Landau Quantization: Classical Particle Motion in a Uniform Magnetic Field

[Sonny Liston]

1. The problem statement
Problem: "Consider a classical particle with charge q and mass m in 2 dimensions (xy-plane) moving in the presence of a uniform magnetic field B=B_z.''

There are a number of parts to this problem, but it's the first three that have me confused.

"(a) Describe the classical trajectory at a given energy E of the particle
(b) Evaluate the classical action along this trajectory
(c) Evaluate the magnetic flux through the classical trajectory as a function of the energy''

The Attempt at a Solution

I'm having more trouble setting this up than anything else. I know what the trajectory should look like, I think -- the particle should obey the Lorentz force law and have a periodic circular trajectory (helical, if we take motion in the z-direction into account). What throws me off is the statement "at a given energy E", since that has nothing to do with the Lorentz force the particle experiences, as far as I know. I'm not sure how to take that information into account, and how it affects the classical trajectory.

Then since I can't figure out how to do part (a), I can't evaluate the action along that trajectory. I know the Lagrangian for a classical charged particle moving in a uniform B-field, but I'm not totally sure how to put it to use in this problem. Similarly, I know how to evaluate a classical magnetic flux, but can't do it without the classical trajectory.

Hopefully someone can point me in the right direction!

 

[Dickfore]

Hint:

The Hamiltonian function of a particle in an external magnetic field is:

<br /> H = \frac{1}{2 m} \, \left(\mathbf{p} - q \, \mathbf{A}(\mathbf{r}) \right)^{2}<br />

where \mathbf{A} is the vector potential of the field that is connected to the magnetic field:

<br /> \mathbf{B} = \nabla \times \mathbf{A}<br />

What are the equations of motions in Hamiltonian dynamics?

 

[Sonny Liston]

Well that's nice. Solving \dot{x}=\partial_{p}H gives m\ddot{x}=p-(q/c)\overrightarrow{A}, and solving \dot{p}=-\partial_{p}H gives something which, if you play around with long enough, gives you the Lorentz force law when combined with the solution to \dot{x}=\partial_{p}H.

So the classical trajectory is, in fact, independent of E and is just the circular trajectory given by the Lorentz force law? And to calculate the action, just use the legendre transform to get (or, more likely, look up) the Lagrangian for a classical particle in a uniform magnetic field?

 

[Dickfore]

Well, \mathbf{A}(\mathbf{r}) is a function of position and your momentum derivative equation is incorrect. You must find such a vector potential that:

<br /> \nabla \times \mathbf{A} = B \, \hat{\mathbf{z}}<br />

Also, the trajectory is indeed a circle, but the radius of the circle depends on the speed of the particle and, thus, on energy.

 

[Sonny Liston]

Oops, typo in the momentum derivative equation. I did solve the right one when I worked it out. Thanks.

As for the latter, I can write the vector potential in Landau Gauge A=(0,B_{x},0).

Then my Lagrangian is L=(1/2)mv^2 + (q/c)v \cdot A, and integrating this over [t_{0}, t_{1}] yields [(1/2)mv^2+ ((q/c)v_{x}B_{x})]\Delta t. This seems really wrong to me.

 

[Dickfore]

You seem to be copying the expressions from some book. What is the meaning of B_{x} in the y-component?

 

[Sonny Liston]

Taking them from my class notes, actually (at least the form of the vector potential). The Lagrangian is written down in Shankar's QM book, and I'm not sure what you mean by "significance"--I thought it was just a way of writing down the vector potential in a gauge that produces the same B_{z}?

I think I am very confused.

 

[Dickfore]

I think I am very confused.

Yes, you are. Do you know what curl is?

 

[Sonny Liston]

Yes.

 

[Dickfore]

Ok. So, can you find the curl of a vector field given by:

I can write the vector potential in Landau Gauge A=(0,B_{x},0).

