# Expectation value for a superposition

1. Oct 28, 2007

### T-7

1. The problem statement, all variables and given/known data

$$u(x) = \sqrt{\frac{8}{5}}\left(\frac{3}{4}u_{1}(x)-\frac{1}{4}u_{3}(x)\right)$$

Determine the time-dependent expectation value of position of this wave function (the particle is in an infinite potential well between x = 0 and x = a).

3. The attempt at a solution

I make it a/2, ie. that the expectation value for this wave function is time-independent - we always expect it at the centre of the well. Does that seem reasonable for that superposition?

On integrating, I end up with three integrals, two of which amount to just the expectation values of the first and third eigenfunction times their respective probabilities (so I say [9/10]*[a/2] + [1/10]*[a/2], without bothering to unpack them) , and one integral which involves the product of u1 and u3 with x times a cosine component. This I evaluate as 0. Hence we have just a/2.

Does that seems stupid to anyone?

Cheers.

2. Oct 29, 2007

### Galileo

The wavefunction is supposed to represent the state of a physical system. Its time-dependence means the observable physical properties change with time. Wouldn't it be rather weird if you could change that by simply shifting your choice of axes???

The wavefunction is only time independent if it is in a stationary state. That is, an eigenstate of the Hamiltonian. (I guess that's what the u_i(x)'s stand for). Actually, a stationary state will pick up a time dependent phase-factor exp(-iwt), where the angular frequenty depends on the energy of the eigenstate.
When you have a linear combination of eigenstates, each with different energy, each term will pick up a different phase factor and it's not a stationary state anymore.

So what is the time dependence of u(x)?

3. Oct 29, 2007

### T-7

I expected time-dependence too for a superposition. But I still seem to be evaluating the relevant integral as zero:

<x> = (a/10).(a/2) + (1/10).(a/2) - (3/10).2cos$$\left(\frac{E_{3}-E_{1}}{\hbar}.t\right)\int^{+\infty}_{-\infty}u1.x.u3.x dx$$

4. Oct 29, 2007

### nrqed

Did you integrate from minus infinity to plus infinity or from 0 to a? You need to do it from 0 to a!

5. Oct 29, 2007

### T-7

Nope. From 0 to a. Tried it with Maple too.

The integral of sin(Pi*x/a) * x * (sin3*Pi*x/a) from 0 to a is zero.

int(sin(Pi*x/a)*x*sin(3*Pi*x/a),x=0..a);
ans: 0

If it was eigenfunction 1 with eigenfunction 4, say, it would be a different matter.

int(sin(Pi*x/a)*x*sin(4*Pi*x/a),x=0..a);
ans: (16a^2)/(-225*Pi^2)

Thinking about it, simply superimposing 1 and 3 does not create a net 'bulge' on either side of a/2 as it does with 1 and 2, say. Try drawing it.

The oscillating component seems to vanish because it is multiplying a zero integral.

Last edited: Oct 29, 2007
6. Oct 29, 2007

### Galileo

Well, it looks good then.

7. Oct 29, 2007

### nrqed

Oh, I did not notice it was u1 and u3! Of course it's zero (for some reason I thought it was u1 and u2). It's clear that it's zero since it's odd with respect to the center of the well, as you say. <x> is zero whenever the complete wave is a sum of two wavefunctions with index differing by an even number.