# Infinite Square Well homework problem

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1. Jan 5, 2017

### Fetchimus

1. The problem statement, all variables and given/known data
A particle of mass m, is in an infinite square well of width L, V(x)=0 for 0<x<L, and V(x)=∞, elsewhere.

At time t=0,Ψ(x,0) = C[((1+i)/2)*√(2/L)*sin(πx/L) + (1/√L)*sin(2πx/L) in, 0<x<L

a) Find C
b) Find Ψ(x,t)
c) Find <E> as a function of t.
d) Find the probability as a function of t, that measurement of energy will yield ħ2π2/2mL2
e) Find <x> as a function of t

2. Relevant equations
Nonrelativistic quantum mechanical model for the infinite square well.
Currently using the Griffith's textbook "Intro to QM"

3. The attempt at a solution
So I'm at part e) and feeling odd about it.
For part a) I got C=1
For b) Ψ(x,t) = c1Ψ1(x)e-iE1t/ħ + c2Ψ2(x)e-iE2t/ħ where c1=(1+i)/2 & c2=√2/2.

Now for part E, here is what I have so far.
<x>=<Ψ(x,t)|xΨ(x,t)> = <c1Ψ1+c2Ψ2|xc1Ψ1+xc2Ψ2>=(1/2)<Ψ1|xΨ1>+(1/2)<Ψ2|xΨ2>+c1*c21|xΨ2>+c2*c12|xΨ1>

Assuming all this is fine and dandy my main problem seems to stem from the terms c1*c21|xΨ2>&c2*c12|xΨ1>

Ignoring the constants with respect to the integral that comes out of the inner product I think the integral becomes ∫xsin(πx/L)sin(2πx/L)dx Perhaps this is wrong?

If it isn't wrong I've been trying to look at the integrand as xsinAsinB then using a trig identity sinAsinB=(1/2)[cos(A-B)-cos(A+B)]. Now I have (1/2)[∫xcos(-πx/l)dx-∫xcos(3πx/L)dx] then of course I take advantage of cos(-x)=cos(x) to get rid of that negative sign in the first integral.

At this point, I wonder if I'm alright as far as my steps go. This integral in particular has been giving me fits. I've worked it out and have answers, would be happy to keep sharing steps if so far it's all good.

Last edited: Jan 5, 2017
2. Jan 5, 2017

### PeroK

I haven't checked all your work, but you should have a contribution here for the out-of-phase time evolution factors $exp[-\frac{iE_n}{\hbar}]$

3. Jan 5, 2017

### Fetchimus

I believe it's ei(E1-E2)t/ħ. I left it out for the sake of pinpointing the integral itself.

4. Jan 5, 2017

### PeroK

The rest looks okay to me. It's just integration by parts now.

5. Jan 5, 2017

### Fetchimus

Okay. So doing integration by parts on the integral and once again ignoring the constants for the time being, I ended up with -(L/π)2+(L/3π)2. Is this the correct result when evaluating the integral from 0 to L?

6. Jan 5, 2017

### PeroK

Let me check...

By the way, what did you get for c) and d)?

7. Jan 5, 2017

### Fetchimus

For part c) I got (5/2)(π2ħ2/2mL2)

For part d) I got (1/2)

Spent a few hours on these because I was uneasy about not seeing t anywhere in my answer, but a buddy of mine that has done this problem before claims that these are correct. Perhaps they are not. Not entirely sure.

8. Jan 5, 2017

### Fetchimus

There is also a part f) where I am asked to find <p>, but seeing as how I could simply use m(d/dt)<x> I was really just hoping to get the proper answer for <x>.

I have math worked out for everything, but the typing would take forever, lol. Btw, thanks for all your help so far. It is much appreciated.

9. Jan 5, 2017

### PeroK

Yes, it's almost a trick question. You have effectively conservation of energy in QM, as the expected value of $E$ does not change with time (for a time-independent potential).

And, yes, you can get the expected value of momentum the quick way.