 

[Sonny Liston]

Yeah, I did so in between my post and yours, and am now confused as to why that is written in both my notes and in Shankar as a gauge where (\nabla \times A) = B_{z}. I was taking their word for it, previously, which was stupid. Confusion is now amplified.

 

[Dickfore]

Yeah, I did so in between my post and yours, and am now confused as to why that is written in both my notes and in Shankar as a gauge where (\nabla \times A) = B_{z}. I was taking their word for it, previously, which was stupid. Confusion is now amplified.

I don't understand what you are trying to say. I just asked you if you could calculate the curl of the A you had posted.

 

[Sonny Liston]

I did do that. I got \hat{z}\partial_{x}B_{x} - \hat{x}\partial_{z}B_{x}. Is that wrong?

 

[Dickfore]

It is not wrong. However, you seem to be having a component in the x-direction. Does the original magnetic field have a component in that direction?

 

[Sonny Liston]

No. That is what I meant by my previous post, the one you said you didn't understand -- that it isn't giving me back the magnetic field it is supposed to give.

(Thank you, by the way, for your help and patience. I've been very frustrated by my inability to understand what seems like a straightforward problem.)

 

[Dickfore]

Well, it would if:

<br /> \frac{\partial B_{x}(x, y, z)}{\partial z} = 0<br />

and:

<br /> \frac{\partial B_{x}(x, y, z)}{\partial x} = B = \mathrm{const.}<br />

Can you find the explicit form of a function B_{x}(x, y, z) that would satisfy these conditions?

 

[Sonny Liston]

It seems like I can find infinitely many functions of that form -- as long as there is no z-dependence and B_{x} is first order in x. Right?

 

[Dickfore]

It seems like I can find infinitely many functions of that form -- as long as there is no z-dependence and B_{x} is first order in x. Right?

YES! So, the point was, that it was not B_{x} as the y-component of the vector potential \mathbf{A}, but it was actually B \cdot x. The arbitrariness that you had found is because a vector field is not uniquely determined if you simply give its curl (as is the case with A). You can always add a gradient of a scalar function to it. This is referred to as gauge invariance. To uniquely determine the vector potential, we need to choose a gauge. The Landau gauge (or Coulomb or radiation gauge) is:

<br /> \nabla \cdot \mathbf{A} = 0<br />

If you use this gauge, you get the potential you cited with the caveat about the typo in your notes explained above.

Now, do you want to use Lagrange's or Hamilton's formalism to proceed further?

 

[Sonny Liston]

Since I am asked to calculate the classical action along the trajectory of the particle, Lagrange's formalism seems natural to me. Is that not the best way?

Edit: I posted something that was wrong.

 

[Dickfore]

Also, in the Landau gauge, does my Lagrangian now become L=\frac{1}{2} m v^{2} - \frac{q}{c} v \cdot A \rightarrow L= \frac{1}{2} m v^{2} - \frac{q B v_{x}}{c} ?

Yes, you are right about the action and Lagrangian. I took the liberty of correcting the formating of your equation (and one minus sign). Can you write the equations of motion (in the xy - plane) for this lagrangian?

 

[Sonny Liston]

Using (d/dt)(\partial_{\dot{x}_{i}} L) - \partial_{x_{i}} L = 0

I ended up with (d/dt)(m\dot{x_{i}} + (q/c)A_{i}) - (q/c) \partial_{\dot{x}_{i}} v_{i} A_{i}= 0, and using the fact that the time derivative of A has a component (v \cdot \nabla)A and combining equations (and letting indices run over x and y only), I ended up with

ma= (q/c)[\nabla (v \cdot A) - (v \cdot \nabla)A] = (q/c)[v \times B], the Lorentz force law in the x-y plane.

So far so good?

 

[Dickfore]

Ok, you just derived Newton's law with a lorent'z force, but, what is the PARTICULAR form of the equations of motion in our case under consideration?

 

[Sonny Liston]

Oh. Right.

