p = \pm\frac{\hbar}{2\Delta x} \end{aligned}

D_1 is the rotation of the last figure: imaginary eigenvalues. Times -i\hbar they turn real; P is \tfrac{\hbar}{2\Delta x} times matrix D above.

Example: is momentum Hermitian?

Show that \hat{p} = -i\hbar\,d/dx satisfies \int \phi^*\,(\hat{p}\psi)\,dx = \int (\hat{p}\phi)^*\,\psi\,dx.

Integrate by parts. The boundary term vanishes because physical \phi, \psi go to zero:

\begin{aligned} \int\phi^*\big(-i\hbar\,\psi'\big)\,dx &= \Big[-i\hbar\,\phi^*\psi\Big]_{-\infty}^{\infty} + i\hbar\int\phi'^*\,\psi\,dx \\ &= 0 + \int\big(-i\hbar\,\phi'\big)^*\,\psi\,dx \quad\checkmark \end{aligned}

Without the i the sign comes out wrong, (d/dx)^\dagger = -d/dx; the -i turns it around.

Recipe: move the derivative to the other function by parts, then check the signs and the conjugates

Order matters

\hat{A}\hat{B}\,f means: apply \hat{B} first, then \hat{A}
For numbers the order never matters, 3 \times 5 = 5 \times 3. For operators it often does

[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}

The commutator is zero exactly when the order does not matter

Can two observables be sharp at once?

A sharp value a means an eigenfunction, \hat{A}\psi = a\psi. Sharp a and sharp b needs \hat{B}\psi = b\psi too. Then

\hat{A}\hat{B}\,\psi = \hat{A}\,(b\psi) = ab\,\psi \qquad\qquad \hat{B}\hat{A}\,\psi = \hat{B}\,(a\psi) = ba\,\psi

\big[\hat{A},\hat{B}\big]\,\psi = \big(\hat{A}\hat{B} - \hat{B}\hat{A}\big)\,\psi = 0 \quad \text{on the shared } \psi

So two observables can be sharp together only if their commutator gives zero on that state

Order matters: flip and swap

F = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad S = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}: \qquad [F, S] = FS - SF = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \neq 0

Both Hermitian, no shared eigenvector: F keeps (1,0), (0,1); S keeps (1,\pm 1). Spin along z, x (Ch. 5)

Example: x and d/dx

Find [x, d/dx] by acting on an arbitrary test function f.

x\,\frac{df}{dx} - \frac{d}{dx}\big(xf\big) = xf' - f - xf' = -f \qquad\Longrightarrow\qquad \left[x,\frac{d}{dx}\right] = -1

Multiply by -i\hbar: [\hat{x},\hat{p}] = i\hbar. The product rule leaves one term, for every f.

Recipe: act on a test function f, use the product rule, then drop f

Picture it: the two orders differ by f

\left[x,\frac{d}{dx}\right] = -1 \qquad\xrightarrow{\ \times\,(-i\hbar)\ }\qquad [\hat{x},\hat{p}] = i\hbar

Example: XP - PX on three points

Three grid points spaced \Delta x apart: X = \operatorname{diag}(x_1, x_2, x_3), and P = -i\hbar D_1 has P_{j,j\pm 1} = \mp\dfrac{i\hbar}{2\Delta x}. Find XP - PX.

Entry by entry, (XP - PX)_{jk} = (x_j - x_k)\,P_{jk} with x_j - x_{j\pm 1} = \mp\Delta x:

XP - PX = \frac{i\hbar}{2}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \qquad \text{averages the neighbors: } \approx i\hbar\,\psi

Its trace is 0, not 3i\hbar: no finite grid gives exactly i\hbar I.

Example: momentum and energy of a free particle

For a free particle \hat{H} = \hat{p}^2/2m. Is e^{ikx} an eigenfunction of \hat{p}? Of \hat{H}?

\begin{aligned} \hat{p}\,e^{ikx} &= -i\hbar\,(ik)\,e^{ikx} = \hbar k\,e^{ikx} \\ \hat{H}\,e^{ikx} &= \frac{1}{2m}\,\hat{p}\,\big(\hbar k\,e^{ikx}\big) = \frac{\hbar^2k^2}{2m}\,e^{ikx} \end{aligned}

Yes, both: this state has a sharp momentum and a sharp energy.

The two operators commute: \hat{H}\hat{p} = \hat{p}^3/2m = \hat{p}\hat{H}

Takeaway

Observables are Hermitian operators, the matrix version of real numbers: equal to their own conjugate transpose, with real eigenvalues and orthogonal eigenvectors. Two observables can be sharp together only when their operators commute, and [\hat{x},\hat{p}] = i\hbar says position and momentum never are.