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Wave functions as vectors with an inner product, observables as Hermitian operators, and why their eigenvalues are real and their eigenvectors orthogonal.
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By the end of this lesson you will be able to test whether an operator is Hermitian, find eigenvalues and eigenvectors of two-state observables, and compute expectation values with matrices.
You know how to find eigenvalues and eigenvectors of a matrix from linear algebra, and from the first unit how to compute expectation values as $\int\Psi^*\hat{Q}\Psi\,dx$. You have seen that a general state is a sum of stationary states, $\Psi = \sum c_n\psi_n$. This lesson shows that these are the same idea: wave functions are vectors, operators are matrices, and measurement is about eigenvectors.
| Term | What it means |
|---|---|
| Hilbert space | The space of square-integrable functions, with inner product $\langle f\vert g\rangle = \int f^*g\,dx$. |
| Inner product | $\langle f\vert g\rangle$, the generalization of the dot product; $\langle f\vert f\rangle$ is the squared length. |
| Orthonormal basis | A set of vectors with $\langle e_m\vert e_n\rangle = \delta_{mn}$ in terms of which every vector can be expanded. |
| Hermitian conjugate | $\hat{Q}^\dagger$, defined by $\langle f\vert \hat{Q}g\rangle = \langle\hat{Q}^\dagger f\vert g\rangle$; for a matrix, the conjugate transpose. |
| Hermitian operator | One equal to its own conjugate, $\hat{Q}^\dagger = \hat{Q}$; the operators of observables. |
| Eigenvalue equation | $\hat{Q}f = qf$: acting with the operator returns the same function times a number. |
| Determinate state | An eigenstate of $\hat{Q}$, in which every measurement of $Q$ gives the same value. |
Wave functions can be added and multiplied by complex numbers and still be wave functions: they form a vector space. The inner product
$$\langle f|g\rangle = \int f^*(x)\,g(x)\,dx$$
plays the role of the dot product, with $\langle f|f\rangle$ the squared length and $\langle f|g\rangle = 0$ meaning orthogonal. The space of functions with finite length is called Hilbert space. The stationary states $\psi_n$ of the square well form an orthonormal basis, $\langle\psi_m|\psi_n\rangle = \delta_{mn}$, and expanding $\Psi = \sum c_n\psi_n$ is just writing a vector in components, $c_n = \langle\psi_n|\Psi\rangle$.
An operator acting on vectors is, in a basis, a matrix, $Q_{mn} = \langle e_m|\hat{Q}e_n\rangle$. Measured values must be real, and $\langle Q\rangle = \langle\Psi|\hat{Q}\Psi\rangle$ must be real for every state. That requires
$$\langle f|\hat{Q}g\rangle = \langle\hat{Q}f|g\rangle \quad\text{for all } f, g,$$
which defines a Hermitian operator; as a matrix, $Q_{nm}^ = Q_{mn}$, equal to its conjugate transpose. Two theorems follow. The eigenvalues of a Hermitian operator are real. And eigenvectors with different eigenvalues are orthogonal*; in a finite space they form a complete basis. Every observable — position, momentum, energy, angular momentum — is represented by a Hermitian operator, and its eigenvalues are the values a measurement can return.
Another way: picture
Picture ordinary arrows in the plane. A symmetric real matrix stretches the plane along two perpendicular directions, its eigenvectors, by amounts equal to its eigenvalues. A Hermitian operator is the complex, possibly infinite-dimensional version of that: a stretching along perpendicular axes by real amounts. The axes are the determinate states, and the stretch factors are the values a measurement can give.
Another way: steps
Checks. The eigenvalues must be real, must add to the trace and multiply to the determinant. Eigenvectors of distinct eigenvalues must have zero inner product. An expectation value must lie between the smallest and largest eigenvalue.
Suppose $\hat{Q}f = qf$ with $\hat{Q}$ Hermitian. Then $\langle f|\hat{Q}f\rangle = q\langle f|f\rangle$, while $\langle\hat{Q}f|f\rangle = q^\langle f|f\rangle$, because the inner product conjugates its first slot. Hermiticity makes the two equal, and $\langle f|f\rangle \ne 0$, so $q = q^$: the eigenvalue is real.
Now let $\hat{Q}f = qf$ and $\hat{Q}g = q'g$ with $q \ne q'$. Then $\langle f|\hat{Q}g\rangle = q'\langle f|g\rangle$ and $\langle\hat{Q}f|g\rangle = q\langle f|g\rangle$, using that $q$ is real. Hermiticity makes these equal, so $(q' - q)\langle f|g\rangle = 0$, and since $q' \ne q$, $\langle f|g\rangle = 0$. When eigenvalues coincide, a degenerate case, the eigenvectors can still be chosen orthogonal by the Gram–Schmidt procedure.
