Particle in an infinite potential box- expected values of energy

[Rorshach]

Homework Statement

I think this is a very easy problem, I will try to show you guys what I tried to come up with:
A particle is in an infinite potential box and is described in a certain moment of the normalized wavefunction $\psi(x)=\sqrt{\frac{8}{3a}}sin^2(\frac{\pi x}{a})$ for (0<x<a) and 0 otherwise.
a) Calculate the expected value of the energy. Compare with the ordinary (stationary) energy levels.
b) What is the probability of finding the particle in the ground state and the first excited state?
c) Determine <p> and Δp by leveraging development coefficients (no integrals to be solved).

Homework Equations

$ΔxΔp≥\frac{\hbar}{2}$
$Δx=\frac{a}{2}$
$<E_kin>=\frac{(Δp_x)^2}{2m}$

The Attempt at a Solution

Ok, so if I understood everything correctly, estimating the value goes like this:
$p_x=\frac{\hbar}{2} \frac{2}{a}=\frac{\hbar}{a}$
$<E_kin>=\frac{(Δp_x)^2}{2m}=\frac{\hbar^2}{2ma^2}$
and comparison with first state: $\frac{\pi^2 \hbar^2}{2ma^2}>\frac{\hbar^2}{2ma^2}$
as for the probability of those levels- I thought I am supposed to calculate the integral
$\int_{-\frac{a}{2}}^{+\frac{a}{2}}\psi^*_0 (x)\psi(x)\,dx$, but now I don't know. What do you guys think?

 

[TSny]

Hello.

What does $p_x$ represent and how did you get a value for it? You want to find the expectation value of the energy: $<E> = \frac{<p^2>}{2m}$, so you need $<p^2>$.

Check the limits on your integral in the last line of your post.

[Edit: Also, I don't think $\Delta x = a/2$]

 

[Rorshach]

$p_x$ represents the momentum of the particle, and I got it from solving the uncertainty equation $ΔxΔp=\frac{\hbar}{2}$, where $Δx$ comes from the equation i found in a book for problems with infinite potential well- it was supposed to be width of the well divided by 2. Also, I checked the formula for the expectation value of the energy, but in the book it is $<E>=\frac{(Δp)^2}{2m}$. I am not sure about the correctness of the ground level probability integral, or the limits of it.

 

[tannerbk]

Remember you are given limits 0<x<a. A quick read through of chapter 2.2 of Griffiths Introduction to Quantum Mechanics will help enormously with this problem. To solve part (b) you first calculate $c_{n}=∫ψ_{n}(x)^{*}ψ(x)dx$. The probability that a measurement of the energy would yield the value $E_{n}$ is $|c_{n}|^2$. Also, listen to TSny to solve part (a), the equation you've given is, although true in this case because <p>=0, in general not true.

 

[TSny]

$p_x$ represents the momentum of the particle,

The momentum of the particle does not have a definite value, so you cannot write down a value for it.

and I got it from solving the uncertainty equation $ΔxΔp=\frac{\hbar}{2}$,

Here, Δp represents the uncertainty in momentum, not the momentum itself. Δp is usually defined as

$\Delta p = \sqrt{<\hat{p}^2> - <\hat{p}>^2}$ where $\hat{p}$ is the momentum operator.

... $Δx$ comes from the equation i found in a book for problems with infinite potential well- it was supposed to be width of the well divided by 2.

Like Δp, Δx is the uncertainty in x and is defined by $\Delta x = \sqrt{<x^2> - <x>^2}$ and generally does not equal a/2. When you looked up Δx in a book, for what quantum state was the book considering? Maybe what you found was a value for <x> for one of the energy eigenstates. Anyway, you will not need Δx or the uncertainty principle for this problem.

Also, I checked the formula for the expectation value of the energy, but in the book it is $<E>=\frac{(Δp)^2}{2m}$.

The generally correct expression for the infinite well is $<E>=\frac{<\hat{p}^2>}{2m}$. If $<\hat{p}^2>$ happens to equal $(Δp)^2$, then what you wrote is ok. But you would need to show $<\hat{p}^2> = (Δp)^2$

I am not sure about the correctness of the ground level probability integral, or the limits of it.

The expression for the integral is correct except for the limits. Note that the potential well extends from x = 0 to x = a. The integral gives the probability amplitude, not the probability.

