1. 9. Electromagnetic Forces
9A. Electric Forces
The Scalar Potential
In E&M, electric fields produce a force F = qE
In quantum mechanics, we always work with the potential
Assume no magnetic field. Can we write electric field in terms of potential?
Faraday’s Law:
If B = 0, then:
Anything with zero curl can be written as a gradient:
U is the electrostatic potential or scalar potential
Then the potential energy will be qU(r,t)
Hamiltonian for electrons:
For other particles, e  –q

2. Is the Scalar Potential Unique?
Can two scalar potentials produce the same electric field?
The difference must be independent of position:
Why write it this way? We’ll see soon
The problem: Schrödinger’s equation will have different solutions for different U’s
Of course, if U' – U is a constant, this is just a shift in energy, and we know how to deal with that:
This suggests:

3. Gauge Transformations With Only E-fields
Assert: if  is a solution of Schrödinger’s equation with potential U, then ' is a solution with potential U'
Note, for particles other than electrons, gauge transformation for charge q is

4. 9B. Electromagnetic Forces
The Vector Potential
Classically, the force on a charge is given by:
Not obvious how to write this in terms of a potential
Is there something akin to the electrostatic potential we can use to write the magnetic field?
Gauss’s law for magnetic fields:
Anything with zero divergence can be written as a curl
A(r,t) is called the vector potential
Faraday’s Law
Anything with zero curl can be written as a gradient

5. Are the Vector and Scalar Potentials Unique?
Can two vector potentials produce the same magnetic field?
Anything with vanishing curl can be written as a gradient
We also need to match the electric field
So we must have
For the wave function, it makes sense to make the same guess as before

6. Gauge Transformations
The three transformation at right collectively describe a gauge transformation
They lead to identical electric and magnetic fields
Choosing exactly which A and U to use is called a gauge choice
Classically, it doesn’t matter which one you pick – they are physically equivalent
Sort of like picking an origin
We would expect Schrödinger’s equation – whatever it is – to be unchanged as well
Goal: Find a Schrödinger’s equation that:
Is unchanged when we perform a gauge transformation
Physically leads to forces like:

7. Sample Problem
A region of space contains no electric field and a constant magnetic field B0 in the z-direction. Find three different vector fields A that could describe this magnetic field, find a gauge transformation relating two of them, and describe the types of problems each of the three would be most useful for.
The fields are:
Seems like a good idea to make U = 0 and A independent of time
To get Bz, write out curl:
First solution: Let Ax = 0, then
Second solution: Let Ay = 0, then
Third solution: Average them

8. Sample Problem (2)
…, find a gauge transformation relating two of them, and describe the types of problems each of the three would be most useful for.
Let’s try to find a gauge transformation relating 1 and 2
What are each of these useful for?
Solution 1 is independent of y
Useful for problems independent of y
Solution 2 is independent of x
Useful for problems independent of x
Solution 3 – plot it:
Useful for rotationally symmetric problems

9. Kinematic Momentum
We want Schrödinger’s equation to be invariant under gauge transformations
How about the equation we already have?
Turns out it doesn’t work
Problem: We would like to have expressions like
What do we actually have?
It looks like we wish we had P + eA instead of P
Define the kinematic momentum  as
As always, e  – q for other particles
Then we see that:

10. Schrödinger’s Equation Revisited
Let’s guess the following Schrödinger’s equation:
Assert, if true for , then true for ’
First note that:
Similarly:
We therefore have:
Hamiltonian is:

11. 9C. Working With Kinematic Momentum
Sample Problem
Which of the expectation values P and  is gauge invariant?
Recall that:
No for P and yes for 

12. Kinematic vs. Conventional Momentum
Note that any measurement must be independent of gauge
Therefore, P must be unmeasurable, but  might be measureable
How would one measure momentum?
One way: measure velocity (change in position over time) and multiply by mass
It will be necessary to revisit Ehrenfest’s theorem
These were derived from
But we assumed we had a Hamiltonian of
But our Hamiltonian is now

13. Commutation Relations
Recall:
Also recall:
We now need to find commutators of  and R
To understand this, consider a specific component, say

14. Ehrenfest’s Theorem Revisited: Position
For any operator:
We therefore have:
Generalize the formula:
Conclusion: Velocity times mass is kinematic momentum

15. Ehrenfest’s Theorem Revisited: Momentum (1)
For kinematic momentum we have

16. Ehrenfest’s Theorem Revisited: Momentum (2)
Electric field is:
The sum is a cross product: consider i = z
Now generalize

17. Ehrenfest’s Theorem Discussion
First equation tells you  is like momentum, p = mv
In second equation, what does this mean classically?
This is exactly what we’d expect
Strong indication that we guessed the right Hamiltonian
However, for particles with spin, we actually missed something

18. 9D. Magnetic Dipole Moments
Why They Should Exist
Think of an elementary particle as a spinning ball of charge q and angular momentum S
Electric current is flowing the same direction it is spinning
It should act like a little dipole magnet
Naïve model then predicts dipole moment:
This model is naïve, but dimensionally correct
For electron, simple Dirac model predicts g = 2
Experiment yields
For protons, a similar formula applies:
Proton not elementary: gp =
Even the neutron has a dipole moment
With gn = –

19. Schrödinger’s Equation Re-Revisited
A magnetic dipole in a magnetic field likes to align with the magnetic field
This implies there is an energy associated with it
This implies a new term in the Hamiltonian
Commonly approximate g = 2 for electron
Schrödinger’s equation:
Is it still gauge invariant?
The old Schrödinger equation was gauge invariant
The new one just contains B, which is gauge invariant
So new expression is also gauge invariant

