# Raising and lowering operators

t_n_p
[SOLVED] raising and lowering operators

## Homework Statement

http://img125.imageshack.us/img125/2923/85098487ch9.jpg [Broken]

## The Attempt at a Solution

I expand a+ and a-, introduce the wavefunction and then substitute the values given at the very end to give:

http://img187.imageshack.us/img187/8224/37075232fm3.jpg [Broken]

This is where I'm stuck. I can't see how epsilon = 2mE/hbar comes into the equation. Are my initial calculations even correct?

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Staff Emeritus
No, I don't think your calculations are right. You are dealing with operators, so you've missed a term when expanding out. What you should do is first apply a- to psi, then apply a+ to that (you'll see any extra term drops out by the product rule).

In order to put the equation in terms of epsilon, I would use the Schroedinger equation that you're given in the question.

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t_n_p
i get the same thing if i apply a- to psi and then a+ to that. Am I still doing something wrong?

http://img237.imageshack.us/img237/719/11483685tq9.jpg [Broken]

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Homework Helper
Gold Member
2022 Award
yes, you do wrong

$$A_-\psi (x) = \frac{d\psi}{dx}+ \tanh (x)\psi (x)$$

then $$\frac{d}{dx}(\frac{d\psi}{dx}+ \tanh (x)\psi (x)) = \text{??}$$

Staff Emeritus
You're not applying the operator a- to psi properly:

$$\hat{A}_-\psi(x)=\left(\frac{d}{dx}+\tanh x\right)\psi=\frac{d\psi}{dx}+\tanh x\cdot\psi$$

Now apply a+ to this.

t_n_p
oh, for some reason I was under the impression the wavefuction is not performed on tanh(x).

I will try again

t_n_p
Ok, cleared that up

http://img409.imageshack.us/img409/5826/67746695rh8.jpg [Broken]

Now I can see from the Schrodinger equation that two of my terms appear similar. I'm not clear as to where to progress from here.

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Staff Emeritus
I think you should look at your last line again. I agree with your first two terms, but not the others. Remember that you are expanding:

$$\left(-\frac{d}{dx}+\tanh x \right)\left(\frac{d\psi}{dx}+\tanh x\cdot\psi\right)$$.

Do this term by term, and then simplify after; it reduces the chance of making mistakes. You should get some cancellations.

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t_n_p
Cristo, in my notes it says this..

"In obtaining this result it is importatnt to note that we are dealing with operators, and hence the order in which we operate is critical, whence: (A + B)(C + D) = AC + BC + AD + BD

Staff Emeritus
Ok, so if you expand the brackets according to that rule you get

$$-\frac{d}{dx}\left(\frac{d\psi}{dx}\right)+\tanh x\cdot\frac{d\psi}{dx}-\frac{d}{dx}\left(\tanh x\cdot\psi\right)+\tanh x\cdot\tanh x\cdot \psi \hskip3cm (1)$$

Can you simplify this? Note that the brackets in the third term is a product.

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t_n_p
Because of the order of the expansion, I am not 100% sure if the second and third terms can cancel. I think that's what you are hinting at, but I have heard conflicting ideas.

Also,two hints are provided:
1) d/dx (tanhx) = sech^2(x)
2) tanh^2(x) - sech^2(x) = 1-2sech^2(x)

Using these two, I substituted to give what is shown in post #7.

Staff Emeritus
The second and third terms will not cancel entirely! The third term is a product; what would the solution to this be:

$$\frac{d}{dx}(u\cdot v)$$ where u and v are functions of x?

t_n_p
using product rule, evaluates to

(d/dx)(tanh(x)(ψ) + (dψ/dx)(tanh(x))

using the equation given above, will yield
ψsech^2(x) + (dψ/dx)(tanh(x))

Staff Emeritus
using product rule, evaluates to

(d/dx)(tanh(x)(ψ) + (dψ/dx)(tanh(x))

using the equation given above, will yield
ψsech^2(x) + (dψ/dx)(tanh(x))

Good. Now plug that into Eq. (1) (post #10)

t_n_p
now this is where the schrodinger equation comes in?

Staff Emeritus
now this is where the schrodinger equation comes in?

Yes.

t_n_p
Got it.

Thanks

Staff Emeritus