Plugging in our B=(0,0,B), I get m(d^2_{t})x = \hat{x}(v_y)B_{z} - \hat{y}(v_x)B_{z}.

 

[Dickfore]

right, let me rewrite these in the form:

<br /> \begin{array}{rcl}<br /> m \, \dot{v}_{x} &amp; = &amp; \frac{q}{c} \, v_{y} \, B \\<br /> <br /> m \, \dot{v}_{y} &amp; = &amp; -\frac{q}{c} \, v_{x} \, B<br /> \end{array}<br />

Can you find a solution for v_{x}(t) and v_{y}(t) of this coupled system of 1st order ODEs?

 

[Sonny Liston]

Letting \frac{qB}{mc}=\omega_{c}, we get that v_{x} = sin(\omega t) + x_{0} and v_{y} = cos(\omega t) + y_{0}.

So these describe the motion of the particle in the xy plane, carving out a circular trajectory. Excellent.

Now how would I go about calculating the classical action along this trajectory? Do I just integrate the Lagrangian over arbitrary [t_{0}, t_{1}]?

 

[Dickfore]

Substitute your solutions for v_{x}(t) and v_{y}(t) in the expression for the Lagrangian L. The action is given by the definite integral:

<br /> S = \int_{t_{0}}^{t}{L \, dt}<br />

 

[Sonny Liston]

Alright, so I integrated \int (1/2)m[1 + 2(x_{0} sin(\omega t) + y_{0} cos(\omega t)) + (x^2)_{0} + (y^2)_{0}] - m\omega(sin(\omega t) + (x_{0})

and get

(1/2)m[1 + (x_{0})^2 + (y_{0})^2 - x_{0} \omega]\Delta t - mx_{0} \omega(cos(\omega t_{0}) - cos(\omega t_{1})) + my_{0} \omega(sin(\omega t_{0}) - sin(\omega t_{1})) + ((\omega)^2)(cos(\omega t_{0}) - cos(\omega t_{1})).

The resulting action can probably be simplified, but does that look in the ballpark of correct? And again, I can't thank you enough for how helpful you've been.

 

[Dickfore]

First of all, the solution of the equations:

<br /> \begin{array}{rcl}<br /> \dot{v}_{x} &amp; = &amp; \omega \, v_{y} \\<br /> <br /> \dot{v}_{y} &amp; = &amp; -\omega \, v_{x}<br /> \end{array}<br />
which can be manipulated to get an equation with respect to a single variable:
<br /> v_{y} = \frac{\dot{v}_{x}}{\omega}<br />

<br /> \ddot{v}_{x} = \omega \, \dot{v}_{y} = -\omega^{2} \, v_{x}<br />

Imagine the initial velocity is directed in the x-direction. This means:

<br /> v_{x 0} = v_{0}, \; v_{y 0} = 0<br />

The equation \ddot{v}_{x} + \omega^{2} \, v_{x} = 0 with the initial conditions is v_{x 0} = v_{0}. \; \dot{v}_{x} = 0 has a solution:

<br /> v_{x}(t) = v_{0} \, \cos{(\omega t)}<br />

and, for v_{y} = \dot{v}_{x}/\omega, we get:

<br /> v_{y} = -v_{0} \, \sin{(\omega t)}<br />

You should use these in the Lagrangian. The integrals are elementary (after some double angle formulas from trigonometry). The actions S(t) is a periodic function of time with the period 2\pi/\omega. I would appreciate if you posted the final expression for S(t) here.

 

[Sonny Liston]

So I'm mildly concerned, because I didn't have to use any double angle formulas, but here is what I got:

My Lagrangian becomes L = (m/2)((v_{x})^2 + (v_{y})^2) - (mv_{0})\omega cos(\omega t) \longrightarrow L = (m/2)[(v_{0})^2((sin(\omega t))^2 + (cos(\omega t))^2] - (mv_{0})\omega cos(\omega t)

which integrates very easily to

S(t) = [(m(v_{0})^2)/2]\Delta t - ((\omega)^2)mv_{0}[sin(\omega t_{1}) - sin(\omega t_{2})].

