# Raising and lowering operators

1. Oct 19, 2011

### v_pino

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

Please see attached pdf. Can anyone please advise on how to approach the questions or provide websites where I can read up on it? I am reading Griffiths but it doesn't seem to cover this much. And I don't know how to google on this topic because I can't type the characters in.

2. Relevant equations

3. The attempt at a solution

For part 1a, I tried to rearrange the second equation given in the problem such that a- and a+ are isolated. I then substituted them back into the first equation. But I don't think that's what the problem is after.

For part 1b, I tried substituting the first euqation of H into the new expression with 'nu'. Am I heading in the right direction?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### PS7-7.pdf
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2. Oct 19, 2011

### grzz

If you show your attempt one can point out any mistakes.

3. Oct 19, 2011

### v_pino

Here is my attempt at part (d) of problem 1. Is that correct?

$$[\widehat{b}_-,\widehat{b}_+]=\widehat{b}_-\widehat{b}_+-\widehat{b}_+\widehat{b}_-$$

$$[\widehat{N}',\widehat{b}_+]=[\widehat{b}_+\widehat{b}_-,\widehat{b}_+]=\widehat{b}_+[\widehat{b}_-,\widehat{b}_+]+[\widehat{b}_+,\widehat{b}_+]\widehat{b}_- = \widehat{b}_+[\widehat{b}_-,\widehat{b}_+]=\widehat{b}_+$$

$$[\widehat{N}',\widehat{b}_-]=[\widehat{b}_+\widehat{b}_-,\widehat{b}_-]=\widehat{b}_+[\widehat{b}_-,\widehat{b}_-]+[\widehat{b}_+,\widehat{b}_-]\widehat{b}_- = \widehat{b}_+[\widehat{b}_-,\widehat{b}_-]=\widehat{b}_-$$

4. Oct 19, 2011

### grzz

Is [$\hat{a}$,$\hat{a^{+}}$] = 1?

5. Oct 19, 2011

### v_pino

In my lecture notes, $$[\widehat{a}_-,\widehat{a}_+]$$ =1. So I assumed it is the same for 'b'.

And I made a mistake in my third equation. The second to last on the right hand side should be b+[b+,b-].

6. Oct 19, 2011

### vela

Staff Emeritus
You're not done here. A better approach would be to calculate
$$[\widehat{b}_-,\widehat{b}_+] = [\alpha\widehat{a}_-+\beta,\alpha\widehat{a}_++\beta]$$Presumably, you know what $[\widehat{a}_-,\widehat{a}_+]$ equals.
You have a typo in the second one. You should have kept the second term, not the first. The final answers for both aren't right because you haven't evaluated $[\widehat{b}_-,\widehat{b}_+]$ correctly yet.

7. Oct 19, 2011

### v_pino

Can you please give me a hint on the earlier parts of the problem like 1a and 1b? I feel that I shouldn't proceed onto part d until I have those solved as part d requires beta. Thanks

8. Oct 19, 2011

### vela

Staff Emeritus
Perhaps I'm missing something, but I think parts a and b are backwards. There are no constraints in the beginning on alpha and beta, so asking you to find them doesn't really make sense. If you read ahead to part b, however, you can see what form of the Hamiltonian you're going for and figure out what the constants have to be to achieve that form.

I would solve for the a's in terms of the b's and plug those expressions into H and compare it to the expression for η. Try that and see what you get.

9. Oct 19, 2011

### v_pino

These are the equations I came up with for part (a) and (b) of the problem. I'm having trouble writing them with a constant on one side.

$$\widehat{H}=\frac{\varepsilon_1}{\alpha^2}( \widehat{b}_--\beta)(\widehat{b}_+-\beta)+\frac{\varepsilon_2}{\alpha}((\widehat{b}_--\beta)+(\widehat{b}_+-\beta))$$

$$\eta =\frac{\widehat{H}}{\varepsilon_1}=\widehat{a}_+\widehat{a}_-+\frac{\varepsilon_2}{\epsilon_1}(\widehat{a}_-\widehat{a}_+)$$

