• Support PF! Buy your school textbooks, materials and every day products Here!

Infinite Square well

  • Thread starter wood
  • Start date
  • #1
23
0

Homework Statement


Consider an infinite square well defined by the potential energy function
U=0 for 0<x<a and U = ∞ otherwise
Consider a superposed state represented by the wave function ## \Psi(x,t)## given at time t=0 by
$$\Psi(x,0) = N \{(-\psi_1(x) + (1+ i)\psi_2(x)\}$$

1. Assume that the constant N is real and positive. Fix N such that Ψ(x, 0) is normalised to one particle.
2. Find Ψ(x, t) for all times t. Is your result correctly normalised? Is this state stationary?
3. In terms of {En}, what is the average energy <E> of the superposed state? Does it depend on t?
4. Calculate <x> as a function of time t.
5. Given ##E_n=(\frac{\hbar^{2}}{2m})( \frac{n\pi}{a})^{2}## , find the average time for an electron to move back and forth once in a well 1 nm wide. Compare this with the classical result for an electron with energy equal to <E>.




Homework Equations




The Attempt at a Solution


I am really struggling with what the lecturer is asking here. I have an answer to 1. ##N=\frac{1}{\sqrt{3}}## then I am not sure where to go.
for 2. I think I get from the TDSE

$$ \Psi(x,0)=\frac{1}{\sqrt{3}} \{(-\psi_1(x)e^{-iE_1t/\hbar} + (1+ i)\psi_2(x) e^{-iE_2t/\hbar}\} $$

I am not sure if it is correctly normalised but It is not stationary as E1≠E2


Could someone please point me in the right direction. I have both Griffiths and Serway and have read through but cannot find anything that helps me understand what is being asked. If someone could either talk me through it or point to relevant sections in the text I would be very grateful.

Thanks
 

Answers and Replies

  • #2
1
0
Your always doing the same problems as me at the same time - we must be in the same class! As far as the problem goes:
1. I got the same N
2. Im a little stuck here too, but i suppose
Ψ(x,t)=∑cn(t)Ψn(t)
We have Ψ1,Ψ2 so we need to find c1,c2 where
cn(t)=cn(0)exp(-i*En*t/ħ)
 
  • #3
blue_leaf77
Science Advisor
Homework Helper
2,629
784
I am not sure if it is correctly normalised but It is not stationary as E1≠E2
Then to be sure if your result is correctly normalized, you need to calculate the inner product with itself and see if it yields unity. 'Non-stationaryness' should not affect the result of any inner product due to the unitarity property of the time evolution operator you have applied to your wavefunction.
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149

Homework Statement


Consider an infinite square well defined by the potential energy function
U=0 for 0<x<a and U = ∞ otherwise
Consider a superposed state represented by the wave function ## \Psi(x,t)## given at time t=0 by
$$\Psi(x,0) = N \{(-\psi_1(x) + (1+ i)\psi_2(x)\}$$

1. Assume that the constant N is real and positive. Fix N such that Ψ(x, 0) is normalised to one particle.
2. Find Ψ(x, t) for all times t. Is your result correctly normalised? Is this state stationary?
3. In terms of {En}, what is the average energy <E> of the superposed state? Does it depend on t?
4. Calculate <x> as a function of time t.
5. Given ##E_n=(\frac{\hbar^{2}}{2m})( \frac{n\pi}{a})^{2}## , find the average time for an electron to move back and forth once in a well 1 nm wide. Compare this with the classical result for an electron with energy equal to <E>.




Homework Equations




The Attempt at a Solution


I am really struggling with what the lecturer is asking here. I have an answer to 1. ##N=\frac{1}{\sqrt{3}}## then I am not sure where to go.
Well how did you arrive at this answer? Presumably you required ##\int \psi^*(x)\psi(x)\,dx = 1## because that's what normalized means.

for 2. I think I get from the TDSE

$$ \Psi(x,0)=\frac{1}{\sqrt{3}} \{(-\psi_1(x)e^{-iE_1t/\hbar} + (1+ i)\psi_2(x) e^{-iE_2t/\hbar}\} $$
That's right.

I am not sure if it is correctly normalised but It is not stationary as E1≠E2
Again, what does it mean for wave function to be normalized?
 
