Momentum eigenfunctions in an infinite well

[dyn]

Hi

For an infinite well , solving the Schrodinger equation gives wavefunctions of the form sin(nπx/L). These are not eigenfunctions of the momentum operator which means there are no eigenvalues of the momentum operator. Does this mean momentum cannot be measured ?

Inside the infinite well the Hamiltonian is p²/(2m) ; this commutes with p so that should mean that momentum and energy share common eigenfunctions but as i stated above there are no momentum eigenfunctions. Where am i going wrong ?

Thanks

 

[anuttarasammyak]

For an infinite well , solving the Schrodinger equation gives wavefunctions of the form sin(nπx/L).

Yes in the well but it is zero outside. That matters.

These are not eigenfunctions of the momentum operator which means there are no eigenvalues of the momentum operator. Does this mean momentum cannot be measured ?

It is superposition of various mometum eigenstates, the coefficiets of which provide probability to observe specific momentum value by product with complex conjugate.

Inside the infinite well the Hamiltonian is p2/(2m) ;

Yes inside but it includes no zero potential energy outside. That matters.

 

[PeroK]

Have you tried looking for momentum eigenstates? Note that momentum is a vector and that any momentum eigenstate ought to be an energy eigenstate, but not vice versa.

 

[anuttarasammyak]

any momentum eigenstate ought to be an energy eigenstate,

In free space system of zero potential energy everywhere, yes.

 

[DrClaude]

In the case of an infinite well, the momentum operator is not self-adjoint.

 

[vanhees71]

This is a very good question. The answer is surprising, but only at first sight: There is no momentum-observable for particle in the infinitely high potential pot, because the operator $\hat{p}=-\mathrm{i} \hbar \partial_x$ is not self-adjoint on the interval $[0,a]$. We had this discussion some times in the forum. Here's one answer

https://www.physicsforums.com/threa...the-infinite-square-well.1004624/post-6509638

Some nice, didactic papers on self-adjointness are

https://arxiv.org/abs/quant-ph/9907069
https://arxiv.org/abs/quant-ph/0103153

 

[anuttarasammyak]

This is a very good question. The answer is surprising, but only at first sight: There is no momentum-observable for particle in the infinitely high potential pot, because the operator p^=−iℏ∂x is not self-adjoint on the interval [0,a].

For this exercise for the biginners I prefer observing what happens in finite well potential V system in the limit of $V \rightarrow \infty$ to discussing mathematical subject of $V = \infty$. We may keep the interval $[-\infty,+\infty]$ and apply usual Fourier transform method to get momentum wave function.
e.g. mometum wave function, Particle in a box wiki https://en.wikipedia.org/wiki/Particle_in_a_box

 

[Vanadium 50]

You got some mathy answers. You can also use your physical intution:

(1) If I solve the time-independent Schroedinger Equation, I ghet energy eigenstates, which tale the from E-p²/2m. You could then say, $p = \pm \sqrt{2mE}$ which obviously has two values.

(2) These two values correspond to the particle moving to the left and moving to the right.

(3) But these solutions cannot be good for all time, because eventually the particle will hit the well boundary. So they can't be solutions of the time-independent Schroedinger Equation: they aren't time-independent.

Therefore, there are no solutions.

 

[anuttarasammyak]

Therefore, there are no solutions.

My complement
A classical perfect reflection analogue of
$$\alpha|p'>+\beta|-p'>$$
where
$$p'=\sqrt{2mE'}$$
isn't a solution neither.

 

[dyn]

Thanks for all your replies. Before i go any further , i need to ask ; is a state of definite momentum the same as a momentum eigenfunction ?

 

[anuttarasammyak]

Each mometum eigenfunction has eigenvalue which is a definite value of momentum of the state.

 

[dyn]

To make a wavefunction that has the form sin( nπx/L) inside the infinite well and is zero outside the well requires a continuous superposition of momentum eigenfunctions performed using a Fourier transform. Momentum eigenfunctions extend from +∞ to -∞. This means that the wavefunction of a particle confined to a well consists of superpositions of momentum states that extend over all space. That is how i understand it but it seems strange !

