9. Electromagnetic Forces

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

Is the Scalar Potential Unique? • Can two scalar potentialsproduce 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 willhave 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:

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

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

Are the Vector and Scalar Potentials Unique? • Can two vector potentialsproduce the same magnetic field? • Anything with vanishing curlcan be written as a gradient • We also need to matchthe electric field • So we must have • For the wave function, it makes sense tomake the same guess as before

Gauge Transformations • The three transformation at right collectivelydescribe a gauge transformation • They lead to identical electric and magnetic fields • Choosing exactly which A and U to use iscalled 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:

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 makeU = 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

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 transformationrelating 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

Kinematic Momentum • We want Schrödinger’s equation to be invariant under gaugetransformations • How about the equationwe 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:

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:

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

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

Commutation Relations • Recall: • Also recall: • We now need to find commutators of  and R • To understand this, consider aspecific component, say

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

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

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

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

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 = 5.5856957 • Even the neutron has a dipole moment • Withgn = –3.826085

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’sequation: 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

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 notuniform, this will produce a force • By creating a non-uniform magnetic field, we can effectively measure the spin • This is the Stern-Gerlach experiment

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 operatorsA1, A2, A3as 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 manyterms as possible with corresponding eigenvalues • Solve the remaining problem

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 foundthree 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:

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

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

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:

Hydrogen Atom in Strong Magnetic Field (2) • Let’s look at the size of the B2 term: • Atoms are around 10-10 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

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

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

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…

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

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,

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:

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

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

Double Slit Experiment With a Magnet Now adda magnet • We can choose gauge AIsuch 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 theymust 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:

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

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:

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”

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

9. Electromagnetic Forces