130_notes.dvi

We use the raising operator (See section 10) equation for an energy eigenstate. A†un= √ n+ 1un+1 Now simply compute the matrix e ...

∣ ∣ ∣ ∣ ∣ ∣ −b 1 0 1 −b 1 0 1 −b ∣ ∣ ∣ ∣ ∣ ∣ = 0 wherea=√ ̄h 2 b. −b(b^2 −1)−1(−b−0) = 0 b(b^2 −2) = 0 There are three solutions ...

Before rotation the state is ψ (x) + ̄h=   1 (^21) √ 2 1 2   The rotated state is. ψ′=   i 0 0 0 1 0 0 0 −i     1 (^21 ...

Now ifψmis an eigenstate ofLz, thenLzψm=m ̄hψm, thus Hψm= μBB ̄h ∗(m ̄hψm) = (mμBB)ψm Hence the normalized eigenstates must be j ...

The beam in apparatus 3 all goes along the same path, the lower one.Apparatus 3 blocks that path. I 3 = 0 The following is a mor ...

I 3 = 3 4 I 2 = 1 16 I 0 Now lets see what happens if we remove the blocking in apparatus 2. Oven(I 0 )→    + 0 | −|    z ...

〈ψ(t)|Lx|ψ(t)〉 = ( e+iμBBt/h ̄ 2 √^1 2 e−iμBBt/ ̄h 2 ) ̄h √ 2   0 1 0 1 0 1 0 1 0      e−iμBBt/ ̄h (^21) √ 2 eiμBBt/h ̄ 2 ...

Recall the standard method of finding eigenvectors and eigenvalues: Aψ=αψ (A−α)ψ= 0 For spin^12 system we have, in matrix notati ...

The last equality is satisfied only ifχ 1 =χ 2 (just write out the two component equations to see this). Hence the normalized ei ...

18.10.12Eigenvectors ofSu As an example, lets take theudirection to be in thexzplane, between the positivexandzaxes, 30 degrees ...

18.10.13Time Development of a Spin^12 State in a B field Assume that we are in an arbitrary spin stateχ(t= 0) = ( a b ) and we h ...

m=-1/2 m=+1/2 Ε=μΒ Ε=−μΒ B field excites spin emittted Pulse of oscillating state if hν=2μΒ As excited state decays back to grou ...

requires the energy of the photons in the EM fieldE= ̄hωto be equal to the energy difference between the two spin states ∆E= 2 ̄ ...

Lx = 1 2 (L++L−) = ̄h √ 2   0 1 0 1 0 1 0 1 0   Ly = 1 2 i (L+−L−) = ̄h √ 2 i   0 1 0 −1 0 1 0 −1 0   What is the dimens ...

18.11.4Compute theℓ= 1 Rotation OperatorRz(θz)*. eiθLz/ ̄h= ∑∞ n=0 (iθLz ̄h )n n! ( Lz ̄h ) 0 =   1 0 0 0 1 0 0 0 1   ( Lz ̄ ...

( Ly ̄h ) 3 = 1 √ 2 i 1 2   0 2 0 −2 0 2 0 −2 0  = ( Ly ̄h ) ... All the odd powers are the same. All the nonzero even power ...

S+ = ̄h ( 0 1 0 0 ) S−χ− = 0 S−χ+ = √ s(s+ 1)−m(m−1) ̄hχ−= ̄hχ− S− = ̄h ( 0 0 1 0 ) We can now calculateSxandSy. Sx= 1 2 (S++S−) ...

Rx(θ) =     ∑∞ n=0, 2 , 4 ... (iθ 2 )n n! ∑∞ n=1, 3 , 5 ... (iθ 2 )n n! ∑∞ n=1, 3 , 5 ... (iθ 2 )n n! ∑∞ n=0, 2 , 4 ... (iθ ...

d dt (beiω^0 t) = iω 1 2 beiω^0 t = iω 1 2 t It appears that the amplitude grows linearly with time and hence the probability wo ...

Particles with angular momentum 1 are passed through a Stern-Gerlach apparatus which separates them according to the z-componen ...