Why does this Bra Ket (with creation and annihilation operators) equal zero?

[unknownuser9]

1. Explain why <n|(a-a+)^3|n> must be zero



2. a and a+ (a dagger) are the raising and lowering operators (creation and annihilation operators).



3. Because it says explain, I am not sure any mathematical proof is needed. I am best answer is that because (ignoring that the bracket has been raised to the power 3) a+ increases the value of n by 1 and a decreases it by 1, the overall operation on n will be zero leaving <n|n>. Because n (bra) and n (ket) must be different for a transition intensity to be observed, the overlap integral of <n|n> equals zero thus the entire thing is zero. Am i on the right tracks?

Im pretty stuck so any help would be useful. Thanks in advance!

 

[fzero]

You've got it backwards. The states $$|n\rangle$$ should be an orthonormal basis, so

$$\langle n'|n\rangle = \delta_{nn'}$$

and $$\langle n|n\rangle = 1$$.

The fact that the power, 3, is odd is important. Think about the kind of terms that appear in the expansion.

 

[unknownuser9]

Firstly, thank you for the reply.

Secondly, does that mean, being orthonormal, <n'|n> = 0?

The kind of terms? All i can think is that they are operators and operate on n to raise or lower is as such:

a|n> = n^1/2|n-1>

a+|n> = (n+1)^1/2|n+1>

The expansion? By expanding the bracket, i get aaa + a+a+a - a+a+a+ - ...etc

Sorry, I'm still confused. Thanks

 

[fzero]

Firstly, thank you for the reply.

Secondly, does that mean, being orthonormal, <n'|n> = 0?

Orthonormal in general means the vectors in a basis for a vector space have zero inner product with each other (so they are orthogonal) and unit product with themselves (unit norm, or normalized). So in this case it's precisely the equation I wrote down.

The kind of terms? All i can think is that they are operators and operate on n to raise or lower is as such:

a|n> = n^1/2|n-1>

a+|n> = (n+1)^1/2|n+1>

The expansion? By expanding the bracket, i get aaa + a+a+a - a+a+a+ - ...etc

Sorry, I'm still confused. Thanks

Since you're still having problems, try to compute $$\langle n|a a^\dagger a^\dagger|n\rangle$$ explicitly. Does it suggest anything about the expectation value of the rest of the terms in the expansion?

 

[unknownuser9]

I have solved the question you gave and got n^1/2.n+1<n|n+1>. (Sorry if this is wrong but I am trying!)

Thinking about it, after expanding and solving all the terms (aaa + a+a+a - ...), it should cancel to zero?

 

[fzero]

You need to be more careful when computing the action of the operators, since your coefficient is wrong. Each operator acts in turn, so

$$\langle n|a a^\dagger a^\dagger|n\rangle = \langle n|a a^\dagger (a^\dagger|n\rangle) = \sqrt{n+1} \langle n|a a^\dagger |n+1\rangle$$

The next operation will bring in a factor of $$\sqrt{n+2}$$.

Assuming that you can go back and get the factors right, what can you say about the value of $$\langle n|n+1\rangle$$?

 

[unknownuser9]

Oh i see so when the second a+ operates, the factor becomes $$\sqrt{n+2}$$ and not $$\sqrt{n+1}$$.$$\sqrt{n+1}$$ = n+1 (it is not multiplied but instead increases the number that is added to n).

<n|n+1> would give an answer of zero therefore making the entire Dirac bra-ket and the factors zero?

This would give (after expansion of the bracket) 0 + 0 + 0 ...?

Thanks

 

[unknownuser9]

I think the penny has just dropped! I understand why the power being odd is important now!

Thanks for your help.