 

[Dickfore]

You are right about the kinetic energy term. The term with the vector potential, however, is:

<br /> -\frac{q}{c} \, (\mathbf{v} \cdot \mathbf{A}) = -\frac{q \, B}{c} \, v_{y} \, x<br />

So, you will need to find the dependence of x(t) before you proceed.

 

[Sonny Liston]

I just integrated v_{x} from 0 to t to get (v_{0})(cos(t\omega) - cos(0)) + x_{0} \longrightarrow (v_{0})cos(t\omega) -v_{0} + x_{0}

although the (-v_{0}) term seems strange to me

 

[Dickfore]

But, x is not a constant:

<br /> x(t) = x_{0} + \int_{t_{0}}^{t}{v_{x}(t&#039;) \, dt&#039;}<br />

 

[Sonny Liston]

Right. Integrating v_{x} = v_{0}cos(t\omega) over [t_{0}, t] gives

v_{0}[\omega(sin(t_{0} \omega) - sin(t \omega))] where (sin(t_{0} \omega) is some constant, unless I'm being an idiot.

Oh jeez, I just saw that I did the wrong integral in my last post. Now I think your comment makes sense.

 

[Dickfore]

No!

<br /> \int{\cos{(a t)} \, dt} = \frac{1}{a} \, \sin{(a \, t)}<br />

 

[Sonny Liston]

Jesus. Embarrassing. Long day.

But we are doing a definite integral, so it should still end up as v_{0}[\frac{1}{\omega}(sin(t_{0} \omega) - sin(t \omega))] shouldn't it? If we set t_{0}=0, then the first sin term vanishes, and we end up with v_{0}[\frac{1}{\omega}(-sin(t \omega))] ?

 

[Dickfore]

Yes, this would be x(t) that you need to substitute in the above expression and integrate again.

 

[Sonny Liston]

Integrating my LagrangianL = (m/2)[(v_{0})^2 +1] - (q/c)B[(x_{0} - \frac{v_{0}}{\omega} sin(t\omega)][-(v_{0})sin(t\omega)]

Gives

S(t) = [(m(v_{0})^2)/2]\Delta t - <br /> <br /> (q/c)B[\frac{(v_{0})x_{0}}{\omega}](cos(\omega t_{1}) - cos(\omega t_{0})) - <br /> <br /> (q/c)B[\frac{(v_{0})^2}{\omega}]((t_{1})/2 - \frac{sin(2\omega t_{1})}{4\omega} - (t_{0})/2) + \frac{sin(2\omega t_{0})}{4\omega}

This looks more complicated than it should be..

 

[Dickfore]

Ok, do this first:

Take the lagrangian:

<br /> L = \frac{m v^{2}_{0}}{2} \, \left(v^{2}_{x} + v^{2}_{y} \right) - \frac{q B}{c} \, x \, v_{y}<br />

substitute:

<br /> \begin{array}{rcl}<br /> v_{x} &amp; = &amp; v_{0} \, \cos{(\omega \, t)} \\<br /> <br /> v_{y} &amp; = &amp; -v_{0} \, \sin{(\omega \, t)} \\<br /> <br /> x &amp; = &amp; \frac{v_{0}}{\omega} \, \sin{(\omega t)}<br />

and perform integration from 0 to t, by keeping in mind that:

<br /> \omega = \frac{q \, B}{m \, c}<br />

What is the lagrangian as a function of t?

What is the integral of this from 0 to t?

 

[Sonny Liston]

I get that L=(v_{0})^2[\frac{m}{2} + (sin(\omega t)^2)

Have I done anything stupid so far?

 

[Dickfore]

First of all, your equation is dimensionally inconsistent. In the large parentheses, the first term has a dimension of mass, and the second term is dimensionless.

