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Momentum eigenstates - particle in a box

  1. Nov 9, 2015 #1

    dyn

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    Hi. I am just starting to self-study QFT. Have been looking at the non-relativistic case of particles in a box. Have come across creation operators in the occupation number representation which create a particle in specific momentum state. But I thought in a rigid box there are no momentum eigenstates so how can a particle be created in a specific momentum state ?
    Also when describing a state in position coordinates which is done by taking the inverse Fourier transform of the state in momentum coordinates the integral is replaced by a sum as the momentum is quantized. I have never seen this before in QM. Is it sometimes assumed or ignored or should it always be written with a sum if the momentum is discrete ?
    Thanks
     
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  3. Nov 10, 2015 #2

    kau

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    If we have periodicity in the wavefunction in position space then momentum is quantized and we can have fourier series expansion of the wavefunction in terms of momentum wavefunction...
    Say f(x) is periodic in x [0,L]...Then we have fourier series expansion ##f(x)=\sum_{k} f_{k}e^{[ikx]}## ..##f_{k}=1/L\int f(x) e^{[-ikx]},dx## Now if we have L-> infinity,meaning very large then spacing between discrete momenta is very small,since k goes like 1/L.. In that case we can replace the sum by an integral..##\sum_{k}=(L/2\pi)\int dk##
    In case of a box if you put periodicity then momentu is well defined and you can have momentum description..Translational symmetry is what all we need.
     
  4. Nov 10, 2015 #3

    dyn

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    So for a rigid box , momentum is quantized but momentum eigenstates do not exist ? But if they don't exist how can the creation operators place particles in states of definite momentum ?
     
  5. Nov 11, 2015 #4

    kau

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    By a rigid box what do you mean? At x=0 and at x=L (for a box of length L) the potential is infinity. If that is the case then imposing the periodic boundary condition in the wavefunction, momentum eigenstates are well defined....I mean I donno why it should not be.... May be you can explain it to me.
     
  6. Nov 11, 2015 #5

    A. Neumaier

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    Talking about momentum presupposes the existence of a translation group, whose infinitesimal generator is the momentum. This is the case iif position space is $R^d$ with either asymptotix zro, or periodic boundary conditions (or a combination of both).

    A finite box (unlike a periodic box or a line) has no translation operators, hence no momentum (which generates translation), hence no corresponding states with definite momentum.

    Thus if you want to think about momentum states you should not consider infinite barriers at the walls but a periodic continuation.
     
  7. Nov 11, 2015 #6

    dyn

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    On a previous thread I was told that momentum eigenstates do not exist in a rigid box ,ie with infinite potential at the walls
     
  8. Nov 11, 2015 #7

    A. Neumaier

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    yes, you cannot have walls and momentum states.

    So if you want to consider a box, you need to assume periodic boundary conditions. (Then the momentum is quantized.)
     
  9. Nov 11, 2015 #8

    dyn

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    So to get momentum eigenstates I need to assume periodic boundary conditions. I presume periodic boundary conditions does not apply to an isolated rigid box with infinite walls ? So with momentum eigenstates and creation/annihilation operators we must assume a regular lattice of boxes ie. like solid state physics ?
     
  10. Nov 11, 2015 #9

    Avodyne

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    In a one-dimensional box of length ##L## with infinite-potential walls, the single-particle states have definite energy, ##E_n = \pi^2\hbar^2 n^2/2mL^2##, ##n=1,2,3,\ldots##. We then define creation and annihilation operators ##a^\dagger_n## and ##a_n## with the commutation relations ##[a_n,a_{n'}]=[a^\dagger_n,a^\dagger_{n'}]=0## and ##[a_n,a^\dagger_{n'}]=\delta_{nn'}##. The hamilonian is then ##H=\sum_n E_n a^\dagger_n a_n##. This describes a system of non-interacting non-relativistic bosons in a one-dimensional box.
     
  11. Nov 11, 2015 #10

    dyn

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    Thanks. But for those commutation relations to exist we need momentum eigenstates which requires periodic boundary conditions ie. ψ(x+L) = ψ(x) but for a truly rigid infinite well ψ = 0 for x greater than L so are periodic boundary conditions a bit of a "fudge" to create quantized momenta ? If the walls are treated as high but not infinite the wavefunction would decay exponentially beyond the walls so how do we achieve ψ(x+L) = ψ(x) ?
     
  12. Nov 12, 2015 #11

    A. Neumaier

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    No. The c/a operators given by Avodyne create energy eigenstates, not momentum eigenstates! Momentum eigenstates need a finite, periodic potential.

