28 May 2025

Quantum Postulates, Wavefunctions, and Operators

quantum-postulates wavefunction probability-current hermitian-operators commutators

Postulates of quantum mechanics

A pure state is represented by a non-zero vector $\lvert\psi\rangle$ in a complex Hilbert space. Vectors differing only by a non-zero complex factor represent the same physical ray. In the position representation, $\psi(\mathbf r,t)=\langle\mathbf r\vert\psi(t)\rangle$.
Each observable $A$ is represented by a linear Hermitian operator $\hat A$. Its possible measured values are its eigenvalues, defined by $\hat A\phi_a=a\phi_a$.
If a normalized state is expanded in an orthonormal eigenbasis as $\lvert\psi\rangle=\sum_a c_a\lvert a\rangle$, a measurement of $A$ gives $a$ with probability $\lvert c_a\rvert^2$. Immediately after an ideal measurement with result $a$, the state lies in the corresponding eigenspace.
The expectation value in a normalized state is $\langle A\rangle=\langle\psi\vert\hat A\vert\psi\rangle$.
Between measurements the state evolves according to $i\hbar\,\partial_t\lvert\psi\rangle=\hat H\lvert\psi\rangle$, where $\hat H$ is the Hamiltonian.

Probability interpretation and normalization

Born’s interpretation identifies

\[\boxed{\rho(\mathbf r,t)=\lvert\psi(\mathbf r,t)\rvert^2}\]

as probability per unit volume. Thus $\rho$ has SI units $\mathrm{m^{-3}}$, and the probability of finding the particle in a volume $\Omega$ is

\[P(\Omega,t)=\int_\Omega \lvert\psi(\mathbf r,t)\rvert^2d^3r.\]

A state representing one particle over all space is normalized by

\[\boxed{\int_{\mathbb R^3}\lvert\psi\rvert^2d^3r=1}.\]

If an unnormalized square-integrable function is $f$, then $\psi=Nf$ with

\[\lvert N\rvert^{-2}=\int_{\mathbb R^3}\lvert f\rvert^2d^3r.\]

The overall phase of $N$ is physically irrelevant because it cancels from $\lvert\psi\rvert^2$ and all expectation values.

Physical acceptability of a wavefunction

A bound-state wavefunction must be single-valued, finite, and square-integrable. It is continuous wherever the potential has no infinite discontinuity. To obtain the derivative condition in one dimension, integrate the stationary Schrödinger equation across $(x_0-\epsilon,x_0+\epsilon)$:

\[-\frac{\hbar^2}{2m}\int_{x_0-\epsilon}^{x_0+\epsilon}\psi^{\prime\prime}dx +\int_{x_0-\epsilon}^{x_0+\epsilon}(V-E)\psi\,dx=0.\]

Therefore

\[\psi^{\prime}(x_0+\epsilon)-\psi^{\prime}(x_0-\epsilon) =\frac{2m}{\hbar^2}\int_{x_0-\epsilon}^{x_0+\epsilon}(V-E)\psi\,dx.\]

If $V$ and $\psi$ remain finite, the integral vanishes as $\epsilon\to0$, so $\psi^{\prime}$ is continuous. At an infinite wall this argument fails and the physical boundary condition is instead $\psi=0$ at the wall; its derivative need not match the identically zero exterior derivative. For a normalizable state on an unbounded domain, $\psi$ must approach zero sufficiently rapidly as $\lvert\mathbf r\rvert\to\infty$.

Linearity and superposition

The Schrödinger equation is linear. If $\psi_1$ and $\psi_2$ satisfy the same homogeneous equation, then for constants $c_1,c_2$,

\[(i\hbar\partial_t-\hat H)(c_1\psi_1+c_2\psi_2) =c_1(i\hbar\partial_t-\hat H)\psi_1 +c_2(i\hbar\partial_t-\hat H)\psi_2=0.\]

Hence $c_1\psi_1+c_2\psi_2$ is also a state after normalization. The probability density contains cross terms $c_1c_2^{\ast}\psi_1\psi_2^{\ast}+\mathrm{c.c.}$; these interference terms distinguish a coherent superposition from a classical addition of probabilities.

