# Expectation value of energy for a quantum system

1. Nov 7, 2013

### CAF123

1. The problem statement, all variables and given/known data
Let $\Psi(x,0)$ be the wavefunction at t=0 described by $\Psi(x,0) = \frac{1}{\sqrt{2}}\left(u_1(x) + u_2(x)\right)$, where the $u_i$ is the $ith$ eigenstate of the Hamiltonian for the 1-D infinite potential well.

The energy of the system is measured at some t - what are are the possible outcomes of such a measurement?
Find the average energy of the system as a function of time. Does this value depend on the initial state of the system?

2. Relevant equations
Average value of energy, $\langle H \rangle$

3. The attempt at a solution
Since the 1D Infinite well represents a discrete spectrum of eigenvalues, I would say that the only possible values of the energy are E1 and E2. However, to see this I considered the probabilities of obtaining such energies: For E1, it is given by $\langle u_1 | \Psi \rangle$and this gives $\frac{1}{2} e^{-Kt}$ where K is some numerical factor. Similarly for probability of E2: $\frac{1}{2}e^{-K't}$. This suggests at t=0, the probability of getting a particular Ei, i either 1 or 2 is 1/2, but that this decays exponentially. How should I interpret this? At some finite t, P(E1) + P(E2) is not 1, so does this mean there exists another possible energy?

I think I may compute the average energy as $\langle H \rangle = \int dx H |\Psi(x,t)|^2$. Since it says '...find this as a function of time, this suggests there is more to the first part since if at any time t, we could get E1 or E2 with equal probability, then this would suggest the average energy would be (E1+E2)/2 ≠ f(t).

Many thanks.

2. Nov 7, 2013

### Staff: Mentor

K should be purely imaginary, so the probability (as squared magnitude) should be constant.

The average energy just follows from the probabilities to find states 1 and 2.

3. Nov 7, 2013

### CAF123

I see, okay it makes sense now.

The average energy, although I could do the integral $\int H |\Psi(x,t)|^2dx$ physically I already know that the answer will be (E1+E2)/2, which is not a function of time. So at all times in the motion, the average energy is time independent. In answering the last question, I think this would depend on the initial state - if you had the initial w. f as a superposition of other e. states then the average energy would not be simply the average of the energy eigenvalues corresponding to the lowest e. states. Is this right?

4. Nov 7, 2013

Right.

5. Nov 7, 2013

### CAF123

Thank you, I have derived the average position of the particle to be the following: $$\langle x \rangle = \frac{a}{2}\left(1 - C \cos (Pt ) \right)$$

This has been confirmed correct. Classically, we expect average position to be a/2. Quantum mechanically, there is this second oscillatory term. Can we gain any insight into why this term is here?

Also, in the calculation of the average energy, $(\int H |\Psi(x,t)|^2dx)$ I take the Hamiltonian to be $-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$ simply right? When I use this, I end up with 0, which I confirmed with Wolfram alpha. However, I am not sure if I should be using $\hat{H} \mapsto i\hbar \frac{\partial}{\partial t}$ or not.

Last edited: Nov 7, 2013
6. Nov 8, 2013

### Staff: Mentor

Classically, the expectation value for a specific t is just the position of the particle at that time. In general, the particle will move inside the potential well, so you get some oscillation as well (just not with a cos, but with linear motion).
I think you mean the average over time - in this case, both classical an QM calculation give a/2.

The Hamiltonian inside the well? How did you get that formula?
What do you mean with $\hat{H} \mapsto i\hbar \frac{\partial}{\partial t}$?

7. Nov 8, 2013

### CAF123

Classically, it is clear that the average is a/2. The particle's position is a uniform random variable with a constant probability density. I don't see yet why this average is the same in the QM calculation. The average of x I got was precisely the form with the cos term.

Generally $\hat{H} = \frac{\hat{P}^2}{2m} + V(\hat{X})$ The second term vanishes inside the well and I use $\hat{P} \mapsto -i\hbar \frac{d}{dx}$. Actually, I found my mistake, the formula I quoted is not quite correct because the Hamiltonian has a derivative so I used $\int \Psi^* \hat{H} \Psi dx$ and obtained the expected average of energy.
I have seen forms of the Hamiltonian where it is represented by $i \hbar \frac{\partial}{\partial t}$ and that as written above. Are these two forms equivalent and where did the first one come from?

8. Nov 8, 2013

### Staff: Mentor

You are comparing two different averages here.
For the classical particle, you average over time, for the QM particle you do not. The results are different, but that shouldn't be surprising.

That looks like it is coming from the Schroedinger equation. They are not the same, but acting on the wavefunction they should give the same result.

9. Nov 8, 2013

### CAF123

So what does my calculation of <x> actually really represent then? I see that it cannot be the average of position over time since time is still a factor in my expression for <x>.

10. Nov 8, 2013

### Staff: Mentor

It is the expectation value of the position at a specific time - the average over many measurements of copies of the same system, if you like.

11. Nov 8, 2013

### CAF123

It makes sense, thanks. So how would I find the average over time, that is how could I show that the average position over time is a/2?

12. Nov 9, 2013

### Staff: Mentor

Average over one period of the cos.

13. Nov 9, 2013

### CAF123

So like $$\frac{1}{P} \int_{0}^{\frac{2 \pi}{P}} \langle x \rangle\,dt?$$

14. Nov 9, 2013

### Staff: Mentor

That looks good.
Here, you don't need an integral, it is sufficient to know that the cos function has an average value of zero.