https://en.wikipedia.org/wiki/Landau_quantization
solves the problem of an electron in a uniform magnetic field. The solution hangs on the fact that the hamiltonian only depends on coordinate x in the Wikipedia article. The solution is then symmetric on translations along the y axis, which means that p_y must be constant, an eigenvalue of the p_y operator.
The operator
p_y = -i d/dy
gives the canonical momentum which is not the measured kinetic momentum. Our earlier discussion in the text below mixed the canonical momentum with the kinetic momentum. This led to confusion in our text.
https://physics.stackexchange.com/questions/281687/why-is-p-y-conserved-in-the-landau-gauge-when-we-know-the-electron-moves-in-ci
At the Physics Stack Exchange there are several questions that rise from the confusion.
If we break the symmetry along the y axis by adding an electric potential which confines the electron into some y interval (y_0, y_1), what happens then?
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The Pauli equation contains a hamiltonian where the kinetic energy term is something like
(p - qA)^2 / (2m).
There p is the momentum vector, q is the charge, and A is the magnetic vector potential.
If we have a wire with a current, then the vector potential points to the direction of the current and is less when we go farther from the wire.
We assume that the wire has no electric field. Its magnetic field lines are circles around the wire.
If we let a charge fly freely closer to the wire, the magnetic force is at a right angle relative to the kinetic momentum of the electron.
---
NOTE: p is the canonical momentum. It does not change. The kinetic momentum p - qA does change, as it should.
The curl of the magnetic vector potential
The magnetic vector potential A is defined as the vector field whose curl is the magnetic field B:
∇ × A = B.
The component of the curl pointing to the direction of the thumb at a point x is defined as the path integral of A on the circular path pointing to the direction of the fingers of the right hand, divided by the area enclosed by the path.
Let us consider an electric wire of a finite length:
• • •
• • • • • • •
-----------------------------
I -->
The dots • mark magnetic field B lines pointing out of the page.
A vector potential A which reproduces the field B is like:
--> --> -->
----> ----> ----> A
--------> -------->
--------------------------------
I -->
|A| is larger close to the wire. A points to the direction of the electric current.
Now, if we have an electron approaching the wire in the diagram from up, the electron will draw a path like:
e- ^
\ /
\ ____/
-----------------------------
I -->
^ y
|
|
------> x
^ y
|
|
------> x
The electron will turn counterclockwise in the magnetic field. Since the magnetic force
F = q v × B
is orthogonal to the velocity vector v, it will not change the absolute value of v but its direction. The electron will leave the magnetic field with the same absolute velocity at which it arrived.
In the diagram above, the quantity
H = (p - eA)^2 / (2m)
is not larger when the electron is close to the wire because p is not the kinetic momentum but the canonical momentum. Note that the charge e of the electron is negative.
Hamilton's equations
The equations are
dp/dt = -dH/dq
dq/dt = dH/dp,
where p is the canonical momentum and q is the (canonical?) position.
where p is the canonical momentum and q is the (canonical?) position.
The lagrangian
The hamiltonian is derived from the lagrangian. Let us check if the lagrangian gives the correct circular orbit for the electron in our example case.
The action is defined as the path integral of the lagrangian over a time interval [t1, t2] over a path q(t). If the path is a correct time evolution of the system, then the action should remain constant under "very small changes" of the path.
The path integral, of course, suffers from the fact that the set of allowed paths does not have an exact mathematical definition. Do we allow fractals? Let us consider just paths q(t) which are built from intervals of analytical functions.
Does the action change under a "very small" deformation of the circle orbit q(t)? We assume that t_1, t_2, q(t_1), and dq(t_1)/dt remain fixed.
https://en.wikipedia.org/wiki/Lagrangian_mechanics#Electromagnetism
The lagrangian of a massive charged particle with charge e in magnetic field is
L = m/2 * v^2 + e v • A,
where v is the velocity vector and A is the magnetic vector potential. The dot marks the inner vector product.
Let us consider a circular path in the diagrams above. Let us assume that the magnetic field B is constant and points out of the page.
