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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?

John M
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3 Answers3

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  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.

joshphysics
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4

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.

nijankowski
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2

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 $n=0$

Joel Klassen
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