 

[Rorshach]

Of course You were right about the limits, Thank You:) I tried to calculate $<p^2>$ by integration, but I get different result from the one I obtained by solving Heisenberg's uncertainty equation, here's the link: http://www.wolframalpha.com/input/?...2/3)+(1/a)^(5/2)+π^2+cos((2+π+x)/a))),[x,0,a]
It should be equal to $\frac{a}{\hbar}$, but it isn't. What is wrong?

 

[Rorshach]

besides, when I calculate the probability according to integration formula, I get value entangled with $a$.
http://www.wolframalpha.com/input/?i=integrate((sqrt(8/3a)sin^2((pi*x)/a))^2),[x,0,a]
Help greatly appreciated:)

 

[tannerbk]

It should be equal to $\frac{a}{\hbar}$, but it isn't. What is wrong?

Remember ΔxΔp ≥ h-bar/2, it doesn't have to be equal - and usually is not!

 

[TSny]

I tried to calculate $<p^2>$ by integration,

Your integration looks correct. From that result you can get <E>.

but I get different result from the one I obtained by solving Heisenberg's uncertainty equation

Why do you want to use Heisenberg's uncertainty relation which deals with uncertainties (Δx and Δp) rather than expectation values?

It should be equal to $\frac{a}{\hbar}$, but it isn't. What is wrong?

What should be equal to $\frac{a}{\hbar}$? This quantity has dimensions of the inverse of momentum.

 

[Rorshach]

The result I obtained for energy is weird, I am not sure if it is correct: $\frac{2\pi^2\hbar^2}{3a^2 m}$. Not sure if it supposed to be like this...

 

[tannerbk]

The result I obtained for energy is weird, I am not sure if it is correct: $\frac{2\pi^2\hbar^2}{3a^2 m}$. Not sure if it supposed to be like this...

Looks perfectly reasonable. Why do you think its 'weird'?

Edit: The source of confusion may be that the wavefunction you are considering is not an energy eigenstate of the ISW.

 

[Rorshach]

I was just expecting something with $2ma^2$ in the denominator and natural number in the nominator, like in the formula $E_n=\frac{n^2 \pi^2 \hbar^2}{2ma^2}$

 

[Rorshach]

but why can't I obtain $a$ free value for probability of the energies?

 

[tannerbk]

I was just expecting something with $2ma^2$ in the denominator and natural number in the nominator, like in the formula $E_n=\frac{n^2 \pi^2 \hbar^2}{2ma^2}$

Read the edit to my last post. This form is true for energy eigenstates of the infinite square well $ψ(x)=√\frac{2}{L}sin(k_{n}x)$. The wavefunction given to you in the problem is not of this form.

 

[TSny]

I am not sure if it is correct: $\frac{2\pi^2\hbar^2}{3a^2 m}$. Not sure if it supposed to be like this...

I believe that's the correct answer. As tannerbk noted, you should not expect the answer to correspond to one of the energy eigenstates because your wavefunction is not an energy eigenstate.

But you can see that your answer is just 4/3 times the ground state energy.

 

[Rorshach]

ok, but I still have the problem with the probabilities- what do I do wrong?

 

[TSny]

besides, when I calculate the probability according to integration formula, I get value entangled with $a$.
http://www.wolframalpha.com/input/?i=integrate((sqrt(8/3a)sin^2((pi*x)/a))^2),[x,0,a]
Help greatly appreciated:)

Check to see if you typed in the formula for the wavefunction correctly. Where should the $a$ appear in the coefficient?

 

[tannerbk]

ok, but I still have the problem with the probabilities- what do I do wrong?

What have you done? Read my first post (post 4). Calculate $c_{n}$ for the ground state and first excited state. My notation was bad in that post, remember you must use both the wavefunction given to you in the problem and the energy eigenstates.

 

[Rorshach]

Those stupid typos... thanks for the notice:) Is it okay that for the ground state the probability is equal to 1 and for the first excited state it is equal to 4/9?

 

[tannerbk]

No, the sum over all the individual probablities should equal 1. Show your work and we can help.

