# Expected energy and viability of a linear combination of wave functions

1. Feb 5, 2013

### xicor

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

Show that the following wave function eigenfunction decomposition is viable and find it's expected energy.

ψ(x, t) = [1/(1+i)]ψ1(x)e^(-iw1t) - (1/√2)ψ2(x)e^(-iw2t)

2. Relevant equations
∫ψ(x)*ψ(x)dx = 1 probability

∫ψm(x)*ψn(x)dx = 0 orthogonality

∫ψm(x)*ψn(x)dx = δmn

Ʃ|C$_{n}$|$^{2}$ = 1

<H> = Ʃ|C$_{n}$|$^{2}$*E$_{n}$

3. The attempt at a solution

From my understanding so far, δmn is value given when you combine the condition for both probability and orthogonality. However I dont think I fully understand why you can just then said that the sum of the constants squared of the wave functions is equal to 1. Using the equation I can find that the first part of wave function is 1/2 and the second part is 1/2 so the sum of that is 1 so the wave function is viable. However I'm not sure what to do with the expected energy. I think that ω = En/h but im unsure as to how the linear combination effects the expression of expected energy because shouldn't <H> =En since the sum of the constants is just 1?

Thanks for everyone that helps.

2. Feb 5, 2013

### Simon Bridge

Well - if you made a measurement of the eigenvalue, what is the probability of finding the particle in one of the eigenstates?
How do you find the expectation value of a discrete random variable in normal statistics?
But each value of the energy does not appear with equal probability does it?

3. Feb 5, 2013

### xicor

Alright, so the first part just requires some understanding of linear algebra. We are still dealing with time independent wave functions but I think we will be learning that notation very soon.

Well, I could see showing the different probability for each value of energy but in this case the expected energy would be <H> = (1/2)E1 + (1/2)E2 where En = ω$_{n}$h

4. Feb 5, 2013

### Simon Bridge

That last equation doesn't work - what is the relationship between frequency and energy?

... no idea what you are talking about.

5. Feb 5, 2013

### xicor

Wouldn't the relationship just be based off planck's equation for quantized energy? That would show that En = hv where "h" is planck's constant and "v" is the frequency. As far as I can see, the last equation is in that form unless I'm not understanding it correctly.

6. Feb 5, 2013

### Simon Bridge

$\omega = 2\pi\nu$ ... for some reason I left the h off the end when I quoted you...

I'm sorry, the way you wrote
I inferred that you had some issue with the resulting relation.

Last edited: Feb 5, 2013
7. Feb 5, 2013

### xicor

That was just because I wasn't sure about using the Cn probabilities before where both E1 and E2 have an equivalent probability. I think the professor was just using ω as a substitute for En/h because he never brought up the term of 2*pi. The ω variable is taking the place of the wave function ψn(x,t) = Cn*ψn(x)e$^{-iEnt/h}$. I was just uncertain if I was correctly using the formula for the expected energy <H>.

8. Feb 5, 2013

### Simon Bridge

Oh OK - well that was exactly how to use them.
$\langle H \rangle = c_1^\star c_1 E_1 + c_2^\star c_2 E_2$

But in the wave-function he(?) writes $e^{-i\omega_n t}$ as the time-dependency ... in that relation, $\omega_n = E_n/\hbar$
If he is using $\omega=E_n/h$ then that would be out by a factor of $2\pi$.

If he uses units of $\hbar=1$ then you can write $\omega_n = E_n$.