# Calculate commutator

1. Mar 2, 2014

### LagrangeEuler

In infinitely potential well problem $V(x)$ is zero inside the box and $\infty$ outside the box. Is it possible to calculate commutator
$[V(x),p]$?
If case that this commutator is zero $\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$ will also be eigenstate of $p$. I am confused with this.

2. Mar 2, 2014

### ChrisVer

the commutator for the space where you define the solutions is zero, because V(x) is independent of x (everywhere zero). And of course the solutions are eigenstates of p, because there you have:
$H=\frac{p^{2}}{2m}$
so the eigenstates of energy are also eigenstates of $p^{2}$

3. Mar 3, 2014

### LagrangeEuler

Well if this is eigenfunction of $p^2$ it is also eigenfunction of $p$. Right? I think you are wrong. When you differentiate sine you get cosine and that functions are linearly dependent.

4. Mar 3, 2014

### hilbert2

An infinite square well that is centered at the origin and has width $L$ can be approximated with the potential

$V(x)=\left(\frac{2x}{L}\right)^{2n}$, where $n$ is a large integer. You can calculate the commutator of this function with $p$ and then take the limit $n\rightarrow\infty$.

The commutation relation $[p,x^{n}]=-i\hbar n x^{n-1}$ might be useful.

5. Mar 3, 2014

### ChrisVer

Well that's still an approximation....and a very "weird" one, considering the fact that it makes the whole problem more complicated.

On the other hand, LagrangeEuler, what do you mean "when you differentiate sine"? cos and sin are linearly independent...
The eigenstates of $p^{2}$ and the states of $p$ are not ALL the same... that you can see from the spectrum of their eigenvalues- the squared has eigenvalues from 0 to infinity, while the unsquared has negative eigenvalues too... that's why half the solutions don't count...(cos is missing)
But the commutator relation isn't telling you that all the eigenfunctions are the same- but the eigenfunctions of the one quantity are contained in the other...

6. Mar 3, 2014

7. Mar 3, 2014

### micromass

Staff Emeritus
You're gonna have domain issues. The operator $P$ is an unbounded operator, and thus we cannot take $P$ of every possible function. We need to be careful that the function is actually in the domain. Here, the function you mention will not be in the domain, so it can't be an eigenstate.

Similarly, the commutator $[V(X),P]$ will have a restricted domain. So it is an operator that is undefined on the function $\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$.

It is true that if $\psi$ is an eigenstate and if it is in the domain of the commutator (which is not the case here) and if the commutator is $0$, then it will be an eigenstate of $P$.

It is also for domain issues that the definition of "commuting" of two operators is much more annoying. We cannot just say that $A$ and $B$ commute if $[A,B]=0$ since the domain of $[A,B]$ might be zero. It is a very touchy business.

8. Mar 3, 2014

### hilbert2

The infinite square well potential is not differentiable, so you can't calculate it's commutator with $p$. Therefore, we have to consider a sequence of functions $V_{n}$ that are differentiable and approach the infinite well potential when $n\rightarrow\infty$, and see what happens to their commutator with $p$ when $n\rightarrow\infty$.

Here's the plot of the function $V(x)=x^{50}$:

Looks a lot like the infinite square well potential with width 2, doesn't it? You can also easily calculate its commutator with $p$ and see what that looks like.

Last edited: Mar 3, 2014
9. Mar 3, 2014

### LagrangeEuler

It is extremely hard to understand. Thank you all for discussion. hilbert2 I'm not sure to understand how you get this shape?

10. Mar 3, 2014

### ChrisVer

That's interesting though... Although it's still confusing for me too...
I mean that for the space defined for the potential, you get what I said due to:
$H=p^{2}/2m$
So in fact you get eigenstates of the hamiltonian (the sines) to be the same as the squared momentum...

Last edited by a moderator: May 6, 2017
11. Mar 3, 2014

### strangerep

Well,... not exactly,... the boundary conditions make a difference.

[Edit: Deleted previous Hamiltonian, since not adequate to the situation, as pointed out by ChrisVer in next post.]

The reason why the usual momentum operator $-id_x$ doesn't work very well in these situations is that it is not a good element (generator) in the dynamical algebra for this problem. It's tempting to think that one can just take the lessons from basic Galilean invariance and apply them elsewhere as-is, but one must also consider the phase space of the system. In the ordinary Galilean case, one has the translation generator $-id_x$ which, when exponentiated, can translate you anywhere in position space. But here, the position part of the phase space is restricted. Hence the exponentiated translation elements do not form a good 1-parameter Lie group on the phase space (meaning that it fails to keep us inside the phase space of the system).

BTW, Hendrik: thanks for mentioning that other thread (which I had not previously read).

Last edited: Mar 4, 2014
12. Mar 3, 2014

### ChrisVer

May I attempt to ask then something?
What is the problem in thinking (for the 1D case to make everything simpler) that the space is not $x \in (-∞,+∞)$ but more like a circle ($S^{1}$)? So translations would be allowed around it, and it would have the property that for x=0 and x=L the sin would still be zero (since they'd be identical points)... Or even simpler just accept that space is in a straight line of length L...

