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.


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}.



Show that an eigenfunction representing a stationary state satisfy eq40.


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,



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.



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


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.



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