Both proofs use only the definition, so they hold for matrices and for differential operators alike. They explain why the square well's sine functions were orthogonal: they are eigenfunctions of a Hermitian Hamiltonian with different energies.
To check that $\hat{p} = -i\hbar\,d/dx$ is Hermitian, integrate by parts:
$$\langle f|\hat{p}g\rangle = \int f^(-i\hbar g')\,dx = -i\hbar\,f^g\Big|_{-\infty}^{\infty} + \int(-i\hbar f')^*g\,dx.$$
For normalizable functions the boundary term vanishes, and the remaining integral is $\langle\hat{p}f|g\rangle$. The factor of $i$ is essential: $d/dx$ alone is anti-Hermitian, picking up a minus sign, and $-i$ turns that sign around.
Hermiticity depends on the space of functions as well as the formula. On the infinite square well, the boundary term vanishes because the functions vanish at the walls. The eigenfunctions of $\hat{p}$, plane waves $e^{ipx/\hbar}$, are not normalizable at all, which is why they sit just outside Hilbert space and need the delta-function normalization of the next lesson. The mathematician's careful version of these ideas, self-adjoint operators, was worked out by John von Neumann in the late 1920s.
Heisenberg's matrix mechanics and Schrödinger's wave mechanics looked, in 1926, like rival theories. Schrödinger showed they are the same: expanding wave functions in a basis turns differential operators into (infinite) matrices. In the harmonic oscillator basis, for example, $\hat{x}$ has entries only next to the diagonal, $x_{n, n+1} = \sqrt{(n + 1)\hbar/2m\omega}$, because the ladder operators connect only neighbors.
Which view to use is a matter of convenience. For a particle in a potential, functions are natural. For spin, the space is finite — two-dimensional for an electron — and matrices are the only view. For quantum computing, where registers of qubits live in spaces of dimension $2^n$, everything is matrices, and algorithms are sequences of unitary matrices acting on vectors. The two-by-two examples in this lesson are not toys; they are the complete description of a single qubit.
Some Hermitian operators, such as position and momentum, have eigenvalues that form a continuum rather than a discrete list, and their eigenfunctions are not normalizable. Position's eigenfunction with eigenvalue $y$ is the Dirac delta $\delta(x - y)$; momentum's is the plane wave $e^{ipx/\hbar}/\sqrt{2\pi\hbar}$. They are normalized to delta functions instead of to one: $\langle f_p|f_{p'}\rangle = \delta(p - p')$.
The rules carry over with sums replaced by integrals. A state expands as $\Psi = \int c(p)f_p\,dp$, and $|c(p)|^2\,dp$ is the probability of finding momentum between $p$ and $p + dp$. The coefficient $c(p)$ is the momentum-space wave function, the Fourier transform of $\Psi(x)$. Energy can have both kinds of spectrum at once: the hydrogen atom has discrete bound states below zero and a continuum of scattering states above.
The same state can be written in many bases, just as an arrow can be described by its components along different axes. Going from one orthonormal basis to another is done by a unitary matrix $U$, one with $U^\dagger U = I$, which preserves every inner product: lengths and angles, and so probabilities, are unchanged. An operator's matrix transforms as $Q' = U^\dagger QU$; its eigenvalues, trace and determinant are the same in every basis.
Diagonalizing an observable means choosing the basis of its own eigenvectors, where its matrix has the eigenvalues on the diagonal and zeros elsewhere. In that basis, measuring the observable is easy to describe: each component's squared magnitude is the probability of the corresponding eigenvalue. Most of the work in quantum mechanics is choosing a good basis — the energy basis for time evolution, the position basis for where a particle is — and translating between them.
The quantum processors built by Google, IBM and others use superconducting circuits cooled to a hundredth of a kelvin. Each qubit's two lowest states form a two-dimensional Hilbert space, and its Hamiltonian is a $2 \times 2$ Hermitian matrix. In an older design, the flux qubit, the basis states are currents circulating clockwise and counterclockwise, and $H = \begin{pmatrix} \varepsilon & \Delta \\ \Delta & -\varepsilon \end{pmatrix}$, where the bias $\varepsilon$ is set by an applied magnetic flux and $\Delta$ is the tunneling between the two currents.
