Infinite potential well/eigenfunction problem

I'd give you all kudos if i knew how to :)In summary, the conversation discusses finding the normalised ground state and first excited state energy eigenfunctions for a particle of mass m inside a specific potential. The conversation then goes on to discuss using the ground state wave function to calculate the expectation value of the momentum operator, which should be zero. Finally, the conversation touches on finding the probability of measuring the ground state energy for a given wave function. Through help from other participants, it is determined that the expectation value of the momentum operator should be zero and the probability of measuring the ground state energy is 0.1.
  • #1
s_gunn
34
0

Homework Statement



Consider a particle of mass, m inside the potential:
V(x) = 0 for 0<x<L and infinite Otherwise

(i) Write down the normalised ground state and first excited state energy eigenfunctions?

(ii) Use the ground state wave function to calculate the expectation value of the momentum operator


The Attempt at a Solution



I did what I normally do to waves(!) ang got the ground state to be:
root(2/L)*sin((pi*x)/L)

and the first excited state to be:
root(2/L)*sin((2*pi*x)/L)

First of all are these right for part (i)?

Then I have no idea how to do the second part so if anyone can point me in the right direction, I'll be very grateful!:smile:
 
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  • #2
You know the wave function, and thus everything there is to ever know about the system, including the momentum.

Look for an operator [tex]\hat p[/tex] that yield the momentum when it operates on the wavefunction. Then use:

[tex]\langle p\rangle=\int\psi_0^{*}(x)\hat p \psi_0(x)\, dx[/tex]
 
  • #3
The momentum operator is proportional to the derivative d/dx. Calculate the average of this operator. It should be zero since on average the particle goes nowhere.
 
  • #4
That all makes sense to me and I think I could do it if I knew what the 'ground state' wave function was!

Do I just use psi = root(2/L)*sin((pi*x)/L)
or do i need to use e^i... as the expectation value includes a complex conjugate?
 
  • #5
s_gunn said:
Do I just use psi = root(2/L)*sin((pi*x)/L)
or do i need to use e^i... as the expectation value includes a complex conjugate?

Whatever. The grounsd state (and all excited states too) wave functions can be choosen real. The complex factor is possible, like exp(iφ) but it disappears in the complex conjugate product.

The stationary state wave function has the following general form:

ψn(x)=√(2/L)*sin(πnx/L), with n=1, 2, 3, ...

The ground state is n=1.
 
  • #6
Ok here goes...

[tex]<p> = -i\hbar[/tex] [tex]\int^{\infty}_{-\infty}\sqrt{\frac{2}{L}}sin(\frac{x\pi}{L})[/tex][tex]\frac{\partial}{\partial x}\sqrt{\frac{2}{L}}sin(\frac{x\pi}{L}) dx[/tex]

[tex]=-i\hbar[/tex] [tex]\int^{\infty}_{-\infty}\sqrt{\frac{2}{\pi}}sin(\frac{x\pi}{L})cos(\frac{x\pi}{L})dx[/tex]

[tex]=-i\hbar\left[\frac{L}{\pi}sin^{2}(\frac{x\pi}{L})-\frac{2x\pi-Lsin(\frac{2x\pi}{L})}{4\pi}\right]^{\infty}_{-\infty}[/tex]

Any help from here - I'm not sure what to do with the infinity's or am i supposed to use 0 and L as the limits?!

Sorry to be a pain - this is one subject that I'm really struggling to get my head around!
 
  • #7
Isn't your wavefunctiuon zero except for [itex]0<x<L[/itex]?:wink:
 
  • #8
s_gunn said:
...Any help from here - I'm not sure what to do with the infinity's or am i supposed to use 0 and L as the limits?!

Yes, use 0 and L as limits. Anyway, the wave function "inside" the potential well is zero: it follows from the preponderant equation term U(x)ψ(x) = 0. The higher U, the smaller ψ.
 
  • #9
silly me! So the answer is: [tex]\frac{-iL\hbar}{2}[/tex]??

Now I have to find the expectation value of the momentum operator squared so I may be back!

Thanks for all your help so far! :smile:
 
  • #10
I promise this is the end!

I now have a value of:

[tex]\frac{-\hbar \sqrt{1/L}\pi^{2}}{L^{\frac{5}{2}}}[/tex]

for the expectation value of the momentum operator squared. Could this be right? It seems very long winded to me!


