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Raising and lowering operators

  • Thread starter t_n_p
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  • #1
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[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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Answers and Replies

  • #2
cristo
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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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  • #3
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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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  • #4
malawi_glenn
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yes, you do wrong

[tex] A_-\psi (x) = \frac{d\psi}{dx}+ \tanh (x)\psi (x) [/tex]

then [tex] \frac{d}{dx}(\frac{d\psi}{dx}+ \tanh (x)\psi (x)) = \text{??} [/tex]
 
  • #5
cristo
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You're not applying the operator a- to psi properly:

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

Now apply a+ to this.
 
  • #6
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oh, for some reason I was under the impression the wavefuction is not performed on tanh(x).

I will try again
 
  • #7
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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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  • #8
cristo
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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:

[tex]\left(-\frac{d}{dx}+\tanh x \right)\left(\frac{d\psi}{dx}+\tanh x\cdot\psi\right) [/tex].

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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  • #9
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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
cristo
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Ok, so if you expand the brackets according to that rule you get

[tex]-\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)[/tex]

Can you simplify this? Note that the brackets in the third term is a product.
 
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  • #11
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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
cristo
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The second and third terms will not cancel entirely! The third term is a product; what would the solution to this be:

[tex]\frac{d}{dx}(u\cdot v)[/tex] where u and v are functions of x?
 
  • #13
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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
cristo
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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)
 
  • #15
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now this is where the schrodinger equation comes in?
 
  • #16
cristo
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now this is where the schrodinger equation comes in?
Yes.
 
  • #17
595
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Got it.

Thanks
 
  • #18
cristo
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Well done; you're welcome.
 

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