130_notes.dvi

joccurs in both tensors (and is summed over) we can simplify things. Take the case thati= 1 andk= 2. We only have a nonzero term ...

Note that~p 6 =m~v. The momentum conjugate to~rincludes momentum in the field. We now time differentiate this equation and write ...

Q.E.D. So we have Fi=−eEi− e c ( ~v×B~ ) i which is the Lorentz force law. So this is the right Hamiltonian for an electron in a ...

( ~r×B~ ) 2 = riBjεijkrmBnεmnk= (riBjriBj−riBjrjBi) = r^2 B^2 − ( ~r·B~ ) 2 − 0 So, plugging these two equations in, we get − ̄h ...

From the symmetry of the problem, we can guess (and verify) that[H,pz] = [H,Lz] = 0. These variables will be constants of the mo ...

withλ=nr+ℓ+ 1. The equations are the same if WE set ourλ 4 =nr+1+ 2 |m| wherenr= 0 , 1 , 2 ,.... Recall that ourλ=^4 eBme ̄hc ( ...

ψ=v(x)eikyyeikzz 1 2 me ( − ̄h^2 d^2 dx^2 + ( eB c ) 2 ( x+ ̄hcky eB ) 2 ) v(x) = ( E− ̄h^2 kz^2 2 me ) v(x) − ̄h^2 2 me d^2 dx^ ...

Now we willapply the differential operator to the exponentialto identify the new terms. Note that∇~eiλ(~r,t)=eiλ(~r,t)i∇~λ(~r,t) ...

0 path 1 path 2 B field r r Consider two different paths from~r 0 to~r. f 1 (~r)−f 2 (~r) = ∮ d~r·A~= ∫ dS~·∇×~ A~= ∫ dS~·B~= Φ ...

is given by ~j= ̄h 2 iμ [ψ∗∇~ψ−(∇~ψ∗)ψ+ 2 ie ̄hc Aψ~ ∗ψ]. Remember the flux satisfies the equations∂(ψ ∗ψ) ∂t + ∇~~j= 0. Consid ...

21 Addition of Angular Momentum Since total angular momentum is conserved in nature, we will find that eigenstates of the total ...

audio Verify the quoted eigenvalues by calculation using the operatorSz=S(1)z+S(2)z. We expect to be able to formeigenstates ofS ...

Since our eigenstates ofJ^2 andJzare also eigenstates ofL^2 andS^2 (but notLzorSz), these are ideal for computing the spin orbit ...

momentum quantum numberℓbut is written asS, P, D, F, ...forℓ= 0, 1 , 2 , 3 ,..., andjis the total angular momentum quantum numbe ...

We cancheck(see section 21.8.5)that the number of states agrees with the number of product states. We have been expanding the st ...

21.7 Examples 21.7.1 Counting states forℓ= 3 Plus spin Forℓ= 3 there are 2ℓ+ 1 = 7 different eigenstates ofLz. There are two dif ...

Since the highest m value is 6, we expect to have aj= 6 state which uses up one state for each m value from -6 to +6. Now the hi ...

Normally we will not bother to include that the spins are one half since that’s always true for electrons. We will (and must) ke ...

21.8.2 Using the Lowering Operator to Find Total Spin States The total spin lowering operator is S−=S−(1)+S(2)−. First lets remi ...

21.8.3 Applying theS^2 Operator toχ 1 mandχ We wish to verify that the states we have deduced are really eigenstates of theS^2 o ...