20. Stern-Gerlach Experiment
Suppose you put a particle in a region with a magnetic field
Presence of magnetic field causes a contribution to the potential energy
If the magnetic field is not uniform, this will produce a force
By creating a non-uniform magnetic field, we can effectively measure the spin
This is the Stern-Gerlach experiment

21. 9E. Simple Problems with Magnetic Fields
The Procedure
How do we solve problems with EM fields?
Step one: Find A and U
Because of gauge invariance, you typically have many choices for A and U
Choose one that is compatible with other symmetries of the problem
Step two: Write the Hamiltonian
Step three: Find as many operators A1, A2, A3 as possible that commute with the Hamiltonian and each other
Write your eigenstates as states with eigenvalues under all the other operators
Step four: In Schrödinger’s equation, replace as many terms as possible with corresponding eigenvalues
Solve the remaining problem

22. Landau Levels
What are the energy levels of an electron in a uniform magnetic field ?
Step One: we need A and U
We set U = 0, then we found three solutions to these equations
I like the first one: it’s positive and less complicated
Step two: Now substitute it into the Hamiltonian:
Step three: Find operators that commute with H:
Write your states as eigenstates of these:

23. Landau Levels (2)
Step four: In Schrödinger’s equation, replace as many terms as possible with corresponding eigenvalues
Solve the remaining problem
Where:
This is a shifted harmonic oscillator with energy

24. Landau Levels (3) Now put it all together
First term is motion along magnetic field
The n term is cyclotron motion
The ms term is precession in the magnetic fields
Note that changing ms by one almost cancels changing n by one
Small frequency difference – can be measured experimentally very well

25. Hydrogen Atom in Strong Magnetic Field
Put a proton in a strong magnetic field, then add one electron
Assume magnetic field is constant (on the scale of the atom)
Electric and magnetic fields are
Need to find U and A:
How about:
Which of the following is best for B?
U has rotational symmetry, not translational, so choose A3
The Hamiltonian is now:

26. Hydrogen Atom in Strong Magnetic Field (2)
Let’s look at the size of the B2 term:
Atoms are around m in size, and magnetic fields are rarely more than a few Tesla
This term is tiny – let’s ignore it
When we say strong magnetic field, we don’t mean too strong, say B < 104 T
Recall that:
So we have

27. Hydrogen Atom in Strong Magnetic Field (3)
The first two terms are the unperturbed hydrogen atom
We know states and energies
These states are also eigenstates of the last two terms
States with the same ml and ms get split in presence of magnetic fields
That’s why these letters are called m – magnetic quantum numbers

28. Why a Strong Magnetic Field?
It seems like we never assumed a strong magnetic field
There are actually additional small terms in the Hamiltonian
Spin-orbit coupling
This means the basis vectors we chose are not eigenstates of the Hamiltonian, even in the absence of magnetic field
But our equations are still okay if the magnetic field is strong enough
This is called the Paschen-Back effect

29. Sample Problem
An electron is in a constant electric field and magnetic field as given at right. Find eigenstates of the Hamiltonian, their energy, and their velocity.
We need to find the scalar and vector potentials
Let’s try
Which one for the vector field?
Since U depends on only x, pick A1
Our Hamiltonian is now:
This commutes with
Eigenstates are
So we have
Rearrange it a little…

30. Sample Problem (2) H' is some sort of shifted harmonic oscillator
Complete the square
Perfect square

31. Sample Problem (3)
This is a harmonic oscillator, shifted in space, and with shifted energy
Standard harmonic oscillator: For all states:
But ours is centered at x0, so
For our states,

32. Sample Problem (4)
An electron is in a constant electric field and magnetic field as given at right. Find eigenstates of the Hamiltonian, their energy, and their velocity.
Recall:
Classical force:

33. 9F. The Aharanov-Bohm Effect Can We See the Vector Potential Itself?
If the magnetic field vanishes, we can choose A = 0, right?
Q: If particles move through a region with B = 0, can we ignore A?
A: If the region is simply connected, then you can prove this is true
Consider an ideal solenoid with current flowing as sketched at right
Consider the integral of A around the black dashed curve
Use Stokes’ theorem
You can only make A = 0 everywhere around a loop if there is no magnetic flux through it
This allows us to conjecture a very interesting situation

34. Double Slit Experiment Without a Magnet
Let the initial wave at the first slit be 0
Part of it follows the upper path
Another part follows the lower path
The total wave is then
Interference fringes

35. Double Slit Experiment With a Magnet
Now add a magnet
We can choose gauge AI such that AI = 0 on upper path
We can choose gauge AII such that AII = 0 on lower path
These cannot be the same gauge, but they must be related by a gauge transformation
We can add a constant to  so (0) = 0 (this makes 0 the same in both gauges)
Consider the following integral:

36. Double Slit Experiment With a Magnet (2)
Write both final waves in gauge I
Recall:
Therefore:
Hence we have:

37. Discussion of Aharanov-Bohm
Shift in phase can be detected
Even though the particle didn’t move through region of magnetic field!
The shift is undetectable if:

38. Magnetic Monopoles
A magnetic monopole is a north pole without a south pole
It would have magnetic field:
Can we find a vector potential A that makes this work?
But wait – Doesn’t this rule preclude monopoles?
Note that A is infinite and discontinuous at  = 
The magnetic flux is actually coming in at  = 
But this is infinitely thin, not classically detectable
Like an infinitely thin solenoid, of “string”

39. Magnetic Monopole Quantization
That “string” is invisible classically
But, suppose we send electrons around the string to do Aharanov-Bohm
Presence of string detectable by shift in interference
Unless:
Prediction: if magnetic monopoles exist, their magnetic charge comes in multiples of h/e
This is in fact the typical monopole charge in extensions of the standard model, like Grand Unified Theories
Magnetic monopoles have never been detected