# The wave function – making energy measurements

Required math: calculus

Required physics: Schrödinger equation, particle in a box

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 2.1.

When we solved the particle in a box problem we saw that one of the steps in the solution was to specify the initial condition ${\Psi(x,0)}$ as a sum over the stationary states:

 $\displaystyle \Psi(x,0)$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\psi_{n}(x) \ \ \ \ \ (1)$

where the constants ${c_{n}}$ can be found as the integrals:

 $\displaystyle c_{n}$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\psi_{n}(x)\Psi(x,0)dx \ \ \ \ \ (2)$

The general solution for the given initial condition is then

 $\displaystyle \Psi(x,t)$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar} \ \ \ \ \ (3)$

Intuitively, it would seem that ${c_{n}}$ is some sort of measure of how important the stationary state ${\psi_{n}(x)}$ is in the makeup of the overall wave function ${\Psi(x,t)}$. One of the postulates of quantum mechanics is that the square modulus ${|c_{n}|^{2}}$ is the probability that a measurement of the energy of a particle in the state ${\Psi(x,t)}$ will yield the value ${E_{n}}$.

In other words, the set of energies that we worked out when finding the stationary states are the only possible outcomes of a measurement on any system that is governed by the infinite square well potential, so all measurements must give one of these energies. Further, since the ${c_{n}}$ are independent of time, the distribution of the initial condition over the stationary states does not change with time, even though the wave function itself can do. So a measurement of the particle’s position may vary over time (since ${|\Psi(x,t)|^{2}}$ is the probability density for the position), but the probability distribution for energy measurements remains constant.

We need to check that this postulate is consistent with the requirements of being a probability distribution. If ${|c_{n}|^{2}}$ represents the probability of an energy measurement yielding ${E_{n}}$ then, since any energy measurement has to yield one of the allowed energies, we should have the condition

 $\displaystyle \sum_{n=1}^{\infty}|c_{n}|^{2}$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (4)$

This can be proved by reverse reasoning, starting with the condition that we know is true: the initial wave function must be normalized. That is

 $\displaystyle \int_{0}^{a}|\Psi(x,0)|^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\left(\sum_{n=1}^{\infty}c_{n}^*\psi_{n}^*(x)\right)\left(\sum_{m=1}^{\infty}c_{m}\psi_{m}(x)\right)dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}c_{n}^*c_{m}\int_{0}^{a}\psi_{n}^*(x)\psi_{m}(x)dx \ \ \ \ \ (7)$

Now the integral in the last line, as we saw when we worked out the stationary states, is ${\int_{0}^{a}\psi_{n}^*(x)\psi_{m}(x)dx=\delta_{nm}}$ (the Kronecker delta) so the only terms in the double sum that are non-zero are those with ${m=n}$, so the sum-integral terms reduce to the required formula:

 $\displaystyle \sum_{n=1}^{\infty}|c_{n}|^{2}$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (8)$

By similar reasoning, we can show that an expression for the mean or average value of the energy is

 $\displaystyle \langle H\rangle$ $\displaystyle =$ $\displaystyle \sum_{m=1}^{\infty}|c_{m}|^{2}E_{m} \ \ \ \ \ (9)$

where ${H}$ is the Hamiltonian operator which is used to calculate the energy. Its form is

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \frac{1}{2m}p^{2}+V\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2m}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)^{2}+V\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V \ \ \ \ \ (12)$

Since the formula for calculating the mean of any operator is to sandwich the operator in the middle of the square modulus of the wave function and integrate, we get

 $\displaystyle \langle H\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\Psi^*(x,0)H\Psi(x,0)dx\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\left(\sum_{n=1}^{\infty}c_{n}^*\psi_{n}^*(x)\right)H\left(\sum_{m=1}^{\infty}c_{m}\psi_{m}(x)\right)dx \ \ \ \ \ (14)$

We have used the complex conjugate in the integral since it is perfectly legal to specify the initial condition as a complex function, which means that the constants ${c_{n}}$ could be complex as well.

Now we know that the Hamiltonian operating on one of the stationary states yields that state multiplied by the energy of that state (${H\psi_{n}=E_{n}\psi_{n}}$) since that is just the spatial part of the original time-independent Schrödinger equation, so

 $\displaystyle \langle H\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\left(\sum_{n=1}^{\infty}c_{n}^*\psi_{n}^*(x)\right)\left(\sum_{m=1}^{\infty}c_{m}E_{m}\psi_{m}(x)\right)dx\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}c_{n}^*c_{m}E_{m}\int_{0}^{a}\psi_{n}^*(x)\psi_{m}(x)dx\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}c_{n}^*c_{m}E_{m}\delta_{mn}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sum_{m=1}^{\infty}|c_{m}|^{2}E_{m} \ \ \ \ \ (18)$

Although we worked these formulas out for the special case of the infinite square well, where we can calculate the orthonormality of the stationary states explicitly, another postulate of quantum mechanics is that the stationary states calculated for any potential are orthonormal and also complete. Completeness of an orthonormal set means that any function defined over the same domain of variables (for the infinite square well, this means over the interval ${0\le x\le a}$ but other potentials will have different domains, of course) can be expressed as a linear combination of the functions in that set, just as we have done with the infinite square well by expressing the initial condition as such a linear combination. This may seem like an enormous leap of logic, and the mathematical arguments involved in proving these postulates are far from simple. From the point of view of physics, however, we are safe in assuming these conditions as postulates of the theory (since they turn out to be true in all but the pathological cases that cannot represent real, physical systems).

One final postulate is made, about which more will be said in another post: after a measurement of the energy is made, the system’s wave function ‘collapses’ from whatever linear combination of stationary states it was before the measurement to the one stationary state corresponding to the measured value of the energy. This ‘collapse of the wave function’ postulate is the origin of many entertaining features of quantum mechanics, one of the first of which was the story of Schrödinger’s cat. But more on that elsewhere.

To summarize:

1. The stationary states of any potential form a complete, orthonormal set of functions over the domain of the system.
2. Any other function defined over the same domain can be expressed as a linear combination of the stationary states.
3. The square modulus ${|c_{n}|^{2}}$ of the coefficient of the stationary state ${\psi_{n}}$ is the probability that a measurement of the energy will yield ${E_{n}}$.
4. The mean (or average or expectation value) of the energy is given by
 $\displaystyle \langle H\rangle$ $\displaystyle =$ $\displaystyle \sum_{m=1}^{\infty}|c_{m}|^{2}E_{m} \ \ \ \ \ (19)$

5. If a measurement of the energy yields ${E_{n}}$, the wave function of the system collapses to the stationary state corresponding to that energy, that is, to ${\psi_{n}e^{-iE_{n}t/\hbar}}$.