Probability current density in three dimensions

For a real potential,

\[i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi,\]

and complex conjugation reverses the sign of $i$:

\[-i\hbar\frac{\partial\psi^{\ast}}{\partial t} =-\frac{\hbar^2}{2m}\nabla^2\psi^{\ast}+V\psi^{\ast}.\]

Multiply the first equation by $\psi^{\ast}$, the second by $\psi$, and subtract the second from the first. The potential terms cancel:

\[i\hbar\frac{\partial\lvert\psi\rvert^2}{\partial t} =-\frac{\hbar^2}{2m} (\psi^{\ast}\nabla^2\psi-\psi\nabla^2\psi^{\ast}).\]

Using

\[\psi^{\ast}\nabla^2\psi-\psi\nabla^2\psi^{\ast} =\nabla\!\cdot(\psi^{\ast}\nabla\psi-\psi\nabla\psi^{\ast}),\]

we obtain the continuity equation

\[\boxed{\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf j=0},\]

with

\[\boxed{ \mathbf j=\frac{\hbar}{2mi} (\psi^{\ast}\nabla\psi-\psi\nabla\psi^{\ast}) =\frac{\hbar}{m}\operatorname{Im}(\psi^{\ast}\nabla\psi) }.\]

Since $\rho$ has units $\mathrm{m^{-3}}$, $\mathbf j$ has units $\mathrm{m^{-2}s^{-1}}$. Integrating over a fixed volume and applying the divergence theorem gives

\[\frac d{dt}\int_\Omega\rho\,d^3r =-\oint_{\partial\Omega}\mathbf j\cdot d\mathbf S,\]

so probability decreases inside $\Omega$ only by flowing through its boundary.

Free-particle wavefunction

For $V=0$, a simultaneous momentum and energy eigenfunction is

\[\psi_{\mathbf p}(\mathbf r,t)=A \exp\!\left[\frac{i}{\hbar}(\mathbf p\cdot\mathbf r-Et)\right], \qquad E=\frac{p^2}{2m}.\]

Substitution gives $\nabla^2\psi=-p^2\psi/\hbar^2$ and $\partial_t\psi=-iE\psi/\hbar$, fixing both signs in the Schrödinger equation. Because $\lvert\psi_{\mathbf p}\rvert^2=\lvert A\rvert^2$ is constant, a plane wave cannot be normalized to unity over infinite space. The standard generalized normalization is

\[\langle\mathbf r\vert\mathbf p\rangle =\frac{1}{(2\pi\hbar)^{3/2}} e^{i\mathbf p\cdot\mathbf r/\hbar},\] \[\langle\mathbf p\vert\mathbf p^{\prime}\rangle =\delta^{(3)}(\mathbf p-\mathbf p^{\prime}).\]

A physical localized free state is a square-integrable superposition of these plane waves.

Eigenfunctions, eigenvalues, and Hermitian operators

An eigenfunction satisfies

\[\hat A\phi_a=a\phi_a.\]

Hermiticity means

\[\langle\phi\vert\hat A\psi\rangle =\langle\hat A\phi\vert\psi\rangle\]

for functions obeying the operator’s boundary conditions. If $\hat A\phi_a=a\phi_a$, then

The integration-by-parts equality is a statement about the operator and its domain. Vanishing boundary terms make a differential operator symmetric on that domain; a quantum observable requires the corresponding self-adjoint domain so that its spectrum and time evolution have the required physical properties.

\[a\langle\phi_a\vert\phi_a\rangle =\langle\phi_a\vert\hat A\phi_a\rangle =\langle\hat A\phi_a\vert\phi_a\rangle =a^{\ast}\langle\phi_a\vert\phi_a\rangle,\]

so $a=a^{\ast}$: an observable eigenvalue is real. For two eigenfunctions with $a\neq b$,

\[b\langle\phi_a\vert\phi_b\rangle =a\langle\phi_a\vert\phi_b\rangle,\]

which implies $\langle\phi_a\vert\phi_b\rangle=0$. Distinct eigenvalues of a Hermitian operator therefore have orthogonal eigenfunctions.

Position, momentum, and energy operators

Acting on the plane-wave phase gives

\[-i\hbar\nabla e^{i(\mathbf p\cdot\mathbf r-Et)/\hbar} =\mathbf p\,e^{i(\mathbf p\cdot\mathbf r-Et)/\hbar},\] \[i\hbar\partial_t e^{i(\mathbf p\cdot\mathbf r-Et)/\hbar} =E\,e^{i(\mathbf p\cdot\mathbf r-Et)/\hbar}.\]

Thus the position representation uses

\[\boxed{\hat{\mathbf r}=\mathbf r},\qquad \boxed{\hat{\mathbf p}=-i\hbar\nabla},\qquad \boxed{\hat E=i\hbar\frac{\partial}{\partial t}}.\]

For one Cartesian component and any differentiable test function $f$,

\[\begin{aligned} [\hat x,\hat p_x]f &=x(-i\hbar f^{\prime})-(-i\hbar)(xf)^{\prime}\\ &=-i\hbar xf^{\prime}+i\hbar(f+xf^{\prime})=i\hbar f. \end{aligned}\]

Hence

\[\boxed{[\hat x,\hat p_x]=i\hbar}.\]

Position and momentum are non-commuting operators and cannot have a complete common eigenbasis. Commuting Hermitian operators, $[\hat A,\hat B]=0$, can be simultaneously diagonalized after resolving any degeneracy.