10. Jan 5, 2017

### Fetchimus

Definitely need to take some time to let that sink in.

So for part e) I have <x> = (1/2)L - (4L/9π2)(√2)[(1-i)ei(E1-E2)t/ħ+(1+i)ei(E2-E1)t/ħ]. If this is correct I'll be on my way. If you have something different I can simply go back and check my math at this point.

Last edited: Jan 5, 2017
11. Jan 5, 2017

### PeroK

Yes, I think that's correct!

12. Jan 5, 2017

### PeroK

You need to simplify that and the complex numbers must disappear for an expected value of $x$. In general, for any potential and any initial state that is a linear combination of two energy eigentstates:

$\psi(x, 0) = c_1 \psi_1(x) + c_2 \psi_2 (x)$

You have:

$\langle x \rangle = |c_1|^2 \langle x \rangle_1 + |c_2|^2 \langle x \rangle_2 + 2Re(c_1 c_2^{*} exp[\frac{i(E_2 - E_1)t}{\hbar}] \langle x \rangle_{12})$

Where $\langle x \rangle_{12} = \langle \psi_1|x|\psi_2 \rangle$

In your expression, you'll see you have a sum of a complex number and its conjugate which simplifies to twice the real part.

Last edited: Jan 5, 2017
13. Jan 5, 2017

### Fetchimus

Are you saying that,
<x>12+<x>21 = <x>12 + <x>12* and that's where my sum of a complex # and it's conjugate is coming from?

I totally forgot about that formula tbh. Thanks for the reminder.
I'm having trouble getting the real part out. I feel as though the exp term goes away by default due to the i in the superscript and the (1-i) part of c1 just drops the i part and so I'm left with (-8l√2/π2)?

14. Jan 5, 2017

### PeroK

Yes, that's where the complex conjugate comes from. It's a result of the properties of the inner product.

If $c_1$ and $c_2$ are real, then the $\sin$ term drops out and you have:

$\langle x \rangle = |c_1|^2 \langle x \rangle_1 + |c_2|^2 \langle x \rangle_2 + 2c_1 c_2 \cos[\frac{(E_2 - E_1)t}{\hbar}] \langle x \rangle_{12}$

But, if $c_1$ and $c_2$ are complex, you are going to get both $\cos$ and $\sin$ terms in your final expression.

15. Jan 5, 2017

### Fetchimus

Okay I think I see what you're saying.

So for the RE term I have [(1-i)/2]√2/2(cosθ+isinθ)<x>12 = [(√2/2L)cosθ+(√2/2L)sinθ](2/L)(-8/9(L/π)2) which simplifies to
2RE = [√2cosθ+√2sinθ](-16L/9π2)

So <x> = (1/2)L + 2RE

16. Jan 5, 2017

### Fetchimus

I have let θ= (E1-E2/ħ)t

17. Jan 5, 2017

### PeroK

Note that it's more usual to let $E_n = n^2 \omega \hbar$ so that you get oscillating frequencies related to $\omega$. In this case:

$\exp(3i \omega t) = \cos(3wt) + i \sin(3wt)$

$\theta = 3 \omega$ in this case, would do just as well. But, it's better not to hide the $t$ in the $\theta$. PS it's also slightly better to have $\theta$ as positive by using $E_2 - E1$.

Note it's worth remembering the sequence of steps with the complex conjugate, as it works out that way for all potentials

18. Jan 5, 2017

### PeroK

PPS The answer I get is:

$\langle x \rangle = \frac{L}{2} - \frac{8 \sqrt{2} L}{9 \pi^2}(\cos(3 \omega t) - \sin(3 \omega t)) = \frac{L}{2} - \frac{16 L}{9 \pi^2} \cos(3 \omega t + \frac{\pi}{4})$

Last edited: Jan 5, 2017
19. Jan 5, 2017

### Fetchimus

Thank you so much! I got the first part of your expression!

Was not able to simplify it down to the second part though where you have cos(3ωt+π/4).