# Homework Help: Having a hard time with Lx

1. Dec 21, 2004

### wizzart

I'm having a little bit of a problem with an excercise from my QM class. I've got the feeling it's really basic, and it probably comes down to the fact that I'm still quite flabbergasted by angular momentum operators | :uhh: . Anyway, the problem:

An electron in the hydrogen atom (neglecting spin) is (using |n,l,m>-notation) in the initial state:
$$|\Psi>_{t=0}=3|1,0,0>+|2,1,1>+i \sqrt{5}|2,1,0>-|2,1,-1>$$

a)Normalise $$\Psi$$
Calculating $$<\Psi|\Psi>$$, shows that the squared weights in front of the eigenfunctions sum up to 16. So the normalising constant A=1/4

b)Find the probability of measuring $$E_2$$
Corresponds with n=2, so P($$E_2$$)=7/16

c)Find the probability of measuring Lz=0
m=0, so P(Lz=0)=7/8

The next to questions are the ones that I can't figure out...not with certainty anyway.
d)Find the probability of measuring Lx=0
e)Find the expectation values of Lz and Lx
f)In what way to these values changes with progressing time

I can prob. get the answers in a few days in class, but I'd rather have 'em earlier, since then I can study on...

Last edited: Dec 21, 2004
2. Dec 21, 2004

### Gamma

For part (a),

$$(i \sqrt{5}) ^2 = -5 [\tex] A = 1/[tex] \sqrt{6} [\tex] gamma ---------------------- I tried to use TEX for the first time and got it wrong. Sorry. correction: (i sqrt(5) ) ^2 = -5 so A = sqrt (6) gamma. Last edited: Dec 21, 2004 3. Dec 21, 2004 ### wizzart I think my initial answer to part a was right, Gamma. Realise that [tex]<\Psi|$$ is a complex conjugate of the given $$\Psi$$, so the minus sign cancels out when calculating the inner product.

Anybody any thoughts or clear/constructive texts on the angular momentum operators, or their apllication to this particular problem?

4. Dec 21, 2004

### dextercioby

Let's take it slow,really slow:
(a)$$<\Psi|\Psi>=1\Rightarrow |\Psi>=\frac{1}{4}(3|1,0,0>+|2,1,1>+i \sqrt{5}|2,1,0>-|2,1,-1>)$$
(b)$$P(n=2)=|<2,l,m|\Psi>|^{2}=\frac{1}{16}(1+5+1)=\frac{7}{16}$$
(c)$$P(m=0)=|<n,l,0|\Psi>|^{2}=\frac{1}{16}(9+5)=\frac{7}{8}$$
(d,e)$$\hat{L}_{x}=\frac{1}{2}(\hat{L}_{+}+\hat{L}_{-})$$
$$\hat{L}_{+}|n,l,m>=\hbar\sqrt{l(l+1)-m(m+1)}|n,l,m+1>$$
$$\hat{L}_{-}|n,l,m>=\hbar\sqrt{l(l+1)-m(m-1)}|n,l,m-1>$$
$$\hat{L}_{+}|\Psi>=\frac{1}{4}(i\hbar\sqrt{10}|2,1,1>-\hbar\sqrt{2}|2,1,0>)$$
$$\hat{L}_{-}|\Psi>=\frac{1}{4}(\hbar\sqrt{2}|2,1,0>+i\hbar\sqrt{10}|2,1,-1>)$$
$$\hat{L}_{x}|\Psi>=\frac{1}{8}i\hbar\sqrt{10}(|2,1,-1>+|2,1,1>)$$
$$<\hat{L}_{x}>=<\Psi|\hat_{L}_{x}|\Psi>=\frac{i\hbar\sqrt{10}}{32} (1-1)=0$$
$$P(eigenvalue(\hat{L}_{x})=0)=...??$$ (?? i'm not sure)

(f)The eigenvalues,the averages and the probabilities do not change in time,because the selfadjoint operators and the orthogonal projectors from their spectral decomposition are time independent (in the Schroedinger picture) and they commute with the time independent Hamiltonian.

Last edited: Dec 21, 2004