0177-94

Newton
Posts: 20
Joined: Tue Nov 06, 2012 6:34 am

0177-94

Postby Newton » Thu Nov 08, 2012 11:34 am

Can someone explain the second to last step in this problem:

http://www.physicsgrad.com/pgre/0177-94


No sure how to compose the resulting (2n+1)V. Thanks!!

dau
Posts: 19
Joined: Thu Nov 01, 2012 10:36 am

Re: 0177-94

Postby dau » Thu Nov 08, 2012 1:11 pm

Newton wrote:Can someone explain the second to last step in this problem:

http://www.physicsgrad.com/pgre/0177-94

No sure how to compose the resulting (2n+1)V. Thanks!!

From the definition of a and a^\dag you can show that [a,a^\dag]=1.
This relation is used in the step you mention.

Best regards.

blighter
Posts: 256
Joined: Thu Jan 26, 2012 6:30 pm

Re: 0177-94

Postby blighter » Thu Nov 08, 2012 1:27 pm

dau wrote:
Newton wrote:Can someone explain the second to last step in this problem:

http://www.physicsgrad.com/pgre/0177-94

No sure how to compose the resulting (2n+1)V. Thanks!!

From the definition of a and a^\dag you can show that [a,a^\dag]=1.
This relation is used in the step you mention.

Best regards.


I don't see why you'd need to use that relation. It's simply the application of the operators.

aa^\dag|n\rangle=a(a^\dag|n\rangle)=a(\sqrt{n+1}|n+1\rangle)=\sqrt{n+1}(a|n+1\rangle)=\sqrt{n+1}(\sqrt{n+1}|n\rangle)=(n+1)|n\rangle

Similarly,

a^\dag a|n\rangle=n|n\rangle

Adding the two you get the result.

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randomnick7
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Joined: Sat Sep 01, 2012 6:26 pm

Re: 0177-94

Postby randomnick7 » Thu Nov 08, 2012 1:35 pm

It's easier with the commutation relation as dau pointed out, you must know that a^\dagger a=\hat N is the number operator and voilà.

blighter
Posts: 256
Joined: Thu Jan 26, 2012 6:30 pm

Re: 0177-94

Postby blighter » Thu Nov 08, 2012 2:12 pm

Right, of course.

Newton
Posts: 20
Joined: Tue Nov 06, 2012 6:34 am

Re: 0177-94

Postby Newton » Thu Nov 08, 2012 2:19 pm

Thanks for the responses. I only took QM up to what was needed for engineering. We didn't do anything with operators or the like. So I'm having trouble following what you did. From studying the ETS exams, I just know a couple of commuter relations. Is there a method for all the adjustments of the format and substituting? Or is it just the way its done? Like how does the |n+1> and |n-1> come into play?

Forever grateful :)

dau
Posts: 19
Joined: Thu Nov 01, 2012 10:36 am

Re: 0177-94

Postby dau » Thu Nov 08, 2012 4:18 pm

Newton wrote:Thanks for the responses. I only took QM up to what was needed for engineering. We didn't do anything with operators or the like. So I'm having trouble following what you did. From studying the ETS exams, I just know a couple of commuter relations. Is there a method for all the adjustments of the format and substituting? Or is it just the way its done? Like how does the |n+1> and |n-1> come into play?

Forever grateful :)

The |n+1> and |n-1> come into play from the definitions of a and a^\dag given in the problem. Even if you don't have time to understand the proof of these relations, you can just use the rules and operate with them. You need to know, for example, that <n|m> = \delta_{nm}, and that a^\dag a |n> = n|n> (because of this, the combination a^\dag a is called the "number" operator). You can show the latter relation from the definition of a and a^\dag given in the problem.

Newton
Posts: 20
Joined: Tue Nov 06, 2012 6:34 am

Re: 0177-94

Postby Newton » Fri Nov 09, 2012 12:30 am

Hmm. K. Sorta making more sense now. So we have the Kronecker delta, so is this formulation is similar to orthogonality? I read that the product of aa^{\dagger} is somewhat like taking the product of a complex number and its conjugate... :? What would be these rules? In blighter solution, taking aa^{\dagger} and a^{\dagger}a yield different results?

dau
Posts: 19
Joined: Thu Nov 01, 2012 10:36 am

Re: 0177-94

Postby dau » Fri Nov 09, 2012 12:08 pm

Newton wrote:Hmm. K. Sorta making more sense now. So we have the Kronecker delta, so is this formulation is similar to orthogonality? I read that the product of aa^{\dagger} is somewhat like taking the product of a complex number and its conjugate... :? What would be these rules? In blighter solution, taking aa^{\dagger} and a^{\dagger}a yield different results?

Yes it is, the relation with the Kronecker delta is an orthonormality relation for the states |n>.
But aa^{\dagger} is just an operator in this space spanned by the |n>, not the scalar product in this space. The scalar product is just <n|m>. In general the product of operators does not commute, so aa^{\dagger} is different than a^{\dagger}a, and that's why commutators are important. The following links may be useful:
http://en.wikipedia.org/wiki/Bra-ket_notation#Inner_products_and_bras
http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Ladder_operator_method

Newton
Posts: 20
Joined: Tue Nov 06, 2012 6:34 am

Re: 0177-94

Postby Newton » Sat Nov 10, 2012 2:26 pm

Thanks for the help dau and everyone else :)

walczyk
Posts: 35
Joined: Fri Jan 11, 2013 5:47 pm

Re: 0177-94

Postby walczyk » Tue Jan 15, 2013 12:06 pm

I think I see where some of the confusion lies. You should know off the bat that orthogonality means bra-kets satisfy the kronecker delta, so when we do this V(a+a^{\dagger})^2 |n\rangle=V(aa+aa^{\dagger}+a^{\dagger}a+a^{\dagger}a^{\dagger})|n\rangle we need to automatically recognize that only the middle two terms contribute because they raise AND lower the ket(wavefunction) so that it ends up back at n. Then when you do the math you have to remember that when you raise n by one then you'll be lowering n+1 by one after, and similarly when you lower n by one you will then raise n-1 by 1 after, so make sure your square roots are all correct (try to do this quickly).

walczyk
Posts: 35
Joined: Fri Jan 11, 2013 5:47 pm

Re: 0177-94

Postby walczyk » Tue Jan 15, 2013 12:20 pm

blighter wrote:
dau wrote:
Newton wrote:Can someone explain the second to last step in this problem:

http://www.physicsgrad.com/pgre/0177-94

No sure how to compose the resulting (2n+1)V. Thanks!!

From the definition of a and a^\dag you can show that [a,a^\dag]=1.
This relation is used in the step you mention.

Best regards.


I don't see why you'd need to use that relation. It's simply the application of the operators.

aa^\dag|n\rangle=a(a^\dag|n\rangle)=a(\sqrt{n+1}|n+1\rangle)=\sqrt{n+1}(a|n+1\rangle)=\sqrt{n+1}(\sqrt{n+1}|n\rangle)=(n+1)|n\rangle

Similarly,

a^\dag a|n\rangle=n|n\rangle

Adding the two you get the result.


I think the cleverness of Dau's answer is that when you write out the commutator
aa^\dag-a^\dag a = 1 so aa^\dag+a^\dag a = 1+2a^\dag a
which you can substitute into the potential. It would be good if you already knew it, but you can just as quickly solve both operators. You can also do the same thing with
[a^\dag,a]=-1




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