In summary, the homework statement is a particle in an infinite square well of width L, and its energy is measured at time t. The equation for the energy is C[((1+i)/2)*√(2/L)*sin(πx/L) + (1/√L)*sin(2πx/L)], and <E> is a function of t. The probability of measuring the energy as ħ2π2/2mL2 is <x>=<Ψ(x,t)|xΨ(x,t)>.
  • #1
Fetchimus
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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*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.
 
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  • #2
Fetchimus said:
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}]##
 
  • #3
I believe it's ei(E1-E2)t/ħ. I left it out for the sake of pinpointing the integral itself.
 
  • #4
Fetchimus said:
I believe it's ei(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.
 
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  • #5
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
Fetchimus said:
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)?
 
  • #7
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
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
Fetchimus said:
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.
 
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  • #10
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.
 
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  • #11
Fetchimus said:
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!
 
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  • #12
Fetchimus said:
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.

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.
 
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  • #13
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
Fetchimus said:
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.
 
  • #15
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
I have let θ= (E1-E2/ħ)t
 
  • #17
Fetchimus said:
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
 
  • #18
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})##
 
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  • #19
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).
 

Related to Infinite Square Well homework problem

What is an Infinite Square Well problem?

An Infinite Square Well problem is a theoretical physics problem that involves a particle confined to a one-dimensional box with infinitely high potential walls. This problem is often used to introduce students to the concepts of quantum mechanics.

How do you solve an Infinite Square Well problem?

To solve an Infinite Square Well problem, you must first set up the Schrodinger equation for the particle in the well. This equation involves solving for the wave function, which represents the probability of finding the particle at a certain position. Once the wave function is found, the energy levels and corresponding wave functions can be calculated using boundary conditions and normalization.

What are the boundary conditions for an Infinite Square Well problem?

The boundary conditions for an Infinite Square Well problem are that the wave function must be zero at the boundaries of the well, and that the wave function and its derivative must be continuous at the boundaries. These conditions reflect the fact that the potential inside the well is infinite, and the particle cannot exist outside of the well.

What is the significance of the energy levels in an Infinite Square Well problem?

The energy levels in an Infinite Square Well problem represent the allowed energies that the particle can have within the well. These levels are quantized, meaning they can only take on certain discrete values. The lowest energy level, called the ground state, has the most probability of being occupied by the particle.

What is the connection between the Infinite Square Well problem and real-world systems?

The Infinite Square Well problem is a simplified model that is used to understand quantum mechanical concepts. While it may not directly correspond to any real-world system, the principles and techniques used to solve this problem can be applied to more complex systems, such as atoms and molecules, to better understand their behavior and properties.

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