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My apologies if this question has already been answered. Using the notation of Sakurai, we have an energy eigenket N and an eigenvalue n.

$$N | n \rangle = n | n \rangle$$

For the number operator $N = a^\dagger a $. My question is precisely why does n have to be an integer. I do understand that it must be positive definite due to the norm condition on $ a | n \rangle $:

$$\langle n | a^\dagger a | n \rangle \geq 0$$

I can see that we can show:

$$ a | n \rangle = \sqrt{n-1} | n-1 \rangle $$ and similarly for $a^\dagger | n \rangle$

Sakurai argues that the sequential operation of $a$ operators leads to the form which he does not write, but I will write here for clarity:

$$ a^k | n \rangle = \left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right) | n - k + 1 \rangle $$

Since

$$ a|0 \rangle = 0 $$

the sequence must terminate, but can only terminate if n is an integer. How can we make that jump? Is there a formal mathematical proof that the sequence $\left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right)$ will terminate if and only if n is an integer?

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You should not say "energy eigenvalue of the number operator", the energy eigenvalue is associated with the energy operator. You just want "eigenvalue of the number operator" of "number eigenvalue". –  dmckee Feb 22 '13 at 1:03
    
Your equation with $a^k |n>=\cdots$ is incorrect. $a$ lowers the number. So you start with some number then subtract a bunch of integers and eventually get zero, what does that say about what you started with...? –  Michael Brown Feb 22 '13 at 1:10
    
Related: physics.stackexchange.com/q/23028/2451 –  Qmechanic Feb 22 '13 at 1:30
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3 Answers

up vote 8 down vote accepted
  1. Show that if $|\nu\rangle$ is an eigenvector of $N$ with eigenvalue $\nu$, and if $a|\nu\rangle\neq 0$, then $a|\nu\rangle$ is an eigenvector of $N$ with eigenvalue $\nu-1$.

  2. Convince yourself that the spectrum of the number operator is non-negative.

Assume, by way of contradiction, that there is some non-integer eigenvalue $\nu^*>0$ and let $m$ denote the smallest integer larger than $\nu^*$. Use property 1 repeatedly ($m$-times) to show that $a^m|\nu^*\rangle$ is an eigenvector of $N$ with eigenvalue $\nu^*-m <0$. This is a contradiction QED.

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Is this reproduced in a textbook somewhere? I feel like I've seen it before. –  Chay Paterson Apr 27 '13 at 20:03
    
@ChayPaterson I'm not entirely sure, but volume 1 chapter 5 of Cohen-Tannoudji has a pretty logically detailed analysis of the spectrum of the number operator. I'd look there first. –  joshphysics Apr 27 '13 at 20:35
    
After reading around, I think (2.) can be made a bit stronger. If $\omega > 0$ , and the Hamiltonian is positive definite in the classical limit that $\hbar \rightarrow 0$, then N must be positive definite. Also, the decomposition into a and a* is essentially a Cholesky decomposition, which can only be done for positive definite operators. –  Chay Paterson Apr 30 '13 at 12:14
    
@ChayPaterson Hmm interesting. What does it mean for the Hamiltonian to be positive definite in the context you mention? I ask because the spectrum of the number operator for the quantum harmonic oscillator contains $0$... –  joshphysics May 29 '13 at 2:01
    
I just checked, and in fact the existence condition for the Cholesky decomposition is that the operator is positive semi definite. This is a weaker requirement, namely that $\langle x| H |x\rangle \geq 0$ for any $|x\rangle$. –  Chay Paterson May 29 '13 at 12:24
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The critical condition here is that the expectation value of the number operator must always be positive $n = \langle n | N | n \rangle \geq 0$

If we consider that the operation of the lowering operator $a$ is to lower $|n\rangle$ by an integer value 1, ie:

$$a |n \rangle = \sqrt{n} |n-1 \rangle$$

we can see that if $n$ were not an integer, then the lowering operator could be applied some number of times such that we could achieve some $|n \rangle$ where $n = \langle n | N | n \rangle \lt 0$

More explicitly, consider $|2.5 \rangle$, $a^2|2.5\rangle=\sqrt{2.5}a|1.5\rangle=\sqrt{2.5}\sqrt{1.5} |-.5\rangle$. This is not allowed, you must have an integer $n$ so that after some finite number of applications of the lowering operator you eventually land precisely on $n=0$

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You can get an answer on why n is an integer, if you solve the quantum harmonic oscillator problem using the power series method. Starting from the Schrödinger eq. with a harmonic oscillator potential, you make a substitution to a dimensionless variable $y=\sqrt\frac{m\omega}{\hbar}x$ and this results in

$\frac{d^2\psi(y)}{dy^2}+(\frac{2E}{\hbar\omega}-y^2)\psi(y)=0$

where its general solution is

$\psi(y)=u(y)e^{\frac{-y^2}{2}}$

Now you have to insert this solution in the equation above and solve for $u(y)$. After some calculation you get

$\frac{d^2u(y)}{dy^2}-2y\frac{du(y)}{dy}+(\frac{2E}{\hbar\omega}-1)u(y)=0$

Now you have to use a power series of $y$ as a general solution for the above equation, here $u(y)$ takes the form

$u(y)=\sum_{n=0}^\infty\alpha_n y^n$

Before inserting this series into the equation above, you have to solve the first and second derivaties of $u(y)$. After some long but easy calculation you will obtain the following

$\sum_{n=0}^\infty[(n+2)(n+1)\alpha_{n+2}+(\frac{2E}{\hbar\omega}-1-2n)\alpha_n]y^{2n}=0$

this can be solved for $\alpha_{n+2}$ because the coefficient for each power of $y$ must equal to zero. This leads you to

$\alpha_{n+2}=\frac{2n+1-\frac{2E}{\hbar\omega}}{(n+2)(n+1)}\alpha_n$

From this you can see that for large values of y, n is very large as well and the ratio of $\alpha_{n+1}$ to $\alpha_n$ is very close to $2/n$. This is a problem, because in the limit, 2/n grows faster than the exponential term in $\psi(y)$. Hence, the series must terminate in order for the solution to have physical meaning. And one way is to set the numerator in the series above equal to zero. which will ultimately give

$E=(n+\frac{1}{2})\hbar\omega$

So, n must be an integer because the energy should be quantized and to find a physical solution.

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