Calc Commutator in Infinitely Potential Well: Is it Possible?

[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.

 

[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 $p^{2}$

 

[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.

 

[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.

 

[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...

 

[vanhees71]

There is no momentum operator for a particle restricted to a finite interval. See, e.g., this thread:

https://www.physicsforums.com/showthread.php?t=694158

 

[micromass]

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.

You're going to 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.

 

[hilbert2]

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

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 $V(x)=x^{50}$:
http://www.4shared.com/download/pkyokC0-ce/plot.gif?lgfp=3000.

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.

 

[LagrangeEuler]

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

 

[ChrisVer]

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 $V(x)=x^{50}$:
http://www4b.wolframalpha.com/Calculate/MSP/MSP117951d7d1a8c526f9c4500001c04gf634biag030?MSPStoreType=image/gif&s=24&w=300.&h=202.&cdf=RangeControl .

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.

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...

 

[strangerep]

[...]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...

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).

 

[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...

 

[strangerep]

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)...

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

Or even simpler just accept that space is in a straight line of length L...

Yes, with caveats.

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

[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.

 

[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.

 

[dextercioby]

@Post #14 by vanhees71: Hi Hendrik, see my comment here: https://www.physicsforums.com/showthread.php?t=741204 below your post in that thread.

 

[LagrangeEuler]

Look this
http://www.youtube.com/watch?v=eyXON6C6lKk&list=PL39758D7C8281EFBB
from 20:20.

 

[LagrangeEuler]

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!

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 $\langle \hat{p} \rangle=0$.

 

[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).

 

[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?

 

[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...

 

[vanhees71]

I can only warn again against the FALSE statement the operator $-\mathrm{i} \partial_x$ can be interpreted as the momentum operator (i.e., an operator representing momentum) for a particle restricted to a finite interval. There is nothing else than this interval. So you can't move it somewhere. Of course, an infinite square well is never realized in nature, so that this is an academic discussion anyway.

@hilbert2: your Gaussian wave packet does not represent a state for a particle restricted to the interval, because it doesn't fulfill the boundary conditions.

 

[hilbert2]

@hilbert2: your Gaussian wave packet does not represent a state for a particle restricted to the interval, because it doesn't fulfill the boundary conditions.

¨

Sorry about that. Maybe I should have talked about a gaussian wavepacket moving back and forth is a harmonic potential. It's just an example of the fact that momentum is generally not conserved when a (quantum or classical) particle is bound by some potential.

http://www.st-andrews.ac.uk/physics/quvis/embed_item_3.php?anim_id=22&file_sys=index_phys

 

[vanhees71]

Sure. The problem with the infinite-square well is that it is pretty academic and quite difficult. Ironically it's often used in introductory quantum-mechanics texts.

 

[LagrangeEuler]

I'm very confused now. More then I used to be. If you have particle in the box (0<x<a), and the case of particle in the box (-a/2<x<a/2) yes you will have same energy. In Griffiths QM he define momentum for particle in the box. Is he wrong?

 

[vanhees71]

How does Griffiths define the momentum operator? If he doesn't give a hint that there are serious problems to do so, it's a bit problematic, I'd say. Of course, it doesn't matter at all whether you define your interval to be $[0,a]$ or $[-a/2,a/2]$. That's the same system, just written with a shifted coordinate.

 

[LagrangeEuler]

At page 29 in my book.

 

[samalkhaiat]

eigenfunction of $p^2$ it is also eigenfunction of $p$. Right?

That is true, but this does not mean that the eigenfunctions of $P$ are also eigenfunctions of $H$. Inside the box, the particle IS NOT FREE and the Hamiltonian IS NOT just the usual free particle Hamiltonian, $P^{ 2 } / 2m$. The condition $V ( x ) = 0, \ \forall x \in ( 0 , L )$ does not mean the particle is free inside the box. This is why the eignfunctions of $H$ are not simultaneous eigenfunctions of $P$: inside the box you have
$$[ H \ , \ P ] = [ P^{ 2 } / 2m + V ( x ) \ , \ P ] = [ V ( x ) \ , \ P ] \neq 0$$

 

[hilbert2]

An eigenfunction of $p^{2}$ is not usually an eigenfunction of $p$. To avoid technical difficulties with the momentum operator, consider a free particle (particle in constant potential). A wavefunction $\psi(x)=A\sin(kx)$ is an eigenfunction of $p^{2}$ but it definitely is not an eigenfunction of $p$.

This can also be demonstrated in a system that has a two-dimensional state space. Let's define the hermitian operator $A$ in matrix form as

$A=\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right)$

Now, the matrix $A^{2}$ is just the identity matrix, and any vector in $\mathbb{R}^{2}$ is its eigenvector. However, this does not mean that all vectors in $\mathbb{R}^{2}$ are eigenvectors of $A$.

