Schrodinger equation with Electric/Magnetic Potential

[PsiPhi]

Homework Statement

Consider a particle of charge e and mass m in crossed E and B fields, given by E = (0,0, E), B = (0,B, 0), r = (x, y, z).

Write the Schrödinger equation.

Homework Equations

Schrödinger's equation: \left[ -\frac{\hbar^{2}}{2m} \nabla^{2} + V(r,t) \right] \Psi(r,t) = i\hbar\frac{\partial}{\partial t} \Psi(r,t)

The Attempt at a Solution

I have obtained the expressions for the electric (scalar) potential and magnetic (Vector) potential and substituted them in as the V(r,t) that appears in Schrödinger's equation. The trouble I'm having at the moment is that would V(r,t) just be a sum of electric and magnetic potential? Also, the question also mentions I should introduce a gauge to simply the equation, but I'm not sure how this gauge transformation would look like.

Thanks.

 

[Gokul43201]

No, you don't add the potentials. Go back to Jackson/Griffiths' E&M: what happens to the canonical momentum of a charge in a vector potential, A? And have you heard of the Landau gauge?

 

[dextercioby]

To reformulate what Gokul said, what is the hamiltonian for a nonrelativistic electrically charged massive particle in a em field. Having got that, obtaining the quantum hamiltonian is a piece of cake.

 

[PsiPhi]

Hey,

Sorry I haven't replied in a few days. But, I've been flat out with other subjects. OK, back to this problem, I have found a solution to this where they used Lagrangian-Hamiltonian to formulate the solution. And it seems like they obtain a Hamiltonian of H = q\phi + \frac{1}{2m} \left[ p - qA\right]^{2}, where q is the charged particle, m is the mass, p is the momentum and A is the vector potential. I'm stepping through each of the steps since I don't want to just hand the answer in like that without understanding it. I'm sort of stuck on the derivation at part [IV-2], where they take the total time derivative of the vector potential. Can someone shed some light on why they do this? In the question specified above the E field and B field are constant, can I still consider a time varying vector potential to solve for constant crossed E and B field?

Oh the link for the derivation is at http://people.seas.harvard.edu/~jones/ap216/lectures/ls_3/ls3_u4/ls3_unit_4.html

Thanks.

 

[dextercioby]

Nope, time dependence is not an issue in any case. Just remember the relations between the fields and the potentials. You need the potentials, having the fields.

 

[PsiPhi]

Hey,

I've got the generalised quantum Hamiltonian for the case above. Now, I want to obtain the relevant scalar/vector potentials given the E/B fields specified above. I have solved the vector potential expression, and related all the corresponding B and A components to their respected Cartesian components.

I get these 3 different equations for the components of B (x, y, z)

B_{x} = \partial_{y}A_{z} - \partial_{z}A_{y}
B_{y} = - \left(\partial_{x}A_{z} - \partial_{z}A_{x}\right)
B_{z} = \partial_{x}A_{y} - \partial_{y}A_{x}

The left hand side of the above equation can be modified to read

0 = \partial_{y}A_{z} - \partial_{z}A_{y}
B = - \left(\partial_{x}A_{z} - \partial_{z}A_{x}\right)
0 = \partial_{x}A_{y} - \partial_{y}A_{x}

Now the problem I'm having is, how can i recombine all these equations together to obtain a value for A, where A is the vector potential as a function of x,y,z and t (time). This is needed since the expression of the Hamiltonian needs A.
Anyone have some tips for this solution?

 

[Gokul43201]

This is where the choice of gauge comes in. In the Landau Gauge, A has only one non-zero component, making it simpler to work with.

 

[PsiPhi]

Have you got a link I could look up on where they use Landau gauge? The gauges I'm familiar with are Coulomb and Lorenz only.

thanks.

 

[dextercioby]

In the Landau gauge the scalar potential \varphi is set to 0. And the vector potential \mathbf{A} is required to obey the Coulomb condition.

 

[Gokul43201]

I guess it's not the Landau gauge that you want, but probably a close extension of it...sorry. I forgot about the E-field in the problem. To cut to the chase, what happens if you choose say, A = (zB,0,0)?

 

[PsiPhi]

Taking the Curl of (zB, 0, 0) it returns B = (0, B, 0), which agrees with the B field specified. But also, we can take the curl of (0, 0, -xB) and this yields B = (0, B, 0). Is the zB one correct or -xB one correct or both are correct? I'm thinking it has something to do with the symmetry of the system.

However, if I do proceed with using A = (zB, 0, 0) and the scalar potential \phi = -zE. (Note: the time partial derivative term in the expression relating E and the scalar potential is canceled out since A isn't not dependent on time since B is constant, there E = -\nabla \phi)

Thus, the generalised time-dep. Schrödinger eq. is:

i\hbar\frac{\partial}{\partial t} \Psi\left( \vec{r}, t\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{2m} \left(\vec{A}.\vec{\nabla}+ \vec{\nabla}.\vec{A}\right) + \frac{e^{2}}{2m} \vec{A}^{2} + e\phi\right]\Psi\left( \vec{r}, t\right)

where e is the charge of the particle, m is the mass of the particle. Using the Coulomb gauge: \vec{\nabla} . \vec{A}\right) = 0. Therefore, the above equation can be reduced to:

i\hbar\frac{\partial}{\partial t} \Psi\left( \vec{r}, t\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{m} \left(\vec{A}.\vec{\nabla}\right) + \frac{e^{2}}{2m} \vec{A}^{2} + e\phi\right]\Psi\left( \vec{r}, t\right)

where \nabla. \left( \vec{A}\Psi \right) = \vec{A}. \left( \vec{\nabla}\Psi \right) + \left(\vec{\nabla} . \vec{A}\right)\Psi = A. \left(\vec{\nabla}} \Psi\right). (This eliminates the 2 in the denominator of e/2m of the middle term of the square brackets).

