## Main Question or Discussion Point

Is there a simple expression for the ladder operators, in terms of x and $-i\hbar\partial_x$, for the infinite potential well? After some attempts, I couldn't figure out any nice operators that would map functions like this

$$\sin\frac{\pi n x}{L} \mapsto \sin\frac{\pi(n\pm 1)x}{L}.$$

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Hurkyl
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$$\sin\frac{\pi n x}{L} \mapsto \sin\frac{\pi(n\pm 1)x}{L}.$$
What's wrong with just using this to define a pair of operators? Such sinusoids form a basis for the state space, don't they? Is expressing it in terms of those operators a necessity?

If you really need to write things in terms of those operators, it seems straightforward by using the angle addition formulas for trigonometric functions. You'll probably have to use operators like $\sin X$ and $\cos X$, though.

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What's wrong with just using this to define a pair of operators?
I'm not sure.

Such sinusoids form a basis for the state space, don't they?
Yes.

Is expressing it in terms of those operators a necessity?
I'm not sure.

Better to know, than to not know! If some nice form for the operators exists, I have no intention to ignore it.

If you really need to write things in terms of those operators, it seems straightforward by using the angle addition formulas for trigonometric functions. You'll probably have to use operators like $\sin X$ and $\cos X$, though.
It didn't seem straightforward when I tried it.

$$\exp\frac{i\pi n x}{L}\; -\; \exp\frac{-i\pi nx}{L} \quad\mapsto\quad \exp\frac{i\pi n x}{L}\exp\frac{i\pi x}{L} \;-\; \exp\frac{-i\pi nx}{L}\exp\frac{-i\pi x}{L}$$

Multiplying by some function doesn't do this. There is always some problems with cross terms. Derivatives seem problematic, because they bring $\propto n$ factors down from the exponents.

Hurkyl
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Gold Member
Incidentally, does $\partial_x$ even count as an operator here? When applied to a state, it usually produces something that isn't a state!

Better to know, than to not know! If some nice form for the operators exists, I have no intention to ignore it.
Well, by far the nicest form of these operators is given by the expression you wrote! Let $\psi_n$ denote those wavefunctions, and write everything in that basis. Then, the raising operator is simply given by

$$A \psi_n := \psi_{n+1}$$

Incidentally, does $\partial_x$ even count as an operator here? When applied to a state, it usually produces something that isn't a state!
It seems natural to work in the space spanned by

$$\exp\frac{i\pi kx}{L},\quad k\in\mathbb{Z}$$

and consider the physical state space as a subspace. It is allowable to go outside temporarily.

I didn't take a closer look at the infinite matrices yet. It could be I'm returning to this later or soon.

hmhmh.... or is it allowable? I'm not sure really... The insistence on writing everything in terms of x and p is getting strange of course...

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Hurkyl
Staff Emeritus
Gold Member
I want to say that you can write the "Number" operator as

$$N = \sqrt{ \partial_x^2 }$$

where the square root one is the one induced by the real number square root. (And this would solve your problems with n appearing) Unfortunately, the reasonability of this expression is on the boundary between my knowledge and my optimism.

reilly
The sines are complete, in configuration space. Further they are eigenstates of the free Hamiltonian. When in doubt, brute force is sometime a good way to go. Lets call one of these eigenstates | S,n> so that <x|S, n> =sin (n pi x/L). Your ladder operator for going up then becomes

Sum over n{|S,n+1><S,n|} -- i've neglected constants

A similar expression holds for the n-> n-1 operator, which is adjoint to the above operator..

You also might be able to do this by converting to an oscillator basis; there's a standard generating function for Hermite polynomials that might do the trick.

Regards,
Reilly Atkinson

reilly
Hurykyl's Number Operator

I want to say that you can write the "Number" operator as

$$N = \sqrt{ \partial_x^2 }$$

where the square root one is the one induced by the real number square root. (And this would solve your problems with n appearing) Unfortunately, the reasonability of this expression is on the boundary between my knowledge and my optimism.
Your operator shows up, for example, in classical E&M. The standard approach is to go to a Fourier representation, and end up with deleta functions and principle parts. Discussed in Mandel and Wolf, Optical Coherence ...., page 223.
Regards,
Reilly Atkinson