On a related note , regarding the continuous momentum basis , the eigenvalue p can take any value in the range -∞ < p < +∞ so in QM infinite momentum is allowed ?

 

[PeroK]

On a related note , regarding the continuous momentum basis , the eigenvalue p can take any value in the range -∞ < p < +∞ so in QM infinite momentum is allowed ?

And the maximum value of momentum in Newtonian physics is ...?

 

[dyn]

Yes ; it is infinite but it seems strange to come across such a simple example of incompatibility between QM and SR

 

[PeroK]

Yes ; it is infinite but it seems strange to come across such a simple example of incompatibility between QM and SR

And the maximum momentum in SR is ...?

 

[dyn]

As far as i know and my SR is very rusty you can't have infinite momentum in SR

 

[PeroK]

As far as i know and my SR is very rusty you can't have infinite momentum in SR

First, you're confusing infinite with having no limit. You can't have infinite momentum in any physics, as infinity is not a number. But, there is no limit in theory to the magnitude of momentum that a particle could have.

You're wrong about SR.

 

[dyn]

I know a particle's velocity can't exceed c so i would presume there is also a limit on momentum

 

[PeroK]

I know a particle's velocity can't exceed c so i would presume there is also a limit on momentum

You presume wrong.

 

[dyn]

Relativistic momentum approaches infinity as u approaches c but u can never equal c so momentum can never be infinite

 

[weirdoguy]

so momentum can never be infinite

Just like in Newtonian mechanics because v also cannot be infinite.

 

[dyn]

Why can v not be infinite in Newtonian mechanics ?

 

[weirdoguy]

Because infinity is not a number. It can be as big as you want, but always finite.

 

[anuttarasammyak]

On a related note , regarding the continuous momentum basis , the eigenvalue p can take any value in the range -∞ < p < +∞ so in QM infinite momentum is allowed ?

For an example in a harmonic oscillator energy eigenstate, position x and momentum p both have Gaussian probability distributions around zero. We have tiny but not zero probality to observe any large p as well as x.

 

[PeroK]

Why can v not be infinite in Newtonian mechanics ?

If a particle is the origin and moving at infinite speed, where is it one second later?

 

[Vanadium 50]

A parade of false statements, hoping to be corrected, is a poor way to learn.

Irrespective of the quality of the argument, why would one be surprised that non-relativistic QM is non-relativistic? That's like being surprised that Burfer King sells burgers,

 

[mattt]

As far as i know and my SR is very rusty you can't have infinite momentum in SR

In Newtonian Mechanics p=m*v and p can be as high as you want because v can be as high as you want.

In Relativistic Mechanics p=m(v)*v and p can be as high as you want because m(v)=mo/sqrt(1-(v/c)^2) is as high as you want if v is close enough to c.

 

[malawi_glenn]

That's like being surprised that Burfer King sells burgers,

I would not call them burgers tbh... garbage at best :)

 

[dyn]

A parade of false statements, hoping to be corrected, is a poor way to learn.

Irrespective of the quality of the argument, why would one be surprised that non-relativistic QM is non-relativistic? That's like being surprised that Burfer King sells burgers,

You spelt Burger incorrectly ! Actually i was hoping to not be corrected. Are you telling me that in QM , CM and SR there is no limit on how big momentum can be ? Surely v not being able to reach c places a limit on momentum in SR

 

[weirdoguy]

Surely v not being able to reach c places a limit on momentum in SR

You've already been told that this is not correct...

 

[PeterDonis]

Are you telling me that in QM , CM and SR there is no limit on how big momentum can be ?

Yes, that's what they have told you.

Surely v not being able to reach c places a limit on momentum in SR

Instead of waving your hands, do the math. If you have questions about the math, you can start a new thread in the appropriate forum. This thread is closed as the question you asked in your OP has been answered.

 

[Nugatory]

Are you telling me that in QM , CM and SR there is no limit on how big momentum can be ? Surely v not being able to reach c places a limit on momentum in SR

It does not.
The momentum of a particle moving at speed $v$ is $p=mv/\sqrt{1-(v^2/c^2)}$.

That value can be made as large as we please by choosing a value of $v$ sufficiently close to but still less than $c$.