# Homework Help: The Normal Zeeman Effect and Hydrogen States

1. May 30, 2013

### mrjohns

I'm studying the hydrogen atom and have this question. Apparently it can be solved without perturbation theory, however I'm having trouble justifying it.

1. The problem statement, all variables and given/known data

2. The attempt at a solution

Avoiding perturbation theory I simply get:

E = E(n) - constant*(mh) where m is the angular momentum number not mass

Which seems a little too easy.

For the second part I can see the initial t=0 wavefunction is normalised no problem, but when I apply the time evolution operator e^(iEt)/h things get messy.

Am I right in thinking the probability for the third state | 2 1 0 > is just zero for all time?

2. May 30, 2013

### TSny

Hello mrjohns. Welcome to PF!

That looks good.

Can you show a little more detail. It shouldn't be messy.

Yes.

3. May 31, 2013

### mrjohns

If I say the energies are E=E(n)-E(b)*mh I get by time evolution:

|psi(t)> = 1/sqr(2) [ e^-i(E(n)+E(b))t | 2 1 -1 > - e^-i(E(n)-E(b))t | 2 1 1 > ]

Which by Eulers formula:

=1/sqrt(2) { [ cos[(-E(n)-E(b))t] + isin[(-E(n)-E(b))t] ] | 2 1 -1 > - [ cos[(-E(n)+E(b))t] + isin[(-E(n)+E(b))t] ] | 2 1 1 > }

Which is normalised for all time, since (cos(x)+isin(x))(cos(x)-isin(x))=1

Now obviously it can never get to the |psi(3)>=|210> state of |nlm> since this evolution doesn't change the value of m.

But I can't see how it can evolve to the |psi(2)> state. Should I have left in m, rather than substituted 1 and -1 in for it?

4. May 31, 2013

### TSny

You have |ψ(t)> = 1/√2 [e-iω1t | 2 1 -1 > - e-iω2t | 2 1 1 > ] where ω1 = E(n)+E(b) and ω2 = E(n)-E(b)

Note that the phase factors in front of | 2 1 -1 > and | 2 1 1 > oscillate with different frequencies. So, there will be instants of time when the phase factors are equal (which gives |ψ1>) and instants of time when the phase factors are exactly out of phase with each other (which gives |ψ2>).

Now that you have an expression for |ψ(t)>, how would you get the probability amplitude for this state to be measured as |ψ1>?

5. Jun 1, 2013

### mrjohns

Using matrix multiplication and putting bras to kets i get:

Prob amp for |ψ1> = 1/2 [e^-iω1t + e^-iω2t] => Probability is one when e^-iω1t = e^-iω2t = 1

Prob amp for |ψ2> = 1/2 [e^-iω1t - e^-iω2t] => Probability is one when e^-iω1t = -e^-iω2t = 1

Prob amp for |ψ3> = 0 for all time

Now these probability amplitudes for |ψ1>,|ψ2>,|ψ3> sum to e^-iω1t which does not equal one. Is this correct?

Or am I reading the question wrong, and is it simply asking whether <ψ(t)|ψ(t)> = 1 for all time? (Which it does.)

[Edit-And I've also got the added condition for the phases e^-iω1t = e^-iω2t = 1 rather than when they're just equal, and e^-iω1t = -e^-iω2t = 1 rather than when they're just out of phase.

I know I want to do <ψ1|ψ(t)> to find the probability amplitude for ψ1

Do I want <ψ1| = 1/√2 [< 2 1 -1 | - < 2 1 1 |] or am I missing something?]

Last edited: Jun 1, 2013
6. Jun 1, 2013

### TSny

A probability amplitude is a product of a bra with a ket. So, I'm not following your statements above. But you are correct in what you state below:

Yes this is all you need to do, find <ψ1|ψ(t)>

where <ψ1| = 1/√2 [< 2 1 -1 | - < 2 1 1 |] and |ψ(t)> = 1/√2 [e-1t | 2 1 -1 > - e-iω2t | 2 1 1 > ]

7. Jun 1, 2013

### mrjohns

So does the methodology here look right?