10. Oct 19, 2011

### v_pino

Do they look correct to you for part (d)? I'm still working on the preceding parts. (I dropped the 'hats' to make typing easier.)

$$[b_-,b_+]=\alpha^2$$

$$[N',b_+]=b_+\alpha^2$$

$$[N',b_-]=b_-\alpha^2$$

11. Oct 19, 2011

### vela

Staff Emeritus
You have the factors reversed in the first term. Correct that and expand it out, and you'll get
$$\hat{H} = \frac{\varepsilon_1}{\alpha^2}\hat{b}_+\hat{b}_-+D(\hat{b}_++\hat{b}_-)+C$$where C and D are constants. If you want it to match up to η, you need $\alpha^2=1$ and D=0. The requirement D=0 will allow you to solve for $\beta$.

Yes, those look fine.

12. Oct 19, 2011

### v_pino

Here are my solutions for (a) and (b):

$$\alpha=1$$
$$\beta=\frac{\varepsilon_2}{\varepsilon_1}$$
$$C=\frac{\varepsilon_2}{\varepsilon_1^2}-2\frac{\varepsilon_2^2}{\varepsilon_1^2}$$

For part (c), do I evaluate:

$$b\Psi_0=(a_-+\frac{\varepsilon_2}{\varepsilon_1})\Psi_0 = 0$$

But I don't know what a_ is, do I? I know it is called the lowering operator. But what should I write it in terms of?

thanks for helping and sorry it's taken me some time to figure things out.

13. Oct 19, 2011

### vela

Staff Emeritus
Have you covered the simple harmonic oscillator in class? I assume you have since you had the commutation relation for a+ and a- in your notes.

14. Oct 19, 2011

### v_pino

I found an equation for a_ in SHO:

$$a_-=\frac{1}{2hm\omega}(ip+m\omega x)$$

So I put 'hats' above of p and x to turn them into operators to get a_ hat. Is that permitted?

$$\widehat{b}_-=\widehat{a}_-+\frac{\varepsilon_2}{\varepsilon_1}=\frac{1}{2hm \omega}(i \widehat{p} + m \omega \widehat{x})+\frac{\varepsilon_2}{\varepsilon_1}$$

And operating this on the ground state wavefunction:

$$(\frac{1}{2hm\omega}(i \widehat{p} +m \omega \widehat{x})+\frac{\varepsilon_2}{\varepsilon_1}) \Psi_0=0$$

Does this mean $$\Psi_0$$ is zero here?

Last edited: Oct 20, 2011
15. Oct 20, 2011

### vela

Staff Emeritus
No, you're looking for a non-trivial solution to that equation.

16. Oct 20, 2011

### v_pino

I got this:

$$\Psi_0=A\exp{\frac{\sqrt{2hm \omega}}{h}(\frac{-m \omega x^2}{2}- \frac{\varepsilon_2}{\varepsilon_1}x)}$$

where A is the normalization constant.

When I evaluate A, I got:

$$A=[2 \sqrt{2} \frac{m \omega}{\sqrt{\pi h}} \exp{- \sqrt{\frac{2}{hm \omega}} (\frac{\varepsilon_2}{\varepsilon_1})^2}]^ \frac{1}{2}$$

Does that look right? It looks a bit messy to me.

17. Oct 20, 2011

### vela

Staff Emeritus
It's probably right. I didn't check the constants. It would be more insightful to write it in the form
$$\Psi'_0(x) = A \exp\left[-\frac{(x-x_0)^2}{2\sigma^2}\right]$$with the appropriate values for x0 and σ. You can see it's just the regular solution shifted by x0, and A won't look so messy.

18. Oct 20, 2011

### v_pino

I found the following SE for SHO:

$$h \omega (a_+a_-+ \frac{1}{2}) \Psi =E \Psi$$

Should I simply replace the a's in the equation with b's or do I have to write a's in terms of b's and sub them back in?

19. Oct 20, 2011

### vela

Staff Emeritus
What are you trying to do?

20. Oct 20, 2011

### v_pino

This is the last part of the problem - to find the energy eigenvalues.