2.4 Properties of Energy Eigenstates

We will now discuss some properties of the energy eigenstates. These apply for the eigenstates of the infinite square well (equation 2.63), but also to the solutions of the TISE for any other problem.

2.4.1 Orthonormality

Orthonormality consists of two properties of a set of functions: being mutually orthogonal, and being all normalised. Orthogonality of wavefunctions in quantum mechanics is measured using an inner product, which is defined by the integral:

\int_{-\infty}^{\infty}\psi_{n}^{\ast}\psi_{m}\operatorname{\mathrm{d}}\!x.

(2.70)

If this integral is zero, the two wavefunctions are orthogonal. If it is non zero, then they are not orthogonal. Sometimes the inner product is referred to as the overlap of the two wavefunctions. A set of wavefunctions being mutually orthogonal means that each wavefunction in the set is orthogonal to every other one.

Taking the two conditions together, a set of wavefunctions $\psi_{n}$ are said to be mutally orthonormal if

\int_{-\infty}^{\infty}\psi_{n}^{\ast}\psi_{m}\operatorname{\mathrm{d}}\!x=% \delta_{n,m}=\begin{cases}1&\text{if }n=m\\ 0&\text{if }n\neq m.\end{cases}

(2.71)

This statement implies that the wavefunctions are mutually orthogonal except in the case where we are taking the inner product of a wavefunction with itself, in which case we have the integral of the magnitude squared which would be one to have the wavefunction normalised. We have introduced the Kronecker delta $\delta_{n,m}$ which is shorthand for the piecewise definition on the right so simplify the notation. In Dirac notation, this is written as

\langle n\mathclose{}|\mathopen{}m\rangle=\delta_{n,m}.

(2.72)

Why is it always the case that the wavefunctions are orthonormal? We will prove later on that it is because the Hamiltonian is a Hermitian operator.

2.4.2 Eigenfunctions form a Basis

For any problem in quantum mechanics, the energy eigenfunctions form what is known as a basis. This means that we can expand any valid wavefunction $\psi$ which fulfills the boundary conditions as a linear superposition of energy eigenstates:

\psi(x,t)=\sum_{n=1}^{\infty}c_{n}\psi_{n}(x,t)=\sum_{n=1}^{\infty}c_{n}u_{n}(% x)e^{-\frac{iE_{n}}{\hbar}t},

(2.73)

where $\psi_{n}(x,t)$ are the energy eigenfunctions, $u_{n}(x)$ are the spatial part, $E_{n}$ are the energy eigenvalues, and $c_{n}$ are the coefficients in the superposition. Note that we have taken $n=1$ to $\infty$ to be the indices for the sum since those are the indices for the eigenfunctions of the infinite square well, but when applying this to other problems we could have different labels.

This is exactly what we saw in section 2.2 when we constructed a normalisable wave packet as an integral over energy/momentum eigenstates. In that case (for the free particle), a finite or even discretely infinite superposition was still not normalisable, but in general it will be because the energy eigenstates are normalisable. Note that the probability density of a superposition does depend on time in almost all cases, because the time-dependent phases in the energy eigenstates rotate at different frequencies. This causes them to have a phase difference which causes the time-dependence.

For the wavefunction $\psi$ to be correctly normalised, we must then have

\sum_{n=1}^{\infty}\lvert c_{n}\rvert^{2}=1.

(2.74)

There is a physical interpretation for the coefficients of the superposition $c_{n}$ . If we look at the expectation value of the Hamiltonian for the above superposition, we get

$\displaystyle\langle\hat{H}\rangle_{\psi}$	$\displaystyle=\int_{-\infty}^{\infty}\psi^{\ast}\hat{H}\psi\operatorname{% \mathrm{d}}\!x$	(2.75)
	$\displaystyle=\int_{-\infty}^{\infty}\sum_{m=1}^{\infty}c_{m}^{\ast}\psi_{m}^{% \ast}\sum_{n=1}^{\infty}c_{n}\hat{H}\psi_{n}\operatorname{\mathrm{d}}\!x$	(2.76)
	$\displaystyle=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{n}c_{m}^{\ast}c_{n}\int% _{-\infty}^{\infty}\psi_{m}^{\ast}\psi_{n}\operatorname{\mathrm{d}}\!x$	(2.77)
	$\displaystyle=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{n}c_{m}^{\ast}c_{n}% \delta_{n,m}$	(2.78)
	$\displaystyle=\sum_{n=1}^{\infty}E_{n}\lvert c_{n}\rvert^{2}.$	(2.79)

If we compare this to equation 1.29, we see that $\lvert c_{n}\rvert^{2}$ is the probability of finding the value $E_{n}$ if the particle’s energy is measured.

This is immediately apparent if we write the superposition using Dirac notation, it takes the form of equation 1.3:

\lvert\psi\rangle=\sum_{n=1}^{\infty}c_{n}\lvert n\rangle=\sum_{n=1}^{\infty}% \langle n\mathclose{}|\mathopen{}\psi\rangle\lvert n\rangle,

(2.80)

where $\langle n\mathclose{}|\mathopen{}\psi\rangle$ , the inner product or overlap between $\lvert\psi\rangle$ and the eigenstate $\lvert n\rangle$ , is the probability that we find the particle in state $\lvert n\rangle$ upon measuring the energy.