 

[ChrisVer]

In general, as i stated, the eigenfunctions of the one can be eigenfunctions of the other... This is due to $[p,p^{n}]=0$...
This however does not mean that ALL eigenfunctions are the same... It's more like saying that the one contains the other but can also have something more... this is a rough another example from some other field- that a subgroup contains some elements of the group -it doesn't necessarily mean that it contains all of them.
Another example from QM on this, is also in the Harmonic Oscillator and Parity operator... Parity commutes with the HO Hamiltonian... The eigenfunctions of parity are either odd or even functions, the hamiltonian of the HO in general though, contains them both...

 

[samalkhaiat]

An eigenfunction of $p^{2}$ is not usually an eigenfunction of $p$.

For FREE PARTICLE, the eigenstate of $p^{ 2 }$ is always an eigenstate of $p$.

consider a free particle. A wavefunction $\psi(x)=A\sin(kx)$ is an eigenfunction of $p^{2}$ but it definitely is not an eigenfunction of $p$.

This is because your wave function is not that of a free particle:

(1) $\psi ( 0 ) = 0$, while the wavefunction of a free particle does not vanish anywhere.

(2) your $\psi ( x )$ gives vanishing probability current, while free particles have non-zero probability current.

(3) (for the boxed particle) free particles DO NOT have DISCRETE spectrum.

This can also be demonstrated in a system that has a two-dimensional state space. Let's define the hermitian operator $A$ in matrix form as

$A=\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right)$

Now, the matrix $A^{2}$ is just the identity matrix, and any vector in $\mathbb{R}^{2}$ is its eigenvector. However, this does not mean that all vectors in $\mathbb{R}^{2}$ are eigenvectors of $A$.

In general, yes. The vanishing commutator [ A , B ] = 0 does not mean that every eigenstate of A is an eigenstate of B

even though they may share a complete set of common eigenstates.

 

[vanhees71]

That's obviously not true. Take a particle on the $x$ axis in a finite potential. Then the generalized momentum eigenfuncrions are
$$u_p(x)=\exp(ipx).$$
This is also a generalized eigenfunction of $p^2$, but so is also $u_{-p}$ and thus also any linear combination,
$$a u_{p}(x)+b u_{-p}(x).$$
The latter is not an eigenfunction of $p$ but of $p^2$.

 

[DrDu]

@Post #14 by vanhees71: Hi Hendrik, see my comment here: https://www.physicsforums.com/showthread.php?t=741204 below your post in that thread.

Well, if you assume the infinite potential outside the box as some limit of a finite potential, the wavefunction will be continuous and hence 0 at the boundaries.

Interestingly there are self adjoint extensions of the operator -d^2/dx^2 which can't be obtained as limits of adding a potential and which lead to wavefunctions with are discontinuous but have continuous derivatives. This is called δ'- interaction.

Sometimes, I prefer to think of the "particle in an infinite box" as a particle moving between two delta functions of infinite strength, as this operator is also defined on the whole real axis as is the free hamiltonian.
This becomes interesting when there is also an additional finite potential, e.g. a finite potential well.
I think it is even possible to do some perturbation expansion in the inverse strength of the delta functions, e.g., to calculate resonance linewidths for a finite potential.

 

[DrDu]

There is a very good article on the topic which was also published in Amer J Phys:

http://arxiv.org/pdf/quant-ph/0103153.pdf?origin=publication_detail

 

[samalkhaiat]

That's obviously not true. Take a particle on the $x$ axis in a finite potential.

My reasoning was based on the premiss that the particle IS a FREE particle.

This is also a generalized eigenfunction of $p^2$, but so is also $u_{-p}$ and thus also any linear combination,
$$a u_{p}(x)+b u_{-p}(x).$$
The latter is not an eigenfunction of $p$ but of $p^2$.

This linear combination DOES NOT represent a FREE PARTICLE:
1)free particles do not just flip the sign of their momenta.
2) it can be made proportional to $\sin ( p x )$ which is not a free particle wavefunction. So to restrict the solutions of $P^{ 2 } \psi = p^{ 2 } \psi$ to a free particle, you should set $b = 0$ or $a = 0$.

The main issue regarding the particle in the box is the following: Inside the box, the particle is never a free particle.

 

[strangerep]

There is a very good article on the topic which was also published in Amer J Phys:

http://arxiv.org/pdf/quant-ph/0103153.pdf?origin=publication_detail

Thanks for the reminder about that paper (Bonneau, Faraut & Valent). I had only skimmed it in the past but this thread motivated me to study it more carefully.

It's interesting how other physical conditions, such as time-reversal invariance, parity invariance, or positivity of energy can play a role in selecting possible self-adjoint extensions, as this highlights how one must consider the entire dynamical problem when attempting quantization.

Most surprising is their remark at the end of sect 7.4: "This discussion shows that an infinite potential cannot be simply described by the limit of a finite one" (because the desired self-adjoint extension does not exist in that limit).

For other readers: if you're not sure what the terms "self-adjoint extension" and "deficiency index" mean in the context of this thread, then studying that paper will help.