Particle in a box in momentum basis

[Ravi Mohan]

I have been thinking about it for sometime but couldn't really get the answer. This is the progress I have made till now.

E |ψ> = H |ψ>
E <p|ψ> = <p|H|ψ>

Now how to evaluate the number <p|H|ψ>? Although I can evaluate this number by introducng identity operator 1 = $\int$|x><x| dx. But for it I need to solve schrodinger's equation in position basis which defeats the purpose (because we will have to evaluate the number <x|ψ>).

 

[vanhees71]

What precisely do you mean by "box"? If you have a infinitely deep potential pot in mind, then forget your idea, because then there is no useful momentum operator due to the boundary conditions. This is a rather deep fact concerning the definition domain of the momentum operator, which is well-defined either in infinite (Euclidean) space or for boxes with periondic boundary conditions.

However the Hamilton operator is well-defined. But to directly use the energy representation (for the one-dimensional infinitely deep potential pot) seems not possible to me, because it's hard to express the boundary conditions. So here, I'd really work in the position representation. Then it's a pretty simple problem.

 

[Ravi Mohan]

Thank you for helping me. Initially I had infinite potential pot in mind but then I would certainly like to do it for finite potential pots.

I coudnt get one thing (regarding infinite potential box)

because then there is no useful momentum operator due to the boundary conditions.

Can you please elaborate how?

 

[vanhees71]

Let's consider motion in only one dimension. For three dimensions everything is analogous.

As said yesterday, it's most convenient to work in position representation. The states are then represented by wave functions $\psi(x)$. since the potential goes to infinity at the boundaries, your particle is strictly restricted to the interval $[-L/2,L/2]$, and the wave function must fulfill the boundary conditions $\psi(-L/2)=\psi(+L/2)=0.$

Now suppose you define the momentum operator as usual as the generator of translations. Then it should be as in entire space represented by the differential operator
$$\hat{p}=-\mathrm{i} \mathrm{d}_x.$$
Here and in the following I set $\hbar=1$ (natural units).

Now, if this was a properly defined self-adjoint operator, it should have a complete set of (generalized) eigenfunctions. These must fulfill the eigenvalue condition
$$\hat{p} u_p(x)=p u_p(x)$$
with real eigen values $p$. Of course, you get
$$u_p(x)=N \exp(\mathrm{i} p x)$$
as the general solution of this equation, but there is no solution fulfilling the boundary conditions, and thus the momentum operator cannot be defined as a self-adjoint operator on the here considered Hilbert space $L^2([-L/2,L/2]).$ The operator $\hat{p}$ is Hermitean but not self-adjoint. The derivative operator is formally defined on the differentiable wave functions which fulfill the boundary conditions, but acting on these functions leads to a result which usually does not fulfill the boundary conditions.

That's different for the Hamiltonian, which is given by
$$\hat{H}=\frac{1}{2m} \hat{p}^2=-\frac{1}{2m} \mathrm{d}_x^2.$$
Here the eigenvalue equation reads
$$\hat{H} u_E(x)=E u_E x, \quad u_E''(x)=-2m E u_E(x).$$
The solutions read
$$u_E(x)=A_1 \cos(k x)+A_2 \sin(k x),$$
where $k=\sqrt{2m E}$. There are only solutions for $E >0$ because otherwise you cannot fulfill the boundary conditions.

There are a set of solutions with even (odd) parity, given by pure cos and sin solutions. For the former you must have
$$u_E(L/2)=u_E(-L/2)=N \cos \left (\frac{k L}{2} \right ) \stackrel{!}{=}0.$$
This leads to
$$\frac{k L}{2} = \frac{2n+1}{2} \pi \; \Rightarrow k_{n}^+=\frac{(2n+1)\pi}{L}.$$
For odd parity you get
$$u_E(L/2)=-u_E(-L/2)=N \sin \left (\frac{k L}{2} \right ) \stackrel{!}{=}0.$$
This leads to
$$\frac{k L}{2}=n \pi \; \Rightarrow \; k_n^-=\frac{2 n \pi}{L}.$$
You have a complete set of eigenfunctions on $L^2([-L/2,L/2]).$
$$u_n^{(+)}(x)=N \cos(k_n^+ x), \quad u_n^{-}(x)=N \sin(k_n^- x).$$
The corresponding energy eigenvalues are
$$E_n^{\pm}=\frac{(k_n^{\pm})^2}{2}.$$

Note that the Hamiltonian maps these functions again to the same functions times the corresponding energy eigenvalue, and thus the operators maps these functions again to functions in the Hilbert space. The Hamiltonian is properly defined as a self-adjoint operator. If you, however, act with the (pseudo-)momentum operator on these states the cos becomes a sin and vice versa, i.e., the image under the momentum operator does not fulfill the boundary conditions and are thus outside of the Hilbert space. The momentum operator thus is not defined as a self-adjoint operator on this Hilbert space.