Finally, for a normalized wavefunction,

\[\boxed{\langle\mathbf r\rangle =\int\psi^{\ast}\mathbf r\psi\,d^3r},\] \[\boxed{\langle\mathbf p\rangle =\int\psi^{\ast}(-i\hbar\nabla)\psi\,d^3r}.\]

Integration by parts makes $-i\hbar\nabla$ symmetric when the surface term vanishes. On the full line, its standard self-adjoint domain consists of square-integrable, absolutely continuous states whose first derivative is also square-integrable; on a finite interval, an appropriate periodic or phase-periodic boundary domain gives a self-adjoint momentum operator.

Solved Problems

1. Normalizing a localized exponential state

In one dimension let $\psi(x)=N e^{-a\lvert x\rvert}$, where $a>0$. Normalize the state, find $\langle\lvert x\rvert\rangle$, and evaluate the probability of $\lvert x\rvert<0.500\ \mathrm{nm}$ when $a=2.00\ \mathrm{nm^{-1}}$.

Solution. Choose $N$ real and positive; any constant phase would be physically irrelevant. Evenness gives

\[1=2N^2\int_0^\infty e^{-2ax}dx =2N^2\left(\frac1{2a}\right),\]

\[\boxed{N=\sqrt a}.\]

This has units $\mathrm{m^{-1/2}}$, as a one-dimensional wavefunction must. The mean absolute position is

\[\begin{aligned} \langle\lvert x\rvert\rangle &=2a\int_0^\infty x e^{-2ax}dx\\ &=\frac1{2a} =\boxed{0.250\ \mathrm{nm}}. \end{aligned}\]

For $b=0.500\ \mathrm{nm}$,

\[\begin{aligned} P(\lvert x\rvert<b) &=2a\int_0^b e^{-2ax}dx\\ &=1-e^{-2ab} =1-e^{-2} =\boxed{0.8647}. \end{aligned}\]

The state is real, so its probability current is zero; there is no preferred direction of flow. As $a$ increases, the normalization amplitude grows while the length $1/a$ and $\langle\lvert x\rvert\rangle$ shrink, consistently describing stronger localization. The probability tends to zero as $b\to0$ and to one as $b\to\infty$.

2. Expectation values for a polynomial trial wavefunction

On $0<x<L$, let $\psi(x)=N x(L-x)$ and let it vanish at the endpoints. Normalize it and find $\langle x\rangle$ and $\langle p_x\rangle$.

Solution. The polynomial is real and symmetric about $L/2$. Direct normalization gives

\[\int_0^L x^2(L-x)^2dx=\frac{L^5}{30},\]

and hence

\[\boxed{N=\frac{\sqrt{30}}{L^{5/2}}}.\]

The factor $Nx(L-x)$ has units $L^{-1/2}$, as required. For position,

\[\langle x\rangle =N^2\int_0^L x^3(L-x)^2dx =\boxed{\frac L2}.\]

For momentum, keep the operator sign $-i\hbar\,d/dx$:

\[\begin{aligned} \langle p_x\rangle &=-i\hbar\int_0^L\psi\frac{d\psi}{dx}dx\\ &=-\frac{i\hbar}{2}\left[\psi^2(x)\right]_0^L =\boxed{0}. \end{aligned}\]

The zero integral is real and reflects equal positive- and negative-momentum content rather than absence of kinetic energy. It does not by itself assert that $-i\hbar,d/dx$ is self-adjoint on the rigid-wall Dirichlet domain; self-adjointness depends on the operator domain. The midpoint value follows independently from reflection symmetry $x\mapsto L-x$ and remains $L/2$ under any overall phase choice.

Descriptive Questions

State the quantum postulates used to predict possible measurement results, their probabilities, and subsequent time evolution.
Why must a physical bound-state wavefunction be square-integrable, and under what circumstances can its first derivative be discontinuous?
Derive the three-dimensional probability-current density and explain the sign of each term in the continuity equation.
How do Hermiticity and commutation determine real eigenvalues, orthogonality, and the possibility of simultaneous eigenstates?

Numerical Problems

An observable has eigenvalues $3.00$ and $7.00$ with probabilities $1/5$ and $4/5$. Find its mean and standard deviation.
Final answer: $\boxed{\langle A\rangle=6.20,\quad\Delta A=1.60}$.
A one-dimensional plane wave has density $\rho=4.00\times10^6\ \mathrm{m^{-1}}$ and positive momentum $2.00\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$. Find its electron probability current.
Final answer: $\boxed{j=+8.78\times10^{12}\ \mathrm{s^{-1}}}$.
A constant wavefunction is normalized only on an interval of length $2.00\ \mathrm{nm}$. Find its amplitude magnitude.
Final answer: $\boxed{\lvert A\rvert=L^{-1/2}=2.24\times10^4\ \mathrm{m^{-1/2}}}$.

Every added normalization, expectation value, and probability-current value is checked in the MJ-11 problem-verification worksheet; every printed residual and check is zero.