Thus, finally substituting the vector and scalar potential into the time-dep. Schrödinger equation with the Coulomb gauge applied, I obtain:

i\hbar\frac{\partial}{\partial t} \Psi\left( \vec{r}, t\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{m} \left(zB \frac{\partial}{\partial x} \right) + \frac{e^{2} z^{2} B^{2}}{2m} - eEz\right]\Psi\left( \vec{r}, t\right)

I was wondering if the logic was correct in obtaining the final Schrödinger equation?

Thanks.

 

[dextercioby]

Am i blind, or i don't see a way to eliminate that 2 in the denominator you speak about.

 

[Gokul43201]

I think it does get eliminated. With \vec{\nabla} . \vec{A}\right) = 0, you have \vec{A}.\vec{\nabla} \psi + \vec{\nabla}.\vec{A}\psi = 2\vec{A}.\vec{\nabla} \psi~~ ... or am I making some mistake too?

 

[PsiPhi]

Yep, you get 2 \vec{A} . \nabla. Then this will cancel out, just wondering if my final equation seems logical? With all the steps taken into account, also how about if A was equal to A = (0, 0, -xB) which still yields B = (0, B, 0)

 

[Gokul43201]

It looks fine to me. The choice of a different vector potential should not affect the physics (gauge invariance).

 

[PsiPhi]

Sweet thanks for all the help Gokul and dexter. There's a couple more parts to the overall question, I'm about to get started on these parts. If I get stuck again, I'll ask you guys.

Thanks again, for all the help.

Cheers.

 

[PsiPhi]

Hello again, I was wondering if you guys could check the logic of what I've done so far for the question I'm about to propose. I'm a little stuck on how to reduce the equation further:

So here's the question: (ii) Separate variables and reduce the Schrödinger equation to a one dimensional problem.

The Schrödinger equation that was shown before can be imbedded into the time-independent case:

E \Psi\left( \vec{r}\right) = \left[ -\frac{\hbar^{2}}{2m}\nabla^{2} + i\hbar\frac{e}{m} \left(zB \frac{\partial}{\partial x} \right) + \frac{e^{2} z^{2} B^{2}}{2m} - eEz\right]\Psi\left( \vec{r}\right)

where E on the LHS is the energy eigenvalue. For this PDE we seek solutions that are of the form

Thus, substituing this solution into the above equation and dividing it by \Psi (x, y, z) = X(x)Y(y)Z(z) again I obtain:

E \Psi\left( \vec{r}\right) = \left[ -\frac{\hbar^{2}}{2m}\left(\frac{1}{X(x)}\frac{\partial^{2}}{\partial x^{2}} X(x) +\frac{1}{Y(y)}\frac{\partial^{2}}{\partial y^{2}} Y(y) + \frac{1}{Z(z)}\frac{\partial^{2}}{\partial z^{2}} Z(z) \right)+ i\hbar\frac{e}{m} \left(zB\frac{1}{X(x)} \frac{\partial}{\partial x}X(x) \right) + \frac{e^{2} z^{2} B^{2}}{2m} - eEz\right]\Psi\left( \vec{r}\right)

Now the problem I'm having is that i split this up further to the individual components (i.e. x, y, z), am I allowed to disregard the terms that aren't dependent on the other components. Say for the x dependence I can eliminate the constants 3rd and 4th term in the square brackets but still keep the 2nd term since there is a partial derivative with respect to x. If i do this i get these 3 equations:

E_{x} = -\frac{\hbar^{2}}{2m} \frac{1}{X(x)} \frac{d^{2}}{d x^{2}} X(x) + i\hbar \frac{e}{m}zB\frac{1}{X(x)}\frac{d}{d x} X(x)
E_{y} = -\frac{\hbar^{2}}{2m} \frac{1}{Y(y)} \frac{d^{2}}{d y^{2}}
E_{z} = -\frac{\hbar^{2}}{2m} \frac{1}{Z(z)} \frac{d^{2}}{d z^{2}} Z(z) + i\hbar \frac{e}{m}zB\frac{1}{X(x)}\frac{d}{d x} X(x) + \frac{e^{2} z^{2} B^{2}}{2m} - eEz

This doesn't look like a 1-D problem to me, I still see the partial derivative with respect x in the z-cartesian equation. Can anyone give me any hints on how to reduce this any further, or if I've gone down totally wrong pathway.

Thanks.

 

[RooperDooper]

I have exactly the same problem, with the same fields. Determining the Schrödinger equation is relatively straight foreword. Now the problem is to solve the equation to determine the wavefunction \Psy. I have a hint to separate variables to reduce it to a one dimensional problem. It is straight foreword to separate out the spatial and temporal parts, but the spatial equation then looks horrible. Can anyone help?
Essentially the equation to be solved is:

a1*grad^2(\Psy(x,y,z))+a2*z*d/dx(\Psy(x,y,z))+(a3-a4*z^2+a5*z)*\Psy=0

where the an's are constants.

 

[Easty]

Try guessing the form of the wavefunction to be

\Psi=exp(ikx)Y(y)Z(z)

Once you substitute this into your equation you should be able to decouple the motion in the y direction and the exp(ikx) terms should cancel.