The last part of my question says that if the wave function is given by:

[tex]\phi(x)=\frac{1}{\sqrt{10}}\phi_{1}(x)+\frac{3}{\sqrt{10}}\phi_{2}(x)[/tex]

what is the probability of measuring the ground state energy?


Now this seems to simple after everything else!
Is it just [tex](\frac{1}{\sqrt{10}})^{2}[/tex]
= 0.1?
 
  • #11
s_gunn said:
silly me! So the answer is: [tex]\frac{-iL\hbar}{2}[/tex]??

No, try again; you have made a simple error somewhere in your calculation.

[tex]\langle p\rangle=\int_{-\infty}^{\infty}\psi_0^*(x)\left(-i\hbar\frac{d}{dx}\right)\psi_0(x)dx=-i\hbar\left(\frac{2}{L}\right)\int_0^L \sin\left(\frac{\pi x}{L}\right) \frac{d}{dx} \sin\left(\frac{\pi x}{L}\right) dx[/tex]
 
  • #12
s_gunn said:
silly me! So the answer is: [tex]\frac{-iL\hbar}{2}[/tex]??
I am afraid, not. You have to get zero. Your integrand is just a product of sine and cosine, and they are "orthogonal".
 
  • #13
s_gunn said:
I now have a value of:

[tex]\frac{-\hbar \sqrt{1/L}\pi^{2}}{L^{\frac{5}{2}}}[/tex]

for the expectation value of the momentum operator squared. Could this be right?
I do not know. Simplify it. You have to obtain something of the right dimension. h_bar should be squared at least. I vote for (pi*h_bar/L)2.
 
  • #14
You can simply find <x> instead. you should find that to be zero. then the time derivative of zero is also zero. so <p> should also be zero
 
  • #15
Ok, thanks for that (it was just a case of multiplying instead of dividing by a fraction!)

Now I have: [tex]<p> = 0[/tex]

and [tex]<p^{2}> = -\frac{\pi^{2}\hbar^{2}}{L^{2}}[/tex]



Please let this be right before I go insane!
 
  • #16
Now I'll try the last part again!

s_gunn said:
If the wave function is given by:

[tex]\phi(x)=\frac{1}{\sqrt{10}}\phi_{1}(x)+\frac{3}{\sqrt{10}}\phi_{2}(x)[/tex]

what is the probability of measuring the ground state energy?


Now this seems to simple after everything else!
Is it just [tex](\frac{1}{\sqrt{10}})^{2}[/tex]
= 0.1?
 
  • #17
s_gunn said:
and [tex]<p^{2}> = -\frac{\pi^{2}\hbar^{2}}{L^{2}}[/tex]
Please let this be right before I go insane!
It should be positive: one "mines" comes from (i)2 of momentum operator, the other comes from the cosine derivative, so the total sign is plus.
 
  • #18
Thank you everyone for your help. I don't know how I'd have got through that without people pointing out my silly little (and bigger!) mistakes.
 

1. What is an infinite potential well?

An infinite potential well is a hypothetical system in quantum mechanics where a particle is confined within a finite region by an infinitely high potential energy barrier at the boundaries.

2. What is the significance of the infinite potential well in quantum mechanics?

The infinite potential well serves as a simple and fundamental model for understanding the behavior of particles in confined systems. It allows for the visualization and calculation of quantum properties such as energy levels and probability distributions.

3. What is the eigenfunction problem in the context of an infinite potential well?

The eigenfunction problem in an infinite potential well refers to finding the solutions to the Schrödinger equation that describe the allowed energy states and corresponding wavefunctions within the well.

4. How is the eigenfunction problem solved for an infinite potential well?

The eigenfunction problem is solved by applying boundary conditions, such as the wavefunction being continuous and zero at the boundaries of the well. This results in a discrete set of solutions, known as eigenfunctions, that correspond to different energy levels.

5. What are some real-world examples of systems that can be modeled using the infinite potential well concept?

The infinite potential well concept can be applied to various physical systems, such as electrons in a semiconductor, particles in a box or trap, and vibrating molecules in a crystal lattice. It can also be used to study the behavior of atoms in a magnetic field or the motion of a pendulum with a limited range of motion.

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