The energy eigenstates are not the current states but superpositions of them, and the splitting is $2\sqrt{\varepsilon^2 + \Delta^2}$. With splittings of $10$ to $30$ μeV, the transition frequencies lie between about $2$ and $8$ GHz, in the microwave band, so the qubits are controlled with microwave pulses much like those in a cell phone. Engineers tune $\varepsilon$ to move each qubit's frequency, bringing pairs into resonance to entangle them — diagonalizing a Hermitian matrix every time they design a gate.
The mathematics of Hermitian matrices is not unique to quantum mechanics. The vibrations of a bridge or a building are found by diagonalizing a real symmetric stiffness matrix: its eigenvectors are the normal modes, and its eigenvalues give the squares of their frequencies. Engineers designing skyscrapers in earthquake zones compute these modes to make sure no natural frequency matches the shaking of the ground.
The same theorem, real eigenvalues with perpendicular eigenvectors, guarantees that every rigid body has three perpendicular principal axes about which it spins smoothly, and that every stress in a material can be described by three perpendicular principal stresses. In data science, principal component analysis diagonalizes a symmetric covariance matrix to find the directions along which data vary most. Quantum mechanics adds complex numbers and infinite dimensions, but the heart of it — perpendicular axes and real stretch factors — is the same.
Any linear operator has eigenvalues, but only Hermitian ones are guaranteed real eigenvalues and orthogonal eigenvectors. The operator $d/dx$ has eigenfunctions $e^{kx}$ with any complex $k$; it cannot represent a measurable quantity until it is multiplied by $-i\hbar$ to make the momentum operator. A matrix like $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ has only one eigenvector and cannot describe a measurement at all.
A second misconception is that symmetric and Hermitian are the same. For complex matrices they differ: $\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$ is symmetric but not Hermitian, and its eigenvalues are $\pm i$. Hermitian requires the mirror entry to be the complex conjugate, as in $\begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$, whose eigenvalues are $\pm 1$.
Test whether $Q = \begin{pmatrix} 2 & 3i \\ -3i & 2 \end{pmatrix}$ is Hermitian.
$Q^\dagger = \begin{pmatrix} 2^ & (-3i)^ \\ (3i)^ & 2^ \end{pmatrix} = \begin{pmatrix} 2 & 3i \\ -3i & 2 \end{pmatrix} = Q$
Transpose, then conjugate every entry.
Write the characteristic equation.
$(2 - \lambda)^2 - (3i)(-3i) = (2 - \lambda)^2 - 9 = 0$
$\det(Q - \lambda I) = 0$; note $(3i)(-3i) = 9$.
Solve for the eigenvalues.
$\lambda = 2 \pm 3 = 5, \ -1$
Real, as the theorem promises.
Find the eigenvector for $\lambda = 5$.
$-3v_1 + 3iv_2 = 0 \quad\Rightarrow\quad v = \tfrac{1}{\sqrt{2}}\begin{pmatrix} i \\ 1 \end{pmatrix}$
The first row of $(Q - 5I)v = 0$, then normalize.
Check orthogonality with the other eigenvector.
$w = \tfrac{1}{\sqrt{2}}\begin{pmatrix} -i \\ 1 \end{pmatrix}: \quad \langle v|w\rangle = \tfrac{1}{2}\left((-i)(-i) + 1\right) = \tfrac{1}{2}(-1 + 1) = 0$
Conjugate the first vector's components before multiplying.
A state is $c = \tfrac{1}{5}(3, 4)$. Check its normalization.
$c^\dagger c = \tfrac{1}{25}(9 + 16) = 1$
The squared length must be one.
Choose the observable to average.
$Q = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$
Real and symmetric, so Hermitian.
Act with the matrix on the state.
$Qc = \tfrac{1}{5}\begin{pmatrix} 3 + 8 \\ 6 + 16 \end{pmatrix} = \tfrac{1}{5}\begin{pmatrix} 11 \\ 22 \end{pmatrix}$
Rows times column.
Take the inner product with the state.
$\langle Q\rangle = \tfrac{1}{25}(3 \times 11 + 4 \times 22) = \tfrac{121}{25} = 4.84$
The sandwich $c^\dagger Qc$.
Find the eigenvalues of the matrix.
$\lambda^2 - 5\lambda + 0 = 0 \quad\Rightarrow\quad \lambda = 0, \ 5$
Trace $5$, determinant $4 - 4 = 0$.
Check that the average lies between them.