I mean this works out fine (in my imagination) for the infinite well potential... of course it can't be done for any arbitrary model... For example, in string theory, you can define momenta on the parameter space (in which the σ-parameter is "bounded")...

Also @strangerep, the step function won't help in describing the infinity, i guess it should me something else in a model....

Last edited: Mar 3, 2014
13. Mar 3, 2014

### strangerep

The circle is different, since one need only assume periodic boundary conditions. (I think Hendrik or someone else mentioned that in the other thread.)

Yes, with caveats.

[Edit:] I suppose one should solve the problem using a limit approach described by hilbert2, (or perhaps just by a sequence of ever deeper square wells) and verify that one gets the same results as restricting the space to finite-length line.

Last edited: Mar 4, 2014
14. Mar 4, 2014

### vanhees71

Ok, I try to explain it again since this nonsense of momentum operators for particles in a finite box come up often, including the homework section. Is there any serious textbook, where they claim a momentum operator exists for a particle in a box? I hope not!

Let's work in the position representation, i.e., with the wave-mechanical formulation of non-relativistic quantum mechanics. The Hilbert space is then the set of square-integrable functions $\mathrm{L}^2([0,L])$ with boundary conditions $\psi(x=0)=\psi(x=L)=0$.

By definition the momentum operator is the generator for translations. This is related to Noether's theorem identifying momentum as the conserved quantity that follows from spatial translation invariance. This implies that in position representation the momentum operator is given by
$$\hat{p}=-\mathrm{i} \partial_x.$$
Suppose now, this is a self-adjoint operator in our Hilbert space. Then there must exist a complete set of generalized eigenvectors in this Hilbert space. This implies
$$-\mathrm{i} u_p(x)=p u_p(x) \; \Rightarrow \; u_p(x)=A \exp(\mathrm{i} p x).$$
Now, the eigenfunction must fulfill the boundary conditions, leading to $A=0$, i.e., there doesn't exist any eigenfunction for the momentum operator, and thus this operator cannot be a self-adjoint operator and thus it doesn't exist as an observable for the particle in a box.

That's, of course, different for the Hamiltonian, which is given by
$$\hat{H}=-\frac{1}{2m} \partial_x^2.$$
The eigen-value problem becomes
$$\partial_x^2 u_E(x)=-2 m E u_E(x) \; \Rightarrow \; u_E(x)=A \cos(\omega_E x)+B \sin(\omega_E x) \quad \text{with} \quad \omega_E=\sqrt{2 m E}.$$
The boundary conditions then lead to
$$A=0, \quad \omega_E \in \frac{\pi}{L} \mathbb{N}.$$
These are nice functions in the Hilbert space with eigenvalues
$$E_n=\frac{\pi^2 n^2}{2m L^2}, \quad n \in \mathbb{N}.$$
The corresponding eigenfunctions are
$$u_n(x)=N \sin \left (\frac{n \pi x}{L} \right ).$$
The normalization constant is fixed by
$$\int_0^L \mathrm{d}x |u_n(x)|^2=1 \; Rightarrow \; N=\sqrt{\frac{2}{L}}.$$
As it is well known that these sine functions build a complete set on the Hilbert space under consideration, the Hamiltonian is indeed a self-adjoint operator and thus represents a proper observable.

Indeed, everything changes if you look at the "rigid rotator", where instead of "rigid boundary conditions" you have "periodic boundary conditions". There the translation operator is a proper self-adjoint operator, representing angular momentum.

15. Mar 4, 2014

### dextercioby

16. Mar 5, 2014

### LagrangeEuler

17. Mar 5, 2014

### LagrangeEuler

It exists definitely in many different seriuous problem books. I think that all problem books in quantum mechanis shows that for particle in the box $\langle \hat{p} \rangle=0$.

18. Mar 5, 2014

### ChrisVer

I am getting really confused here now...
Of course the momentum commutes with the hamiltonian for an infinite well potential... That means that momentum is a conserved quantity, and as such you have the translation symmetry...
What happens when you translate the system? nothing... in fact translation will not just move your particle, but it will also move the potential (you move the axis)... that's the reason why the infinite well potential is equally defined in the region $[-L/2, +L/2]$ as it does for $[0,L]$ (by that I mean that the energies remain the same), and so I could take it from $[a,a+L], a\in R$ (that's why at first I proposed the circle geometry).

19. Mar 5, 2014

### hilbert2

^ Suppose an electron described by a gaussian wavepacket $\psi(x)=N\exp(ik(x-x_{0}))\exp(-a(x-x_{0})^{2})$ is shot towards an infinite potential wall and it bounces back from it. Is momentum conserved here? Doesn't the expectation value of momentum have opposite sign after the collision?

20. Mar 5, 2014

### ChrisVer

you can't really talk about incoming particles (that are described by exp[ikx] ) for that potential...
The current model is more like a "trap" that you cannot escape or enter...
If I'd have to think of just a wall (like a step potential but with infinite value to avoid transitions), then indeed the outcoming wave would have opposite momentum...
But the same you can say for light bouncing off a perfect conductor...