Eigenfunctions and eigenvalue equations

An eigenfunction in a vector space of function is a non-zero function, $f$ , that when acted upon by an operator $\hat{O}$ , returns the same function but multiplied by a constant $c$ , which is known as an eigenvalue.

$\hat{O}f=cf$

The above equation is called an eigenvalue equation. The time-independent Schrodinger equation

$\hat{H}\psi=E\psi\; \; \; \; \; \; \; \; 40$

where $\hat{H}$ is a Hermitian operator called the Hamiltonian, is also an eigenvalue equation. We say that the eigenfunction $\psi$ of a stationary state is a solution to the Schrodinger equation. For a one-dimensional, one-particle system in a stationary state, $\hat{H}=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x)$ , and

$\left [-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x)\right ]\psi=E\psi\; \; \; \; \; \; \; \; 41$

where $V(x)=\frac{Ze^{2}}{4\pi\varepsilon _0x}$ .

Question

Show that an eigenfunction representing a stationary state satisfy eq40.

Answer

Stationary states of $\hat{H}$ are determinate states, meaning a measurement of the energy of a particle in that state results in the same value of $E$ . Hence, the variance $\sigma^{2}$ of the observable of $\hat{H}$ for a stationary state would be zero, i.e. $\sigma^{2}=\frac{\sum_{i=1}^{N}(x_i-\mu)^{2}}{N}=0$ . Since $\frac{\sum_{i=1}^{N}(x_i-\mu)^{2}}{N}$ is also the average value of $(x_i-\mu)^{2}$ ,

$\sigma^{2}=\langle(H-\langle H\rangle)^{2}\rangle=\langle(H-E)^{2}\rangle=0$

where $H$ is the observable of $\hat{H}$ and $\langle H\rangle=\langle\psi\vert\hat{H}\vert\psi\rangle$ denotes the average value of $\hat{H}$ , which is $E$ since $\psi$ is a stationary state of $\hat{H}$ . Next, we assume that the operator $(\widehat {H-E})^{2}$ for the observable $(H-E)^{2}$ is $(\hat{H}-E)^{2}$ . Our assumption is only valid if $(\hat{H}-E)^{2}$ is Hermitian, which can be proven as follows:

$\langle(H-E)^{2}\rangle=\langle\psi\vert(\hat{H}-E)^{2}\vert\psi\rangle=\int \psi^{*}(\hat{H}-E) ^{2}\psi d\tau$

$\small =\int \psi^{*}(\hat{H}-E)[(\hat{H}-E)\psi]\, d\tau=\int \psi^{*}\hat{H}[(\hat{H}-E)\psi]\, d\tau-\int \psi^{*}E[(\hat{H}-E)\psi]\, d\tau$

Using the property of the Hermitian operator $\hat{H}$ and the condition that $E$ is real ( $E=E^{*}$ ),

$\langle(H-E)^{2}\rangle=\int \psi\{\hat{H}[(\hat{H}-E)\psi]\} ^{*}d\tau-\int \psi\{E[(\hat{H}-E)\psi]\} ^{*}d\tau$

$=\int \psi\{(\hat{H}-E) [(\hat{H}-E)\psi]\}^{*} d\tau=\int \psi[(\hat{H}-E)]^{2}\psi]^{*}\, d\tau$

This implies that $(\hat{H}-E)^{2}$ is a Hermitian operator and so,

$\langle(H-E)^{2}\rangle=\langle\psi\vert(\hat{H}-E)^{2}\vert\psi\rangle=\langle\psi\vert(\hat{H}-E)[(\hat{H}-E)\psi]\rangle =0$

Since $(\hat{H}-E)\psi$ produces another ket, the property of a Hermitian operator gives

$\langle\psi\vert(\hat{H}-E)[(\hat{H}-E)\psi]\rangle =\langle[\hat{H}-E)\psi]\vert[(\hat{H}-E)\psi]\rangle ^{*}=0$

From the positive semi-definiteness property of the inner product space, where $\langle f\vert f\rangle \geq 0$ , with $\langle f\vert f\rangle = 0$ if $f=0$ ,

$\hat{H}\psi=E\psi$

Examples of functions that are eigenfunctions of $\hat{H}$ are $\psi=sinx$ and $\psi=e^{-kx}$ , and examples of functions that are not eigenfunctions of $\hat{H}$ are $\psi=sin(ax)+sin(bx)$ and $\psi=e^{-k_1x}+e^{-k_2x}$ . We can easily verify whether a function is an eigenfunction or not one by substituting it into eq41.

Question

Why is an eigenfunction defined as a non-zero function?

Answer

Since $\hat{H}0=E0=0$ , an eigenfunction that is zero would have an infinite number of eigenvalues (any value of $E$ multiplied by zero is zero), which contradicts the Born interpretation that an acceptable eigenfunction representing a stationary state in quantum mechanics must be single-valued.

Eigenfunctions and eigenvalue equations

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