 

[Sonny Liston]

All I did was cancel the omega in the x term and the omega in the v_{y} term, which I got from plugging in the provided equations.

Sigh. How did I manage to mess THAT up?

 

[Dickfore]

The vector potential term from the Lagrangian has the product:

<br /> x v_{x} = \frac{v_{0}}{\omega} \, \sin{(\omega t)} \, v_{0} \, \cos{(\omega t)} = \frac{v^{2}_{0}}{\omega} \sin{(\omega t)} \, \cos{(\omega t)}<br />

using the double angle formula:

<br /> \sin{(2 \alpha)} = 2 \, \sin{(\alpha)} \, \cos{(\alpha)}<br />

this product can be further simplified as:

<br /> \frac{v^{2}_{0}}{2 \omega} \, \sin{(2 \omega t)}<br />

However, this product is also multiplied by q \, B/c in the Lagrangian. Using the definition of the cyclotron frequency \omega = q \, B/(m \, c), what is:

<br /> \frac{q B}{\omega c} = ?<br />

The kinetic energy term has the sum of squares:

<br /> v^{2}_{x} + v^{2}_{y} = ?<br />

What is this sum equal to? What is it multiplied by in the Lagrangian?

 

[Sonny Liston]

<br /> \frac{q B}{\omega c} = ?<br />\frac{qB}{\omega c} = \frac{qBmc}{Bc} = qm[/tex]<br /> <br /> The kinetic energy term has the sum of squares:<br /> <br /> &lt;br /&gt; v^{2}_{x} + v^{2}_{y} = ?&lt;br /&gt;<br /> <br /> What is this sum equal to? What is it multiplied by in the Lagrangian?<br /> <br /> &lt;br /&gt; v^{2}_{x} + v^{2}_{y} = ((v_{0})^2)((cos(\omega t))^2 + (sin(\omega t))^2) = ((v_{0})^2)&lt;br /&gt;<br /> <br /> This is then multiplied by \frac{m}{2}

 

[Sonny Liston]

Whaddya mean?

 

[Dickfore]

<br /> \frac{q B}{\omega c} = ?<br />


\frac{qB}{\omega c} = \frac{qBmc}{Bc} = qm[/tex]<br />

<br /> This is incorrect.<br /> <br /> <blockquote data-attributes="" data-quote="Sonny Liston" data-source="post: 2985775" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Sonny Liston said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The kinetic energy term has the sum of squares:<br /> <br /> &lt;br /&gt; v^{2}_{x} + v^{2}_{y} = ?&lt;br /&gt;<br /> <br /> What is this sum equal to? What is it multiplied by in the Lagrangian?<br /> <br /> &lt;br /&gt; v^{2}_{x} + v^{2}_{y} = ((v_{0})^2)((cos(\omega t))^2 + (sin(\omega t))^2) = ((v_{0})^2)&lt;br /&gt;<br /> <br /> This is then multiplied by \frac{m}{2} </div> </div> </blockquote>This is correct.

 

[Sonny Liston]

I didn't see the charge term in the original cyclotron frequency. In that case, we don't get qm, we get m.

 

[Dickfore]

I didn't see the charge term in the original cyclotron frequency. In that case, we don't get qm, we get m.

Yes. So, what is the final expression for the Lagrangian L?

 

[Sonny Liston]

L = \frac{m(v_{0})^2}{2} - \frac{m(v_{0})^2}{2\omega}sin(2t\omega)

 

[Dickfore]

L = \frac{m(v_{0})^2}{2} - \frac{m(v_{0})^2}{2\omega}sin(2t\omega)

Where did the extra \omega come from in the second term? C'mon man. Concentrate on your algebra. This is taking too long.

 

[Sonny Liston]

S(t) = [\frac{m((v_{0})^2)}{2}]t + [\frac{m((v_{0})^2)}{2}]\frac{(cos(2t\omega) + 1}{2\omega}

Extra omega eliminated here. Sorry about that.