 

[Ravi Mohan]

Thank you for the demonstration. It was insightful and I will keep that in mind.
I understood it but I am not able to connect it to my question.

As said yesterday, it's most convenient to work in position representation.

But I don't want to work in this representation. I want to work in momentum representation <p|ψ>. Actually I want to question this "convenience" itself. Why do we always try to work in position representation. One might answer "because this way we can convert $V(\hat{x})$ to $V(x)$". But I am beginning to believe that we can't work in momentum basis everytime.

The states are then represented by wave functions $\psi(x)$.

Thus the states I want to work with must be represented by $\widetilde{\psi}(p)$ (the Fourier transform of $\psi(x)$). And I want to find these energy eigen states using schrodinger's equation. Also we know the answer in advance. It should be linear combination of Dirac-delta functions.

So how come "momentum operator thus is not defined as a self-adjoint operator on this Hilbert space" is restricting me to work in position basis only?

 

[dipstik]

i may be way off base here, but could you use the lower and raising operator momentum equivilent?

http://digitalcommons.uconn.edu/cgi/viewcontent.cgi?article=1025&context=chem_educ

 

[Jano L.]

vanhees71, is the self-adjointness that important? It seems that one can still find "momentum representation" of $\psi(x)$ and of the Schroedinger equation by Fourier transform.

For infinite potential well, we have

$$
\frac{p^2}{2m} \psi(p) = E\psi(p)
$$

plus the conditions

$$
\int \psi(p) dp = 0,
$$
$$
\int \psi(p) e^{ipL/\hbar} dp = 0,
$$

which are however quite awkward.


General Schroedinger equation

$$
\frac{{\hat p}^2}{2m}\psi(x) + U(x)\psi(x) = E\psi(x)
$$

changes into

$$
\frac{p^2}{2m}\psi(p) + \int U(p') \psi(p-p')\frac{dp'}{2\pi\hbar} = E\psi(p),
$$

where

$$
U(p) = \int U(x) e^{-ipx/\hbar}\,dx.
$$

The difficulty seems rather in the fact that we obtain equation with convolution which does not seem to be much easier to solve than the original differential equation.

 

[Ravi Mohan]

Jano L. there is one more problem which one encounters. What boundary conditions shall we put on <p|ψ>? If it is quantization (which will give dirac-delta) then it is only because we have solved in position basis and applied the result form there. Again it will defeat the purpose!

 

[Jano L.]

Yes. The condition

$$
\lim_{p\rightarrow \pm \infty} \psi(p) =0
$$

seems reasonable, as we do not expect infinite momenta, and also

$$
\int |\psi(p)|^2\, dp/(2\pi\hbar) = 1,
$$

to have normalized probability distribution for momenta $\rho(p) = |\psi(p)|^2/(2\pi\hbar)$. But it is hard to solve the equation in this way, I would not know how to do it without going back to equation for $\psi(x)$.

 

[vanhees71]

What's $U(p)$ concretely in the case of the infinitely deep potential well? I don't know, how to describe this in momentum representation, at all. The simple reason is that, as described, the momentum representation doesn't exist ;-).

Let's see, what we can do for the finitely deep well. In position representation we have
$$U(x)=\begin{cases} -U_0 & \text{for} \quad -L/2<x<L/2,\\ 0 & \text{elsewhere}. \end{cases}$$
The matrix element in momentum space is thus
$$\tilde{U}(p,p')=\langle p|U(\hat{x})| p' \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p | x\rangle U(x) \langle x | p' \rangle = U_0 \int_{-L/2}^{L/2} \mathrm{d} x \frac{1}{2 \pi} \exp[\mathrm{i} x (p'-p)]=\frac{U_0}{\pi (p-p')} \sin \left [\frac{L}{2}(p-p') \right ].$$
The time-independent Schrödinger equation then reads
$$\frac{p^2}{2m} u_E(p) + \int_{\mathbb{R}} U(p,p') u_E(p')=E u_E(p).$$
Have fun with the solution ;-)).

 

[Jano L.]

Indeed, it seems that the original equation for $\psi(x)$ is more tractable. However, it is interesting that we can always find $\psi(p)$ from $\psi(x)$, even for the infinite well.

 

[Ravi Mohan]

So how come "momentum operator thus is not defined as a self-adjoint operator on this Hilbert space" is restricting me to work in position basis only?

