Quantum physics: past exam questions

[Prometheus]

I have an exam coming up and have some past exam papers to guide my revision. But i don't have any answers, so i thought i would post some of the questions here with my answers and see if anyone wants to have a crack at them too and check my answers. I'm aiming for at least one a day.

 

 

 

This question concerns a particle in a one-dimensional infinite square well, described by the potential energy function:

 

[latex] V(x) = \begin{cases}0 & \mbox{if} \; 0\leq x \leq L \\\infty & \mbox{otherwise}\end{cases} [/latex]

 

The normalized energy eigenfunctions for the particle in the well take the form:

 

[latex] \psi_n(x) = \begin{cases}\sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) & \mbox{if} \; 0\leq x \leq L \\0 & \mbox{otherwise}\end{cases} [/latex]

 

for [latex]n = 1,2,3,...[/latex]

 

(a) Write down the time-independent Schrodinger equation for the region inside the well and, by substituting ψn(x) into the equation, determine an expression for the corresponding energy eigenvalues, En.

 

 

(b) Determine the expectation values 〈x〉 and 〈x2〉 for the state with n = 2. Hence derive the uncertainty Δx for a measurement of position in this state.

 

 

(c) Write down Schrodinger’s time-dependent equation for the region inside the well and demonstrate that it has solutions of the form:

 

[latex]\Psi_n(x,t) = \psi_n(x)e^{\frac{-iE_n t}{\hbar}}[/latex]

 

for [latex]n = 1,2,3,...[/latex]

 

 

My attempt:

 

a).

 

Using schrodinger's equation and the fact V(x) = 0 we get:

 

[latex] \frac{\hbar^2}{2m} \frac{d^2}{dx^2} \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) = E_n \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) [/latex]

 

Which means,

 

[latex] E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}[/latex]

 

b).

 

Using

 

[latex] E[x] = \int_x x|\psi_n(x)|^2 dx [/latex]

 

we get:

 

[latex] E[x] = \frac{2}{L}\int_0^L x sin^2(\frac{n \pi x}{L}) dx [/latex]

 

Using the substitution:

 

[latex] u = \frac{n \pi x}{L} [/latex]

 

i get

 

[latex] \frac{2}{L}\int_0^{2\pi} u sin^2(u) du = \frac{2}{L} \pi^2 [/latex]

 

the last step coming from a table of integrals.

 

Very similar workings and another given integral give me:

 

[latex] E[x^2] = \frac{2}{L}\int_0^{2\pi} u^2 sin^2(u) du = \frac{8 \pi^3}{3L} - \frac{\pi}{L}[/latex]

 

Hence:

 

[latex] \Delta x = (E[x]^2 - E[x^2])^{\frac{1}{2}} = (\frac{4 \pi^4}{L^2} - \frac{8\pi^3}{3L} + \frac{\pi}{L})^{\frac{1}{2}} [/latex]

 

c).

 

Inside the well the TDSE will be:

 

[latex] i\hbar \frac{\partial }{\partial t} \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) e^{\frac{-i E_n t}{\hbar}} = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial t^2} \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) e^{\frac{-i E_n t}{\hbar}} [/latex]

 

which after differentiating gives:

 

[latex] E_n \Psi_n(x) = \frac{\hbar^2 \pi^2 n^2}{2mL^2} \Psi_n(x) [/latex]

 

Is this enough to show that the given solutions are OK?

[KipIngram]

I think (a) looks right. It matches the value given in the Wikipedia particle in a box article, except they use k = n.pi/L.

 

Your answer to (b) looks kind of fish to me; if I read it right you're saying that <x> = 2.pi^2 / L? That says that by making L large you can wind up expecting your particle to be as far over to the left edge as you wish. I thought <x> was the center of the box. Maybe I misunderstood what you were trying to write down.

 

I didn't have time to look at ©.

[Prometheus]

So you're saying the expectation should be [latex]L/2[/latex].

 

Yes, that should be true for all symmetrical probability amplitudes, a handy quick check during exams, thanks.

 

I probably made a mistake when i changed the variable from x to u, will check in a bit.

 

Edit: Yeah, somehow flipped a fraction on it's head while changing the variable. I now get:

 

[latex] E[x] = \frac{L}{2} [/latex]

 

[latex] E[x^2] = \frac{L}{2}(\frac{4\pi^2}{3} - \frac{1}{2}) [/latex]

 

Edited May 17, 2017 by Prometheus

[Prometheus]

So this one has me scratching my head:

 

In the Coulomb model of a hydrogen atom, a normalized energy eigenfunction takes the form

[latex] \Psi(r, \theta, \phi) = Are^{\frac{-r}{2a_0}} cos(\theta) [/latex]

where A = (1/32πa5)1/2 is the normalization constant and a0 is the Bohr radius.

 

a.) show that ψ(r,θ,φ) is an eigenfunction of both L 2 and L z, and find the corresponding eigenvalues.

