130_notes.dvi

Anelectron in the Coulomb field of a protonis in the state described by the wave function^16 (4ψ 100 + 3ψ 211 −ψ 210 + √ 10 ψ ...

The angular part of the integral can be done. All the terms of the wavefunction contain aY 00 andrdoes not depend on angles, so ...

17 3D Symmetric HO in Spherical Coordinates* We have already solved the problem of a 3D harmonic oscillator by separation of var ...

Explicitly put in this behavior and use a power series expansion to solvethe full equation. R=yℓ ∑∞ k=0 akyke−y (^2) / 2 ∑∞ k=0 ...

Now as usual, the coefficient for each power ofymust be zero for this sum to be zero for ally. Before shifting terms, we must ex ...

ak+1= (k−nr) (k+ 1)(ℓ+k+ 3/2) ak Rnrℓ= ∑∞ k=0 akyℓ+2ke−y (^2) / 2 E= ( 2 nr+ℓ+ 3 2 ) ̄hω The table shows the quantum numbers for ...

18 Operators Matrices and Spin We have already solved many problems in Quantum Mechanics using wavefunctions and differential op ...

This is exactly the formula for the product of two matrices.      (OP) 11 (OP) 12 ... (OP) 1 j ... (OP) 21 (OP) 22 ... (OP) ...

eigenstates ofLzas a basis for our states and operators. Ignoring the (fixed) radial part of the wavefunction, our state vectors ...

computer utilities are readily available to help solve this problem.    A 11 A 12 A 13 ... A 21 A 22 A 23 ... A 31 A 32 A 33 ...

18.4 Anℓ= 1 System in a Magnetic Field* We will derive the Hamiltonian terms added when an atom is put in a magnetic field in se ...

Gradient in B-field N S Lets assume the field gradient is in the z direction. In the Stern-Gerlach experiment, abeam of atoms(as ...

We can use this apparatus toprepare an eigenstate. The apparatus below picks out them= 1 state N S S S N N    + 0 | −|    ...

If we block only them= 1 beam, the apparatus would be represented by    +| 0 −    z →=| 0 〉〈 0 | + |−〉〈−|. See Example 18 ...

Note that both of the aboverotation matricesreduce to the identity matrix for rotations of 2π radians. For a rotation ofπradians ...

18.8 Spin Earlier, we showed that both integer and half integer angular momentum could satisfy (See section 14.4.5) the commutat ...

S~ = ̄h 2 ~σ σx= ( 0 1 1 0 ) σy = ( 0 −i i 0 ) σz= ( 1 0 0 − 1 ) [σi,σj] = 2iǫijkσk σ^2 i = 1 They alsoanti-commute. σxσy=−σyσx ...

A beam of spin one-half particles can also be separated by a Stern-Gerlach apparatus (See section 18.5) which uses a large gradi ...

or we can just use the projection operator ( 1 √ 2 √^1 2 ) ( 1 √ 2 √^1 2 ) = ( 1 2 1 1 2 2 1 2 ) . ( 1 2 1 1 2 2 1 2 ) N 2 ( 0 1 ...

So the eigenstates are E=E 0 −A ( 1 √ 2 √^1 2 ) E=E 0 +A ( 1 √ 2 −√^12 ) The states are split by the interaction term. Feynman g ...