Infinite Square Well homework problem

[Fetchimus]

Homework Statement

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

Homework Equations

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

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^*c₂<Ψ₁|xΨ₂>+c₂^*c₁<Ψ₂|xΨ₁>

Assuming all this is fine and dandy my main problem seems to stem from the terms c₁^*c₂<Ψ₁|xΨ₂>&c₂^*c₁<Ψ₂|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.

 

[PeroK]

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?

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

 

[Fetchimus]

I believe it's e^{i(E₁-E₂)t/ħ}. I left it out for the sake of pinpointing the integral itself.

 

[PeroK]

I believe it's e^{i(E₁-E₂)t/ħ}. I left it out for the sake of pinpointing the integral itself.

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

 

[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?

 

[PeroK]

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?

Let me check...

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

 

[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.

 

[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.

 

[PeroK]

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.

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.

 

[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)e^{i(E₁-E₂)t/ħ}+(1+i)e^{i(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.

 

[PeroK]

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?

Yes, I think that's correct!

 

[PeroK]

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)e^{i(E₁-E₂)t/ħ}+(1-i)e^{i(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.

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.

 

[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)?

 

[PeroK]

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)?

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.

 

[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

 

[Fetchimus]

I have let θ= (E₁-E₂/ħ)t

 

[PeroK]

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

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

 

[PeroK]

PPS The answer I get is:

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

 

[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).