$0 \le 4.84 \le 5$
The state is mostly the $\lambda = 5$ eigenvector, $(1, 2)/\sqrt{5}$.
Start from an eigenvalue equation.
$\hat{Q}f = qf, \qquad \hat{Q}^\dagger = \hat{Q}$
Any Hermitian operator and any of its eigenfunctions.
Let the operator act on the right.
$\langle f|\hat{Q}f\rangle = \langle f|qf\rangle = q\langle f|f\rangle$
The inner product is linear in its second slot.
Let the operator act on the left.
$\langle\hat{Q}f|f\rangle = \langle qf|f\rangle = q^*\langle f|f\rangle$
The inner product conjugates numbers in its first slot.
Use Hermiticity to equate them.
$q\langle f|f\rangle = q^*\langle f|f\rangle$
By definition, the operator can act on either side.
Cancel the norm.
$\langle f|f\rangle > 0 \quad\Rightarrow\quad q = q^*$
An eigenfunction is never the zero function.
Apply the same idea to two eigenfunctions.
$\hat{Q}g = q'g: \quad q'\langle f|g\rangle = \langle f|\hat{Q}g\rangle = \langle\hat{Q}f|g\rangle = q\langle f|g\rangle$
Both eigenvalues are now known to be real.
Conclude that they are orthogonal.
$(q' - q)\langle f|g\rangle = 0 \quad\Rightarrow\quad \langle f|g\rangle = 0 \text{ if } q' \ne q$
Distinct eigenvalues force a zero inner product.
Act on the symmetric vector.
$(1, 1): \ \lambda = 7 + 2 = 9$
Each row sums to nine.
Act on the antisymmetric vector.
$(1, -1): \ \lambda = 7 - 2 = 5$
The off-diagonal entry subtracts.
Check against the trace.
Which of these matrices could represent an observable, that is, which is Hermitian?
Complete the worked solution: find the eigenvalues of $\begin{pmatrix} 10 & 1 \\ 1 & 10 \end{pmatrix}$ and check them against the trace.
Act on the symmetric vector $(1, 1)$.
$\lambda_+ = 10 + 1 =$ p
Each row sums to the same number.
Act on the antisymmetric vector $(1, -1)$.
$\lambda_- = 10 - 1 =$ q
The off-diagonal entry now subtracts.
Add the two eigenvalues together.
$\lambda_+ + \lambda_- =$ t
The sum of the eigenvalues equals the trace.
Match each idea from the formalism to its statement.
| $\int f^*g\,dx$ | $\langle f|\hat{Q}g\rangle = \langle\hat{Q}f|g\rangle$ | real | orthogonal | |
|---|---|---|---|---|
| the inner product | ||||
| Hermitian | ||||
| eigenvalues of a Hermitian operator | ||||
| eigenfunctions with distinct eigenvalues |
For the matrix $\begin{pmatrix} 8 & 2 \\ 2 & 8 \end{pmatrix}$, fill in the two eigenvalues, their sum and their product.
| value | |
|---|---|
| larger eigenvalue | |
| smaller eigenvalue | |
| sum | |
| product |
A system is in the normalized state $\tfrac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix}$, and an observable is represented by $\begin{pmatrix} 9 & 2 \\ 2 & 1 \end{pmatrix}$. What is its expectation value?
Answer:
An observable is represented, in a two-state basis, by the matrix $\begin{pmatrix} 16 & 5 \\ 5 & -8 \end{pmatrix}$. What is the largest value a measurement of it can give?
Answer:
A superconducting qubit's Hamiltonian, in the basis of its two current states, is $\begin{pmatrix} \varepsilon & \Delta \\ \Delta & -\varepsilon \end{pmatrix}$ with bias $\varepsilon = 12$ μeV and tunneling $\Delta = 5$ μeV. At what frequency, in GHz, must a microwave pulse drive it to flip it between its energy eigenstates? Use $h = 4.1357$ μeV/GHz.
Answer: GHz
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For the matrix $\begin{pmatrix} 6 & 7 \\ 7 & 6 \end{pmatrix}$, fill in the two eigenvalues, their sum and their product.
| value | |
|---|---|
| larger eigenvalue | |
| smaller eigenvalue | |
| sum | |
| product |
You can treat wave functions as vectors and observables as Hermitian matrices. Explain to someone why a measured value can never be a complex number.
17. Your turn: find the eigenvalues of $\begin{pmatrix} 7 & 2 \\ 2 & 7 \end{pmatrix}$., step 3
$9 + 5 = 14 = 7 + 7$
The eigenvalues add to the trace.