Ok! I changed the starter of my tubelight and now, I think, it is working.
If momentum operator is not defined as self-adjoint operator, it means that the eigen-vectors of momentum operator don't form complete set of basis . And therefore one is not advised to work in momentum basis for infinite potential well.

Or maybe it is something else. I really need to work on this thread...

P.S Can anyone provide me a reliable source from where I can study difference between hermitian and self-adjoint operators. Thank you.

 

[Jano L.]

Not focused on self-adjointness, but discusses it briefly among other interesting things:

http://arxiv.org/abs/quant-ph/9907069

 

[Ravi Mohan]

Nice find. Should keep me busy for a week or two. Thanks a ton.

 

[HomogenousCow]

Hmmm..interesting, what would happen if instead of using the rather wonky square well we took some smooth function which approached the infinite square well when one of the parameters are taken to infinity. The momentum operator would have to be defined for any "normal" value of the parameter, I'm curious how the momentum operators breaks down as we take that limit.

Two step functions multiplied together with their arguments shifted either ways should give us a square well, we can approximate this with an error function, from there we can get closer and closer to the square well..hmmm

 

[vanhees71]

Jano, what do you think is $\tilde{\psi}(p)$ for the infinite square well? Again, there is no momentum observable on $L^2([-L/2,L/2])$. So what should the physical meaning of some function $\tilde{\psi}(p)$ be?

This is discussed in the very nice paper by Gieres as Example 4!

 

[Jano L.]

For some considerations we do not need this technical notion of "observable". It would be strange if momentum just lost its meaning just because we take the limit of infinite walls.

Quite generally, for any normalized function $\psi(x)$, we can find the corresponding Fourier transform
$$
\tilde{\psi}(p) = \int \psi(x) e^{-ipx/\hbar}\,dx,
$$
which satisfies

$$
\int \frac{|\tilde{\psi}(p)|^2}{2\pi\hbar} dp = 1
$$
and
$$
\langle p \rangle = \int \psi^*(x) \hat{p} \psi(x) dx = \int p \frac{|\tilde\psi(p)|^2}{2\pi\hbar} dp.
$$
These equations suggest that we can view the function
$$
\rho(p) = \frac{|\tilde{\psi}(p)|^2}{2\pi\hbar}
$$
as the probability distribution of momenta.

For the infinite well, we have
$$
\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)~~~\text{for}~x\in \langle 0,L\rangle
$$
$$
\psi(x) = 0~~~~~~~~~~~~~~~~~\text{for}~x\in ℝ - \langle 0,L\rangle,
$$
so the momentum function is
$$
\tilde{\psi}(p) = \sqrt{\frac{2}{L}} \int_0^L \sin\left(\frac{\pi x}{L}\right) e^{-ipx/\hbar}\,dx,
$$
which gives probability distribution of momentum

$$
\rho(p) = \frac{4\pi\hbar^3 L}{(p^2L^2-\pi^2\hbar^2)^2}\cos^2 \frac{pL}{2\hbar}
$$

(I copied the result from Landau&Lifgarbagez, sec. 22, problem 1.)

The plot of $\rho(p)$ for $\hbar,L =1$ is here:

So the zero momentum is most probable, and with increasing magnitude of the momentum probability decreases, which seems quite reasonable.

 

[vanhees71]

That's the momentum distribution for a particle in free space (or in a finite potential) prepared in the state represented by the sine function (and 0 outside the interval [0,L]). That's ok, but for the particle in the infinite square well it doesn't have a physical meaning in the sense of momentum measurements, because there is no momentum observable that you could measure!

 

[HomogenousCow]

I take that to mean that the boundary conditions just aren't physical

 

[Jano L.]

That's ok, but for the particle in the infinite square well it doesn't have a physical meaning in the sense of momentum measurements, because there is no momentum observable that you could measure!

$\rho(p)\Delta p$ can be ascribed the meaning of probability that particle has momentum in the interval $p..p+\Delta p$. I do not see why such technical detail as vanishing of $\psi$ outside of the well should invalidate the concept of momentum.

The above calculation of $\rho(p)$ is simplest for the infinite well. The meaning of the calculation is that we can use the result for deep but finite well. We expect that $\rho(p)$ should be good approximation to the actual distribution there.

 

[Joker93]

For some considerations we do not need this technical notion of "observable". It would be strange if momentum just lost its meaning just because we take the limit of infinite walls.

