can be used tofind the ground state wavefunction. WriteAin terms ofxandpand try it.
(√
mω
2 ̄hx+i
p
√
2 m ̄hω)
u 0 = 0
(
mωx+ ̄h
d
dx)
u 0 = 0du 0
dx=−
mωx
̄hu 0This first order differential equation can be solved to get the wavefunction.
u 0 =Ce−mωx(^2) /2 ̄h
We could continue with the raising operator to get excited states.
√
1 u 1 =
(√
mω
2 ̄hx−√
̄h
2 mωd
dx)
u 0Usually we will not need the actual wave functions for our calculations.
10.6 Examples
10.6.1 The expectation value ofxin eigenstate
We can compute the expectation value ofxsimply.
〈un|x|un〉 =√
̄h
2 mω〈un|A+A†|un〉=√
̄h
2 mω(〈un|Aun〉+〈un|A†un〉)=
√
̄h
2 mω(
√
n〈un|un− 1 〉+√
n+ 1〈un|un+1〉) = 0We should have seen that coming. Since each term in thexoperator changes the eigenstate, the dot
product with the original (orthogonal) state must give zero.
10.6.2 The expectation value ofpin eigenstate
(See the previous example is you want to see all the steps.) The expectation value ofpalso gives
zero.
〈un|p|un〉=−i√
m ̄hω
2〈un|A−A†|un〉= 0