 

[Rorshach]

ok, so the probability for the ground level was supposed to be equal to the square of the integration:
http://www.wolframalpha.com/input/?i=integrate((sqrt(8/(3a))sin^2((pi*x)/a))^2),[x,0,a]
the same goes for the first excited level, for which the probability looks normal:
http://www.wolframalpha.com/input/?...)/a))(sqrt(8/(3a))sin^2((2*pi*x)/a))),[x,0,a]

 

[tannerbk]

You are still confused about what wavefunctions you need to integrate over. Your integral should be the wavefunction given to you multiplied by the energy eigenstate, $c_{n} = ∫√\frac{8}{3a}sin^{2}(\frac{{\pi}x}{a})√\frac{2}{a}sin(\frac{n{\pi}x}{a})$. For n=1 this will give you the probability that a measurement of the energy would yield $E_{1}$. You need to calculate this for $c_{1}$ and $c_{2}$. Again, I suggest you read Chapter 2 in Griffths Introduction to Quantum Mechanics.

 

[Rorshach]

ok, value for the first integral-ground level came out as 0.98014, which squared is equal to more or less 0.96, the second one first excited level is equal to zero. Is this correct now?

 

[TSny]

I think those values are correct. Good.

 

[Rorshach]

great! Now it is the last part of this task- so far I can only think of the equation $\Delta p=\sqrt{<p^2>-<p>^2}$, but then again it is said that no integral is supposed to be solved, and $<p^2>$ came out of the integral... however I am not sure if I understand the term "development coefficients"...

 

[tannerbk]

I believe they want you to use $c_{1}$ and $c_{2}$ to calculate $<p>$ and use that to calculate Δp, since you have already calculated $<p^{2}>$. Hint: $<H>=Ʃ|c_{n}|^{2}E_{n}$ would be an analogous formula to calculate the expecation value of the energy.

 

[Rorshach]

should it look like this? $<p^2>=\Sigma |c_n|^2 (-\hbar^2)$?

 

[Rorshach]

Anybody?

 

[Rorshach]

could someone please point me in the right direction?

 

[Rorshach]

I decided to refresh one of my problems, since I still don't know how to calculate the values without integrals. Any suggestions how should I approach this?

 

[Rorshach]

Could anybody please respond?

 

[TSny]

Hello, Rorshach. I don't understand "by leveraging development coefficients" in part (c), so I don't think I can help much.

You've already calculated $<\hat{p}^2>$, so once you determine $<p>$ you will easily be able to get $\Delta p$.

Note that the wavefunction $\psi$ is an even function about $x = a/2$. Since $\hat{p}$ is proportional to the derivative operator, you should be able to argue that $\hat{p} \psi$ is an odd function about $x = a/2$. So, if you set up the integral for $<\hat{p}>$ you should be able to conclude what the answer is without actually carrying out the integration. But that doesn't appear to be the way you are expected to do it.

I will bring this problem to the attention of some of the other helpers.

I guess you don't have access to the person who made the question. It would be nice if you could get clarification on what he/she meant for you do do here.

 

[Rorshach]

Unfortunately, I have no contact with the person who made this question. I also don't want to integrate anything, since it was clearly stated in the problem to not to do it, however user tannerbk gave a formula a few replies above for expectated value for hamiltonian <H>, so maybe there is an analogic one for <p>?

 

[TSny]

Maybe they want something like this. You know that you can expand your wave function in terms of the energy eigenstates:

$\psi(x) = \sum c_n \sin(k_n x)$ where $k_n = n\pi/a$.

Write each $\sin(k_n x)$ as $\frac{1}{2i}(e^{ik_nx} - e^{-ik_nx})$.

Suppose you could show that each exponential $e^{ik_nx}$ is a momentum eigenstate corresponding to momentum $\hbar k_n$ while each exponential $e^{-ik_nx}$ is a momentum eigenstate corresponding to momentum $-\hbar k_n$.

Can you then construct an argument showing $<\hat{p}> = 0$?

Note, however, that this type of argument might not be sound. There are difficulties with defining a momentum operator for the particle in a box. The functions $e^{ik_nx}$ don't satisfy the boundary conditions of the particle in a box, so I'm not sure they qualify as legitimate states for the particle in a box.

See for example http://academic.reed.edu/physics/fa...ous Essays/Generalized Momentum Operators.pdf

or this thread https://www.physicsforums.com/showthread.php?t=285694