<-- v e-
______
/ \ --> --> -->
| |
\______/ -----> -----> A
^ y
|
|
------> x
The electron e- does a perfect circle. It leaves position x at a velocity v at time t_1, and returns back at time t_2.
We assume that there is no electromagnetic radiation out of the system. The radius of the circle is determined by the absolute speed |v| of the electron and the strength of the magnetic field B = ∇ x A.
What happens to the path integral of the lagrangian L over the time interval [t_1, t_2]? The action is
t_2
S = ∮ m/2 * v^2 + e v • A,
t_1
where the path integral is done counter-clockwise.
We may assume that |A| is zero at the upmost point of the circle. Let us increase the speed of the electron by 1% for the whole circle, so that its radius is 1% larger. The starting point of the electron is kept constant. How does the action S change?
The kinetic part grows 2 %. The circle extends further down. In the lower half of the circle, |A| is 1% larger and |v| is too. The contribution is negative and grows by about 2%. There is no contradiction in these numbers. The lagrangian seems to work ok.
https://en.wikipedia.org/wiki/Lagrangian_mechanics#Electromagnetism
The lagrangian of a massive charged particle with charge e in magnetic field is
L = m/2 * v^2 + e v • A,
where v is the velocity vector and A is the magnetic vector potential. The dot marks the inner vector product.
Let us consider a circular path in the diagrams above. Let us assume that the magnetic field B is constant and points out of the page.
<-- v e-
______
/ \ --> --> -->
| |
\______/ -----> -----> A
^ y
|
|
------> x
The electron e- does a perfect circle. It leaves position x at a velocity v at time t_1, and returns back at time t_2.
We assume that there is no electromagnetic radiation out of the system. The radius of the circle is determined by the absolute speed |v| of the electron and the strength of the magnetic field B = ∇ x A.
What happens to the path integral of the lagrangian L over the time interval [t_1, t_2]? The action is
t_2
S = ∮ m/2 * v^2 + e v • A,
t_1
where the path integral is done counter-clockwise.
We may assume that |A| is zero at the upmost point of the circle. Let us increase the speed of the electron by 1% for the whole circle, so that its radius is 1% larger. The starting point of the electron is kept constant. How does the action S change?
The kinetic part grows 2 %. The circle extends further down. In the lower half of the circle, |A| is 1% larger and |v| is too. The contribution is negative and grows by about 2%. There is no contradiction in these numbers. The lagrangian seems to work ok.
If the energy would vary, could a hamiltonian work for the Pauli equation?
If the energy of the electron would be larger in the lower part of the circle in the diagram, then the phase of its wave function would rotate faster in the lower part. The phase difference between different points in the circle would tend to infinity, which would mean an infinite momentum.
The Schrödinger equation under a scalar potential works beautifully because the total energy is constant, and the phase of the wave function rotates at the same rate everywhere.
The Biot-Savart law and magnetic interaction
For slowly moving charges, the magnetic force is
F = μ_0 / (4π) * q_1 q_2 / r^2 * v_1 x (v_2 x r),
where r is the vector from charge q_1 to q_2. The formula is a consequence of the Biot-Savart law.
To which direction does the force point? In our example above, we have electrons moving to left inside the wire, and our test electron does a circular path. That is, the force is always orthogonal to the velocity vector v_2 of the test electron.
Can we model the force simply as a force between two point objects? The forces on the charges q_1 and q_2 will generally exert a torque on the system q_1 & q_2. Conservation of angular momentum requires that there is an opposite torque on something. Apparently, the magnetic field can store angular momentum.
If we stop both charges with some device, the magnetic field disappears. If there is no radiation out, then the angular momentum stored in the magnetic field has to be returned back to the system q_1 & q_2 & the device.
The minimal coupling is approximately right for the classical electron?
In our example case of an electron circling in a loop, some angular momentum is stored in the combined electromagnetic field of the wire and the electron. Can we neglect this effect in the classical treatment of the electron?
Probably yes, because a hypothetical self-interaction of a single electron is much weaker than the the force which is caused by the external magnetic field B.
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