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How can we show that the eigenvalues of the number operator: $a^Ha$ are indeed $n$. ie: $$a^Ha |n \rangle = n | n \rangle$$

Similarly, how can we show that: $ a|n\rangle = \sqrt{n}| n-1\rangle$ and $ a^H|n\rangle = \sqrt{n+1}| n+1\rangle$

This question bugged me when I was first learned about raising/lowering operators.

Qmechanic
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1 Answers1

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Starting from:

  1. $a| n \rangle = c_1(n) | n-1\rangle$ and $a^H | n \rangle = c_2(n)| n+1\rangle$
  2. $ a| 1 \rangle = |0 \rangle $
  3. $ [a, a^H] = 1 $

We have: $ \langle n | a^Ha | n \rangle = c_1(n)c_2(n-1) = c_1(n)c_1(n)$

So: $ c_2(n) = c_1(n+1) $.

And:

$ \langle n | aa^H | n \rangle = \langle n |(1 + a^Ha)| n \rangle$

$ c_2(n)^2 = 1 + c_1(n)^2 $

$ c_1(n +1)^2 = 1 + c_1(n)^2 $

From: $ c_1(0) = 0 $ we get the pattern:

$$c_1(1)^2 = 1, \ \ c_2(2)^2 = 2, \ \ \text{etc.}$$

So: $$ c_1(n) = \sqrt{n} \ \ \text{and} \ \ c_2(n) = \sqrt{n+1} \ \ \text{and} \ \ a^Ha | n\rangle = n| n\rangle$$

Dedados
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