Quite generally, for any normalized function $\psi(x)$, we can find the corresponding Fourier transform
$$
\tilde{\psi}(p) = \int \psi(x) e^{-ipx/\hbar}\,dx,
$$
which satisfies

$$
\int \frac{|\tilde{\psi}(p)|^2}{2\pi\hbar} dp = 1
$$
and
$$
\langle p \rangle = \int \psi^*(x) \hat{p} \psi(x) dx = \int p \frac{|\tilde\psi(p)|^2}{2\pi\hbar} dp.
$$
These equations suggest that we can view the function
$$
\rho(p) = \frac{|\tilde{\psi}(p)|^2}{2\pi\hbar}
$$
as the probability distribution of momenta.

For the infinite well, we have
$$
\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)~~~\text{for}~x\in \langle 0,L\rangle
$$
$$
\psi(x) = 0~~~~~~~~~~~~~~~~~\text{for}~x\in ℝ - \langle 0,L\rangle,
$$
so the momentum function is
$$
\tilde{\psi}(p) = \sqrt{\frac{2}{L}} \int_0^L \sin\left(\frac{\pi x}{L}\right) e^{-ipx/\hbar}\,dx,
$$
which gives probability distribution of momentum

$$
\rho(p) = \frac{4\pi\hbar^3 L}{(p^2L^2-\pi^2\hbar^2)^2}\cos^2 \frac{pL}{2\hbar}
$$

(I copied the result from Landau&Lifgarbagez, sec. 22, problem 1.)

The plot of $\rho(p)$ for $\hbar,L =1$ is here:

So the zero momentum is most probable, and with increasing magnitude of the momentum probability decreases, which seems quite reasonable.

Isn't zero momentum invalid(unphysical) for the infinite well(since it would give E=0 and thus Ψ=0)?

 

[Joker93]

$\rho(p)\Delta p$ can be ascribed the meaning of probability that particle has momentum in the interval $p..p+\Delta p$. I do not see why such technical detail as vanishing of $\psi$ outside of the well should invalidate the concept of momentum.

The above calculation of $\rho(p)$ is simplest for the infinite well. The meaning of the calculation is that we can use the result for deep but finite well. We expect that $\rho(p)$ should be good approximation to the actual distribution there.

So, is the distribution that you gave valid for the infinite well or is it just an approximation for very deep finite wells?

 

[Joker93]

That's the momentum distribution for a particle in free space (or in a finite potential) prepared in the state represented by the sine function (and 0 outside the interval [0,L]). That's ok, but for the particle in the infinite square well it doesn't have a physical meaning in the sense of momentum measurements, because there is no momentum observable that you could measure!

This is because there are no momentum eigenvalues that satisfy the boundary conditions? I mean, for the momentum representation to have a physical meaning, the momentum operator's eigenfunctions must satisfy the boundary conditions? If so, why does this particular thing make the momentum operator self-adjoint?
Lastly, because the momentum eigenfunctions don't satisfy the boundary conditions, that makes it not be self-adjoint, but does that also mean that it is not hermitian?

 

[vanhees71]

The point is indeed that there's no eigenfunction to any eigenvalue of $\hat{p}=-\mathrm{i} \mathrm{d}_x$, because if you fulfill the boundary conditions for $u_p(x)$ then $\hat{p} u_p(x)$ doesn't fulfill them. This shows that, although $\hat{p}$ is a Hermitean operator it is not self-adjoint, and thus it cannot define a proper representation of an observable. Note that, however, the sin and cos functions with the appropriate boundary condistions are true eigenfunctions of $\hat{H}=\hat{p}^2/(2m)$ and thus $\hat{H}$ is not only Hermitean but also self-adjoint and thus represents an observable (here the Hamiltonian for a particle confined in an infinitely high square well).

 

[Joker93]

The point is indeed that there's no eigenfunction to any eigenvalue of $\hat{p}=-\mathrm{i} \mathrm{d}_x$, because if you fulfill the boundary conditions for $u_p(x)$ then $\hat{p} u_p(x)$ doesn't fulfill them. This shows that, although $\hat{p}$ is a Hermitean operator it is not self-adjoint, and thus it cannot define a proper representation of an observable. Note that, however, the sin and cos functions with the appropriate boundary condistions are true eigenfunctions of $\hat{H}=\hat{p}^2/(2m)$ and thus $\hat{H}$ is not only Hermitean but also self-adjoint and thus represents an observable (here the Hamiltonian for a particle confined in an infinitely high square well).

Thanks for the reply. Sorry if I am being annoying for asking too many things, but how can one find out that the momentum operator is hermitian but not self-adjoint in this case? I mean, I understand what you are saying, but how would one technically find out what the operator is(what computations must he do?)?

 

[Ravi Mohan]

Post #13 cites a very good source which demonstrates that computation (page 9-14).