<ψ1|ψ(t)> = {1/√2 [< 2 1 -1 | - < 2 1 1 |]}{1/√2 [e-iω1t | 2 1 -1 > - e-iω2t | 2 1 1 > ] }
<ψ1|ψ(t)> = 1/2 [< 2 1 -1 | e-iω1t | 2 1 -1 > + < 2 1 1 | e-iω2t | 2 1 1 >]
<ψ1|ψ(t)> = 1/2 [e^-iω1t + e^-iω2t] by orthonormality

And similarly:

<ψ2|ψ(t)> = 1/2 [e^-iω1t - e^-iω2t] by orthonormality

<ψ3|ψ(t)> = 0

To get the probability density I need to square by complex conjugate, so get:

Prob:(ψ1) = 1/4 [e^-iω1t + e^-iω2t][e^+iω1t + e^+iω2t]
Prob:(ψ1) = 1/4 [2+2e^(-i^2)ω1ω2(t^2)]
Prob:(ψ1) = 1/2 [1+e^ω1ω2(t^2)]

And similarly:

Prob:(ψ2) = 1/2 [1-e^ω1ω2(t^2)]

Thus Prob:(ψ1) + Prob:(ψ2) + Prob:(ψ3) = 1/2 [1+e^ω1ω2(t^2)] + 1/2 [1-e^ω1ω2(t^2)] = 1 for all time.

And we get:

Prob:(ψ1)=1 if e^ω1ω2(t^2)=1
Prob:(ψ2)=1 if e^ω1ω2(t^2)=-1

Does this seem ok? One part that's bugging me is e^ω1ω2(t^2) - how do I know it is bound between -1 and 1 in order for the probability density equations to make sense?

Last edited: Jun 1, 2013
8. Jun 1, 2013

### TSny

Good up to the blue line. Rethink the product e-iω1te+iω2t

9. Jun 1, 2013

### mrjohns

Whoops, so I should have typed:

Prob:(ψ1) = 1/4 [e^-iω1t + e^-iω2t][e^+iω1t + e^+iω2t]
Prob:(ψ1) = 1/4 [1+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t+1]
Prob:(ψ1) = 1/4 [2+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t]

And similarly:

Prob:(ψ2) = 1/4 [2-e^-i(ω1-ω2)t-e^-i(ω2-ω1)t]

Thus Prob:(ψ1) + Prob:(ψ2) + Prob:(ψ3) = 1/4 [2+e^-i(ω1-ω2)t+e^-i(ω2-ω1)t] + 1/4 [2-e^-i(ω1-ω2)t-e^-i(ω2-ω1)t] = 1 for all time.

And we get:

Prob:(ψ1)=1 if e^-i(ω1-ω2)t+e^-i(ω2-ω1)t=2
Prob:(ψ2)=1 if e^-i(ω1-ω2)t+e^-i(ω2-ω1)t=-2

So now I need | e^-i(ω1-ω2)t+e^-i(ω2-ω1)t | <= 2 - how do I know it is bound in this way? And isn't the complex part an issue?

Last edited: Jun 1, 2013
10. Jun 1, 2013

### TSny

Use the identity eix + e-ix = 2cos(x)

11. Jun 2, 2013

### mrjohns

Ah, so:

Prob:(ψ1) = 1/4 [2+e^-i(ω1-ω2)t+e^+i(ω1-ω2)t]
Prob:(ψ1) = 1/2 {1+cos[(ω1-ω2)t]}

Prob:(ψ2) = 1/4 [2-(e^-i(ω1-ω2)t+e^+i(ω1-ω2)t)]
Prob:(ψ2) = 1/2 {1-cos[(ω1-ω2)t]}

Now I can see that looks a lot better - the probabilities for each are bound between 0 and 1, and Prob:(ψ1)=1 at t=0.

When cos[(ω1-ω2)t=1, Prob:(ψ1)=1, and when cos[(ω1-ω2)t]=-1, Prob:(ψ2)=1.

12. Jun 2, 2013

### TSny

Yes. Good.

13. Jun 2, 2013

### mrjohns

Cheers, thanks so much for your help.