0177-94

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

0177-94

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

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

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

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

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

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.

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

Re: 0177-94

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

Right, of course.

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

Re: 0177-94

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

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

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

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_no ... s_and_bras
http://en.wikipedia.org/wiki/Quantum_ha ... tor_method

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

Re: 0177-94

Thanks for the help dau and everyone else

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

Re: 0177-94

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

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

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
$$[a^\dag,a]=-1$$