# Raising and lowering operators

1. May 17, 2008

### t_n_p

[SOLVED] raising and lowering operators

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

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

3. 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?

Last edited by a moderator: May 3, 2017
2. May 18, 2008

### cristo

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.

Last edited: May 18, 2008
3. May 18, 2008

### 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]

Last edited by a moderator: May 3, 2017
4. May 18, 2008

### malawi_glenn

yes, you do wrong

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

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

5. May 18, 2008

### cristo

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.

6. May 18, 2008

### t_n_p

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

I will try again

7. May 18, 2008

### 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.

Last edited by a moderator: May 3, 2017
8. May 18, 2008

### cristo

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.

Last edited: May 18, 2008
9. May 18, 2008

### 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

10. May 18, 2008

### cristo

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.

Last edited: May 18, 2008
11. May 18, 2008

### 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 thats 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.

12. May 18, 2008

### cristo

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?

13. May 18, 2008

### 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))

14. May 18, 2008

### cristo

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

15. May 18, 2008

### t_n_p

now this is where the schrodinger equation comes in?

16. May 18, 2008

### cristo

Staff Emeritus
Yes.

17. May 18, 2008

### t_n_p

Got it.

Thanks

18. May 18, 2008

### cristo

Staff Emeritus
Well done; you're welcome.

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