Regarding Ladder Operators and Quantum Harmonic Oscillators

Question

When dealing with the Quantum Harmonic Oscillator Operator $H=-\frac{d^{2}}{dx^{2}}+x^{2}$, there is the approach of using the Ladder Operator:

Suppose that are two operators $L^{+}$ and $L^{-}$ and define $f_{n}$ such that

$$L^{+}(f_{n})=\sqrt{n+1}f_{n+1}$$

$$L^{-}(f_{n})=\sqrt{n}f_{n-1}$$

then $$L^{+}L^{-}(f_{n})=L^{+}(\sqrt{n}f_{n-1})=nf_{n}$$ $$L^{-}L^{+}(f_{n})=L^{-}(\sqrt{n+1}f_{n+1})=(n+1)f_{n}$$

Thus

$$L^{+}L^{-}+L^{-}L^{+}=(2n+1)f_{n}$$

Now since $$H=-\frac{d^{2}}{dx^{2}}+x^{2}=\left(x+\frac{d}{dx}\right)\left(x-\frac{d}{dx}\right)$$

$$=\frac{1}{2}\left[\left(x+\frac{d}{dx}\right)\left(x-\frac{d}{dx}\right)+\left(x+\frac{d}{dx}\right)\left(x-\frac{d}{dx}\right)\right]$$

$$=\frac{1}{\sqrt{2}}\left(x+\frac{d}{dx}\right)\frac{1}{\sqrt{2}}\left(x-\frac{d}{dx}\right)+\frac{1}{\sqrt{2}}\left(x-\frac{d}{dx}\right)\frac{1}{\sqrt{2}}\left(x+\frac{d}{dx}\right)$$

If we let $$L^{+}=\frac{1}{\sqrt{2}}\left(x+\frac{d}{dx}\right)\quad\text{and}\quad L^{-}=\frac{1}{\sqrt{2}}\left(x-\frac{d}{dx}\right)$$

then $$H=L^{+}L^{-}+L^{-}L^{+}$$

Since $$L^{+}(f_{n})=\sqrt{n+1}f_{n+1}\implies$$

$$f_{n}=\sqrt{n!}L^{n}(f_{0})$$

Thus if we can a $f_{0}$ such that

$$f_{n}=\sqrt{n!}L^{n}(f_{0})\implies L^{-}(f_{n})=\sqrt{n}f_{n-1}\tag{1}$$ would be satisfied, then we have found a set of eigenfunctions for H, which is $f_{n}$ and the corresponding eigenvalue is 2n+1.

Now my instructor says that if we let $f_{0}$ be in the kernel of $L^-$, meaning that:

$$L^{-1}f_{0}=0\tag{2},$$ then (1) would be true.

Specifically, this means that

$$f_{0}(x)=e^{-\frac{1}{2}x^{2}}.$$

As it is already verified (and commonly used in quantum mechanics) that this $f_{0}(x)$ does satisfy (1). But I can't really see how does (2) leads to (1). Is this choice $f_0$ really a lucky guess or actually a choice the generally works for all operators that can be expressed as $H=L^{+}L^{-}+L^{-}L^{+}$? A different way of asking this question is: are the ladder operator really just operators that tells us how to get from one state to another, or it is actually something that would be used as a general method of solving some differential equations?

Answer 1

I think the confusion begins where you write "and define $f_n$ such that" at the very beginning. The two lines after that are two ways of defining $f_n$, and part of what needs to be shown is that they're equivalent.

Also you seem to have got $L^\pm$ mixed up -- the one that contains the plus sign is the annihilation operator and the one that contains the minus sign is the creation operator, as you can tell from the fact that your $f_0$ is actually in the kernel of what you called $L^+$. I'll use your definitions so as not to confuse things further, but keep in mind that that means that $L^+$ is now the annihilation operator and $L^-$ is now the creation operator.

I'd proceed as follows. The commutator of the ladder operators is

$$[L^+,L^-]=1\;.$$

If $f_n$ is an eigenfunction of $H$ with eigenvalue $E_n$, then $L^-f_n$ is an eigenfunction of $H$ with eigenvalue $E_n+2$:

$$ \begin{eqnarray} HL^-f_n&=&(L^+L^-+L^-L^+)L^-f_n \\ &=& (L^+L^-L^-+L^-L^+L^-)f_n \\ &=& (L^-L^+L^-+L^-+L^-L^-L^++L^-)f_n \\ &=& L^-(L^+L^-+1+L^-L^++1)f_n \\ &=& L^-(H+2)f_n \\ &=& L^-(E_n+2)f_n \\ &=& (E_n+2)L^-f_n\;. \end{eqnarray} $$

In the same way we can show that $L^+f_n$ is an eigenfunction of $H$ with eigenvalue $E_n-2$.

Now the eigenvalues of $H$ are non-negative. (You can show this for instance by taking the expectation value of $H$ and integrating by parts.) Thus successively applying $L^+$ to lower the eigenvalue must at some point annihilate the eigenfunction. Thus we've shown a) that any eigenfunction gives rise to an entire succession of eigenfunctions with equally spaced eigenvalues and b) this succession must end with a function that is annihilated by $L^+$, and this yields a differential equation whose solution is your $f_0$.

Then it only remains to normalize the resulting eigenfunctions, and to derive the non-recursive forms with the factorial factors becomes a straightforward exercise in induction.

Answer 2

I would say everything you need for (1) is given in (2): From $L^-$ you get $L^+$ and $f_0$ is given by (2).

If it's possible to express your Hamiltonian as ladder operators, this approach would also help to simplify the way to get the solution of your differential equation. They are for example also used in (finite dimensional) spin physics.

Concerning your question, if $f_0$ is just a lucky guess, you can check here, that the Hermite function $f_n$ fulfill the recursion relations $$ x\;f_{n}(x) = \sqrt{\frac{n+1}{2}}f_{n+1}(x) + \sqrt{\frac{n}{2}}f_{n-1}(x) $$ and $$ f_n'(x) = -\sqrt{\frac{n+1}{2}}f_{n+1}(x) + \sqrt{\frac{n}{2}}f_{n-1}(x) . $$ and

Together with definition of $L^-=\frac{1}{\sqrt{2}}(x+\frac{d}{dx})$, you see that $L^- f_0$ exactly cancels to $0$.