 

[Joker93]

Yeah ,right.
So, the momentum operator being Hermitian means that its eigenvalues are real But, what does it mean if it's also self-adjoint? I mean, in my introductory QM course, they just said that if an operator is Hermitian, it has real eigenvalues and orthogonal eigenfunctions. So, what does the lack of self-adjointness mean in general? I can see that there are problems with the momentum operator in this case, but how can we state more generally what the lack of self-adjointness means?

 

[Ravi Mohan]

A quote from that article:

One may wonder whether it is possible to characterize in another way the little “extra” that a Hermitian operator is lacking in order to be self-adjoint. This missing item is exhibited by the following result which is proven in mathematical textbooks. If the Hilbert space operator A is self-adjoint, then its spectrum is real [6, 8][13]-[18] and the eigenvectors associated to different eigenvalues are mutually orthogonal; moreover, the eigenvectors together with the generalized eigenvectors yield a complete system of (generalized) vectors of the Hilbert space4 [19, 20, 8]. These results do not hold for operators which are only Hermitian. As we saw in section 2.1, this fact is confirmed by our previous example: the Hermitian operator P does not admit any proper or generalized eigenfunctions and therefore it is not self-adjoint (as we already deduced by referring directly to the definition of self-adjointness)

And that is what you need for the observable postulate of Quantum Mechanics. The observables should correspond to Self-Adjoint operators and not to the operators that are only Hermitian. Only-Hermitian operators don't have complete set of eigenvectors.

You may want to read a blog post I wrote few years ago https://ravimohan.net/2013/07/25/particle-in-a-box-infinite-square-well/

 

[Joker93]

A quote from that article:And that is what you need for the observable postulate of Quantum Mechanics. The observables should correspond to Self-Adjoint operators and not to the operators that are only Hermitian. Only-Hermitian operators don't have complete set of eigenvectors.

You may want to read a blog post I wrote few years ago https://ravimohan.net/2013/07/25/particle-in-a-box-infinite-square-well/

Oh, I missed that. So, by not being self-adjoint, it does not have proper eigenfunctions. But, this is valid for this particular case. Because for the free particle for example, the momentum operator has well-defined eigenfuctions(plane waves). So, what's up with that?

P.S. Your blog seems useful! Thanks

 

[vanhees71]

For a more formal treatment, see

http://arxiv.org/abs/quant-ph/0103153

 

[Ravi Mohan]

Oh, I missed that. So, by not being self-adjoint, it does not have proper eigenfunctions. But, this is valid for this particular case. Because for the free particle for example, the momentum operator has well-defined eigenfuctions(plane waves). So, what's up with that?

P.S. Your blog seems useful! Thanks

Actually the existence of the well defined eigenfunctions is not enough. They have to form a complete set and satisfy the boundary conditions. The enginefunctions of momentum operator in this problem don't form a complete set. You may want to read post #4.

 

[Joker93]

For a more formal treatment, see

http://arxiv.org/abs/quant-ph/0103153

Please correct me if I am wrong, but from the paper that you provided a link to, I concluded that the self-adjointness of the momentum operator depends on the problem that we examine, right? Because the momentum operator is hermitian(and this is independent of the problem-again correct me if I am wrong) and for it to be self-adjoint it's domain must be the same as the domain of its hermitian conjugate. In the case of the free particle, their domains are the same, so the momentum operator is also self-adjoint, right?

 

[Joker93]

Actually the existence of the well defined eigenfunctions is not enough. They have to form a complete set and satisfy the boundary conditions. The enginefunctions of momentum operator in this problem don't form a complete set. You may want to read post #4.

Yes, I understand that, but in the case of the free particle, all these conditions are satisfied, so the momentum operator is self-adjoint. So, am I right to conclude that the self-adjointness of an operator depends on the given problem?

 

[Ravi Mohan]

Yes, I understand that, but in the case of the free particle, all these conditions are satisfied, so the momentum operator is self-adjoint. So, am I right to conclude that the self-adjointness of an operator depends on the given problem?

Ah, I am sorry. I didn't read the "free particle". In that case you are correct. Also, as you said, the nature of an operator depends on the given physical system.

 

[Joker93]

Ah, I am sorry. I didn't read the "free particle". In that case you are correct. Also, as you said, the nature of an operator depends on the given physical system.

If you have some extra time, could you please show me how to conclude if the momentum operator is self-adjoint or not in the case of the harmonic oscillator? Is it easy to show if it's self-adjoint? I am guessing that it is since there are no boundary conditions, so again the domains of P and P+(dagger) are the same, right?