How to Approach Raising and Lowering Operators in Homework?

[v_pino]

Homework Statement

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.

Homework Equations

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?

 

[grzz]

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

 

[v_pino]

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

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

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

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

 

[grzz]

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

 

[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-].

 

[vela]

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

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

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.

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

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

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.

 

[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

 

[vela]

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.

 

[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.

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


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

 

[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.)

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

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

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

 

[vela]

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.

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

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 $\beta$.

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.)

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

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

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

Yes, those look fine.

 

[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.

 

[vela]

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.

 

[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:

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

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

 

[vela]

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

 

[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.

 

[vela]

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.

 

[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?

Thank you for your replies.

 

[vela]

What are you trying to do?

 

[v_pino]

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

 

[vela]

You have to use the Hamiltonian given to you in this problem.

 

[v_pino]

I think I got E_0:

$$E_0=- \frac{\varepsilon_2^2}{\varepsilon_1}$$

How do I turn this into E_n?

I know the following:

$$\Psi_n (x) = A_n (a_+)^n \Psi_0(x)$$

A_n is a normalization constant for this equation.