2.4.3 Expanding in the Energy Eigenbasis

What about if we have a wavefunction that is not clearly written as a superposition of energy eigenstates. How can we find its time evolution? For example, consider an initial state for the infinite square well

\psi(x,0)=N\sin^{3}\left(\frac{\pi x}{L}\right)\quad\text{for }0<x<L,

(2.81)

where $\psi(x,0)=0$ outside the well. This is a valid wavefunction for the system (it is a valid wavefunction which can be normalised for a suitable choice of $N$ and it goes to zero at the boundaries of the well), but it is not written in the form 2.73.

Since it is not of the form of one of the energy eigenfunctions, given by equation 2.61, it must be a superposition state. The challenge is finding the right coefficients $c_{n}$ to write it in the form 2.73. This is basically a generalised Fourier series problem, with the energy eigenfunctions taking the place of sine and cosine. Notice that the energy eigenfunctions share the same essential properties for solving this problem that sine and cosine have, namely that they for a complete and orthonormal basis for the set of functions we are interested in.

Let us suppose that $\psi(x,0)$ from above takes the form

\psi(x,0)=\sum_{m=1}^{\infty}c_{m}u_{m}(x),

(2.82)

and then, inspired by the connection to Fourier series, let us multiply by $u_{n}(x)*\ast$ and integrate over all space.

$\displaystyle\int_{-\infty}^{\infty}u_{n}(x)^{\ast}\psi(x,0)\operatorname{% \mathrm{d}}\!x$	$\displaystyle=\int_{-\infty}^{\infty}u_{n}(x)^{\ast}\sum_{m=1}^{\infty}c_{m}u_% {m}(x)\operatorname{\mathrm{d}}\!x$	(2.83)
	$\displaystyle=\sum_{m=1}^{\infty}c_{m}\int_{-\infty}^{\infty}u_{n}(x)^{\ast}u_% {m}(x)\operatorname{\mathrm{d}}\!x$	(2.84)
	$\displaystyle=\sum_{m=1}^{\infty}c_{m}\delta_{n,m}$	(2.85)
	$\displaystyle=c_{n}.$	(2.86)

So we have found that for any initial wavefunction $\psi(x,0)$ , we can expand in energy eigenfunctions with coefficients given by the inner product of the spatial part of the energy eigenfunction with the wavefunction:

c_{n}=\int_{-\infty}^{\infty}u_{n}^{\ast}\psi(x,0)\operatorname{\mathrm{d}}\!x.

(2.87)

Of course, we know this already from equation 2.80, but it is nice to prove it another way. Once we have the coefficients $c_{n}$ , the time evolution is straightforwardly given by 2.73.

This derivation makes sense because it shows that, in the same way that Fourier series work, if a given wavefunction looks “similar” to one of the eigenstates, the inner product between them, and therefore the corresponding coefficient, will be larger than the others. This means that the time evolution of the superposition will be mostly similar to that of the most similar eigenstate.

2.4.4 Time Evolution of Superpositions

Let us consider, as a simple example, the following superposition of two energy eigenstates:

\psi(x,t)=\frac{1}{\sqrt{2}}(u_{1}(x)e^{-\frac{iE_{1}}{\hbar}t}+u_{2}(x)e^{-% \frac{iE_{2}}{\hbar}t}).

(2.88)

The time evolution of the probability density is given by

\lvert\psi(x,t)\rvert^{2}=\frac{1}{2}\lvert u_{1}(x)\rvert^{2}+\frac{1}{2}% \lvert u_{2}(x)\rvert^{2}+u_{1}^{\ast}(x)u_{2}(x)\cos\left(\frac{(E_{2}-E_{1})% t}{\hbar}\right).

(2.89)

If no measurement of the particle is made then its wavefunction evolves deterministically like this, governed by the Schrodinger equation. If the particle’s energy is measured, its state will instantly and non-deterministically collapse into one of the energy eigenstates. For example when measuring the energy of the two-energy state above, we can tell from the coefficients that we have $50\%$ probability to measure $E_{1}$ and $50\%$ probability to measure $E_{2}$ . Suppose we measure $E_{2}$ , then the particle’s wavefunction becomes

\psi(x,t)=u_{2}(x)e^{-\frac{iE_{2}}{\hbar}t},

(2.90)

i.e. the particle is now in an energy eigenstate and the probability density is constant over time. All subsequent measurements of the energy will therefore return $E_{2}$ .

If we measured an observable $\hat{O}$ whos eigenfunctions are not also eigenfunctions of the Hamiltonian, then the collapsed state after measurement will evolve as a superposition state. For example, if we had a particle in an infinite square well and we measured the particle’s momentum, then its wavefunction would collapse into a momentum eigenstate. As we have seen, these are not energy eigenstates, so they must be superpositions of energy eigenstates. Such observables with different sets of eigenfunctions are called incompatible, and we will study them more in section 2.6.

Note that when an observable operator $\hat{O}$ acts on a state $\psi$ which is not an eigenstate of $\hat{O}$ , the result $\hat{O}\psi$ is not what we measure. $\hat{O}\psi$ is a superposition of eigenstates of $\hat{O}$ multiplied by eigenvalues and probability amplitudes of measuring each value. The wavefunction after measurement collapses into one of the eigenstates of $\hat{O}$ .