Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Particle in a box in momentum basis

  1. May 28, 2013 #1
    I have been thinking about it for sometime but couldnt 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 = [itex]\int[/itex]|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>).
     
  2. jcsd
  3. May 28, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
  4. May 28, 2013 #3
    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)
    Can you please elaborate how?
     
  5. May 29, 2013 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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 [itex]\psi(x)[/itex]. since the potential goes to infinity at the boundaries, your particle is strictly restricted to the interval [itex][-L/2,L/2][/itex], and the wave function must fulfill the boundary conditions [itex]\psi(-L/2)=\psi(+L/2)=0.[/itex]

    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
    [tex]\hat{p}=-\mathrm{i} \mathrm{d}_x.[/tex]
    Here and in the following I set [itex]\hbar=1[/itex] (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
    [tex]\hat{p} u_p(x)=p u_p(x)[/tex]
    with real eigen values [itex]p[/itex]. Of course, you get
    [tex]u_p(x)=N \exp(\mathrm{i} p x)[/tex]
    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 [itex]L^2([-L/2,L/2]).[/itex] The operator [itex]\hat{p}[/itex] 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
    [tex]\hat{H}=\frac{1}{2m} \hat{p}^2=-\frac{1}{2m} \mathrm{d}_x^2.[/tex]
    Here the eigenvalue equation reads
    [tex]\hat{H} u_E(x)=E u_E x, \quad u_E''(x)=-2m E u_E(x).[/tex]
    The solutions read
    [tex]u_E(x)=A_1 \cos(k x)+A_2 \sin(k x),[/tex]
    where [itex]k=\sqrt{2m E}[/itex]. There are only solutions for [itex]E >0[/itex] 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
    [tex]u_E(L/2)=u_E(-L/2)=N \cos \left (\frac{k L}{2} \right ) \stackrel{!}{=}0.[/tex]
    This leads to
    [tex]\frac{k L}{2} = \frac{2n+1}{2} \pi \; \Rightarrow k_{n}^+=\frac{(2n+1)\pi}{L}.[/tex]
    For odd parity you get
    [tex]u_E(L/2)=-u_E(-L/2)=N \sin \left (\frac{k L}{2} \right ) \stackrel{!}{=}0.[/tex]
    This leads to
    [tex]\frac{k L}{2}=n \pi \; \Rightarrow \; k_n^-=\frac{2 n \pi}{L}.[/tex]
    You have a complete set of eigenfunctions on [itex]L^2([-L/2,L/2]).[/itex]
    [tex]u_n^{(+)}(x)=N \cos(k_n^+ x), \quad u_n^{-}(x)=N \sin(k_n^- x).[/tex]
    The corresponding energy eigenvalues are
    [tex]E_n^{\pm}=\frac{(k_n^{\pm})^2}{2}.[/tex]

    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.
     
  6. May 29, 2013 #5
    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.
    But I dont 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 [itex]V(\hat{x})[/itex] to [itex]V(x)[/itex]". But I am begining to believe that we cant work in momentum basis everytime.


    Thus the states I want to work with must be represented by [itex]\widetilde{\psi}(p)[/itex] (the fourier transform of [itex]\psi(x)[/itex]). 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?
     
    Last edited: May 29, 2013
  7. May 30, 2013 #6
  8. May 30, 2013 #7

    Jano L.

    User Avatar
    Gold Member

    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.
     
  9. May 30, 2013 #8
    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!
     
  10. May 31, 2013 #9

    Jano L.

    User Avatar
    Gold Member

    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)##.
     
  11. May 31, 2013 #10

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    What's [itex]U(p)[/itex] 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
    [tex]U(x)=\begin{cases}
    -U_0 & \text{for} \quad -L/2<x<L/2,\\
    0 & \text{elsewhere}.
    \end{cases}
    [/tex]
    The matrix element in momentum space is thus
    [tex]\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 ].[/tex]
    The time-independent Schrödinger equation then reads
    [tex]\frac{p^2}{2m} u_E(p) + \int_{\mathbb{R}} U(p,p') u_E(p')=E u_E(p).[/tex]
    Have fun with the solution ;-)).
     
  12. May 31, 2013 #11

    Jano L.

    User Avatar
    Gold Member

    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.
     
  13. May 31, 2013 #12
    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 dont form complete set of basis :approve:. 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.
     
    Last edited: May 31, 2013
  14. May 31, 2013 #13

    Jano L.

    User Avatar
    Gold Member

  15. Jun 1, 2013 #14
    Nice find. Should keep me busy for a week or two. Thanks a ton.
     
  16. Jun 1, 2013 #15
    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
     
  17. Jun 1, 2013 #16

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

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

    This is discussed in the very nice paper by Gieres as Example 4!
     
  18. Jun 1, 2013 #17

    Jano L.

    User Avatar
    Gold Member

    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.
     

    Attached Files:

  19. Jun 2, 2013 #18

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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!
     
  20. Jun 2, 2013 #19
    I take that to mean that the boundary conditions just aren't physical
     
  21. Jun 2, 2013 #20

    Jano L.

    User Avatar
    Gold Member

    ##\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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Particle in a box in momentum basis
Loading...