  • #5
23
0
I think I am getting somewhere...(thanks)

Again, what does it mean for wave function to be normalized?
for a function to be normalized I need ##\int \psi^*(x)\psi(x)\,dx = 1## so it is normalised and not stationary

for 3
3. In terms of {En}, what is the average energy <E> of the superposed state? Does it depend on t?
Here $$\langle E\rangle = \int \Psi^*(x)E\Psi(x)\,dx$$

spits out an answer that doesn't depend on t

4. Calculate <x> as a function of time t.
Here i need ## \langle x\rangle =\int \langle x\rangle\Psi^*(x)\Psi(x)\,dx##

The integrals involving cross terms ψ1ψ1= 0 and the ψn2= 1 leaving
$$ \langle x\rangle =\frac{1}{3} \langle x\rangle+\frac{2}{3} \langle x\rangle$$

I think I should have a value for x somewhere to give me an actual answer but am not sure what that should be...
 
  • #6
blue_leaf77
Science Advisor
Homework Helper
2,629
784
spits out an answer that doesn't depend on t
That's what one should expect when the Hamiltonian is time independent.
Here i need ## <x>=\int x\psi^∗(x)\psi(x)dx \langle x\rangle =\int \langle x\rangle\Psi^*(x)\Psi(x)dx ##

The integrals involving cross terms ψ1ψ1= 0 and the ψn2= 1 leaving
<x>=13⟨x⟩+23⟨x⟩​
##\langle x\rangle =\frac{1}{3} \langle x\rangle+\frac{2}{3} \langle x\rangle##

I think I should have a value for x somewhere to give me an actual answer but am not sure what that should be...
Since the question asked you to calculate the time dependency of <x>, you should have used the wavefunctions at time t to do the calculation.
 
  • #7
23
0
Since the question asked you to calculate the time dependency of <x>, you should have used the wavefunctions at time t to do the calculation.
I thought I had by calculating <x> using $$ \Psi(x,0)=\frac{1}{\sqrt{3}} \{(-\psi_1(x)e^{-iE_1t/\hbar} + (1+ i)\psi_2(x) e^{-iE_2t/\hbar}\} $$
 
  • #8
blue_leaf77
Science Advisor
Homework Helper
2,629
784
Alright then, just wondered why t didn't appear in your expression.
 
  • #9
23
0
From ##\psi_1^*\psi_1## I get ##-\psi_1^*(x)e^{+iE_1t/\hbar} \psi_1(x) e^{-iE_2t/\hbar}## which gives ##\psi_1^*(x)\psi_1(x) ## and as they are normalized to one particle ##\int\psi_1^*(x)\psi_1(x) = 1 ## leaving me with just the coefficients.

Yes??
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
12,743
2,907
Yes
Note that your post #7 is a bit confusing: You write ##\Psi(x,0)=##, followed by ##\Psi(x,t)## but you don't state what you do to calculate ##<x>## ...

And I forget the important part: what about the cross terms ? Check carefully if they cancel ...

[edit 3] No ! Sorry:

The Yes above is for the normalization, but it doesn't hold for ##\int \psi_1^* x \psi_1## !
 
Last edited:
  • #11
BvU
Science Advisor
Homework Helper
2019 Award
12,743
2,907
Could you oblige me and elucidate on post #3: your "spits out an answer that doesn' t depend on t " sounds as if it's obvious to you. What exactly do you do to calculate ##< E > ## ?
 
  • #12
23
0
To calculate ##\langle x \rangle## I used should have been $$<x>=\int \langle x\rangle\Psi^*(x)\Psi(x)dx$$

And I didn't check if the cross terms canceled I just 'presumed' they did as they were normalised and therefore ##\int\psi_1^*(x)e^{+iE_1t/\hbar} \psi_2(x) e^{-iE_2t/\hbar} =0## I think you are saying I presumed wrong.
 
  • #13
BvU
Science Advisor
Homework Helper
2019 Award
12,743
2,907
Can't do that. x is an operator (a simple multiplication), ## <x> ## is a number. You can pull the number in front of the ##\int##, but not the unadorned x !
 
  • #14
23
0
So it should read ##<x>=\int x\Psi^*(x)\Psi(x)dx## and I and doing the correct thing?

(thanks for helping by the way...)
 
  • #15
BvU
Science Advisor
Homework Helper
2019 Award
12,743
2,907
Better to write $$
<x>=\int \Psi^*(x)\; x \;\Psi(x)dx$$although for this simple multiplication factor case it doesn't matter. (But for the momentum, for example it does matter!)
 

Related Threads for: Infinite Square well

  • Last Post
Replies
19
Views
3K
  • Last Post
Replies
3
Views
759
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
18
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
15
Views
3K
Top