 

So i've been trying to use [latex] L_z = -ih \frac{\partial}{\partial \phi} [/latex] and [latex] L^2 = -\hbar^2 [\frac{1}{sin\theta}\frac{\partial}{\partial \theta}(sin\theta \frac{\partial}{\partial \theta}) + \frac{1}{sin^2 \theta} \frac{\partial^2}{\partial \phi^2}] [/latex].

 

But as the energy eigenfunction does not depend on φ i get a big fat zero for [latex] L_z \Psi [/latex], which can't be right. For the squared angular momentum i get [latex] L^2 \Psi = -2\hbar^2\Psi[/latex] and i'm not sure that is right either.

 

Hopefully i'm just missing something simple, pointers appreciated.

 

 

[Mordred]

Start with the Hamiltonian of the hydrogen atom.

[Prometheus]

Do we need the Hamiltonian for this question? We just want to check that ψ(r,θ,φ) is an eigenfunction of both L2 and Lz. It's pretty much just a maths question question isn't it?

 

Btw, i now think my answer could be correct - 0 could be an eigenvalue Lz if m = 0. I need to revise eigenvalues of zero though, something about them makes me feel uncomfortable.

[Mordred]

yes but I'm hoping to provide you a clue on how to arrive at your eugenvalues without giving away the answer. The L_2 and L_z are part of the separation of all variables procedure from the Hamilton coupled with the Schrodinger equation. The missing relation you need is the coulomb potential. to get your eugenvalues.

 

 

recall there are is 3 main constituents to the hydrogen atom. the proton, electron and coulomb potential.. The Hamilton includes all 3 Hint [latex]a_0[/latex] is the Bohr radius

 

Next hint what does the energy eugenfuction depend on ?

Edited May 24, 2017 by Mordred

[Prometheus]

My understanding is that the Hamiltonian for the coulomb model of hydrogen is:

 

[latex] \bar{H} = \frac{-\hbar^2}{2\mu} \nabla^2 - \frac{e^2}{4\pi\varepsilon_0 r}[/latex]

 

where [latex] \mu [/latex] is the reduced mass of the proton and electron.

 

The energy eigenfunction is given and depends on the radial distance and the spherical harmonics (which depends only on [latex] \theta [/latex] in this case).

 

I can then find the energy eigenfunction, which i make to be (using [latex] a_0 [/latex]) [latex] \frac{-E_R}{4} [/latex] where [latex] E_R [/latex] is the Rydberg energy.

 

 

 

I'm still not sure how this helps with the angular momentum operators though. To find their eigenvalues isn't it enough to solve [latex] L_z \Psi [/latex] and [latex] L^2 \Psi [/latex], neither of which depend on the form of the Hamiltonian, only the energy eigenfunction?

 

 

Hmm...

 

You are talking about

 

[latex] \Psi(r, \theta, \phi) = R (r_) Y(\theta, \phi)[/latex]

 

from which we can show

 

[latex] L_z \psi = mh \psi [/latex]

 

[latex] L^2 \psi = l(l+1)\hbar^2 \psi [/latex]

 

which is consistent with my answer if m = 0 and l = 1 (assuming my -2 was just a simple mistake and the actual answer is 2).

 

Am i getting closer?

 

[Mordred]

Yes your getting much closer remember now your working in R^3. Your goal is to show the compatibility theorem between l_2 and l_z. Which is the first part of a) in your question.

 

So take your L^2 equation in your first post.

 

then break out your cartesian components with the following in mind.

 

[latex][\hat{L}^2\hat{L}_x]=[\hat{L}^2\hat{L}_y]=[\hat{L}^2,\hat{L}_z]=0[/latex]

 

The hint I was trying to give you from the Hamilton was to remind you of the system your working in. Was also hoping you would have picked up the R^3 elements.

 

You had the right idea to start with the equations you did but couldn't go from there so I was trying to remind you of the angular momentum components that are involved in your L^2 equation.

 

write down you operator expressions for [latex]\hat{L}_x,\hat{L}_y[/latex] if that helps. ie the spherical harmonics you mentioned above.( You already posted two of the 4.) l-z and L^2 the square of the angular momentum. Remember you are working with a central potential system. ie center of mass equivalent.

 

( the instructor is asking for the proof in the first part. The second part is an application of that proof)

Edited May 25, 2017 by Mordred

[Prometheus]

Are you saying part of the question is asking me to show explicitly that [latex] L^2 L_z - L_z L^2 = 0 [/latex]?

[Mordred]

 

a.) show that ψ(r,θ,φ) is an eigenfunction of both L2 and Lz, and find the corresponding eigenvalues.

change show that to prove that

Are you saying part of the question is asking me to show explicitly that [latex] L^2 L_z - L_z L^2 = 0 [/latex]?

In essence yes.

Edited May 25, 2017 by Mordred