    Note that c/a operators cannot be defined only in a momentum representation but also in a position representation, or an energy representation.
     
  13. Nov 12, 2015 #12

    vanhees71

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    Let's have a look at a particle in 1 dimension in a box with (a) periodic boundary conditions (that's equivalent for a particle on a circle) and (b) with rigid boundary conditions (i.e., a particle in a flat potential with very high walls; simplified by a "box potential", ##V(x)=0## for ##x \in [-L/2,L/2]## and ##V(x) \rightarrow \infty## for ##|x|>L/2##) and look for momentum eigenstates. For simplicity I use natural units with ##\hbar=1##.

    First of all one should define the momentum operator. It's by definition the generator of spatial translations and thus in the position representation is given by the operator
    $$\hat{p}=-\mathrm{i} \partial_x.$$
    That's not enough to know whether this operator makes sense as the representative of an observable. For that purpuse it should be self-adjoint. I'm not using the rigorous way to maximally extend this operator on Hilbert space but use the usual physicists' arguments.

    Let's start with case (a) periodic boundary conditions. This means we work on the representation of the separable Hilbert space ##\mathcal{H}=\mathrm{L}^2([-L/2,L/2])## of periodic functions, i.e., ##\psi(-L/2)=\psi(L/2)##. The first step to see, whether the momentum operator makes sense, is to check whether it's Hermitean. That means for any functions in its domain we should have
    $$\int_{-L/2}^{L/2} \mathrm{d} x \psi_1^*(x) \hat{p} \psi_2(x) = \int_{-L/2}^{L/2} \mathrm{d} x [\hat{p} \psi_1^*(x)]\psi_2(x).$$
    It is very easy to see that this is really true by integrating by parts.
    Further for any wave function also ##\hat{p} \psi(x)## should also be in this function space, and that's the case, because the derivative (if existent) of a periodic function is again periodic. So indeed for this case ##\hat{p}## is a very good candidate for a self-adjoint operator.

    If it is a self-adjoint operator, there should be a complete set of eigenstates with real eigenvalues. That also easy to check. The eigenstate equation reads
    $$\hat{p}u_p(x)=-\mathrm{i} \partial_x u_p(x) \stackrel{!}{=} p u_p(x) \; \Rightarrow \; u_p(x)=A \exp(\mathrm{i} p x), \quad A=\text{const}.$$
    Next we have to fulfill the periodic boundary conditions,
    $$u_p(-L/2)=u_p(L/2) \;\ Rightarrow \; \exp(\mathrm{i} p l)=1 \; \Rightarrow \; p=\frac{2 \pi}{L} n, \quad n \in \mathbb{\Z}.$$
    To normalize the state we calculate
    $$\int_{-L/2}^{L/2} |u_p(x)|^2=|A|^2 L \stackrel{!}{=}1 \; \Rightarrow \; A=\frac{1}{\sqrt{L}}.$$
    Now obviously the set of eigenfunctions
    $$u_n(x)=\frac{1}{\sqrt{L}} \exp(2 \pi \mathrm{i} x/L)$$
    is a complete set of orthnormal states, because you can express any periodic function in terms of its Fourier series
    $$\psi(x)=\sum_{n=-\infty}^{\infty} \psi_n u_n(x) \; \Leftrightarrow \; \psi_n=\langle u_n|\psi \rangle = \frac{1}{\sqrt{L}} \int_{-L/2}^{L/2} \exp(-2 \pi x/L) \psi(x).$$
    Also ##\hat{p} u_n(x)## is obviously again in the considered Hilbert space. So for the periodic boundary conditions we have a well-defined self-adjoint momentum operator. Note that this can be made rigorous, using both the conventional and the modern rigged-Hilbert space approach. See, e.g.,

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

    Now for the case of "rigid" boundary conditions. Now we have the realization of the Hilbert space as ##\mathrm{L}^2([-L/2,L/2])## with boundary conditions ##\psi(-L/2=\psi(L/2)=0##. Now already the first condition doesn't work out, because in general if ##\psi(x)## is in the domain of ##\hat{p}##, ##\hat{p} \psi(x)## does not fulfill the boundary conditions. On the other hand, for an ##\mathrm{L}^2## function you can alter the function's values on any countable set of points in the interval ##[-L/2,L/2]##, because it's defined only up to Lebesgue null functions. The Hermitezity check also works out in the same way as in the previous case.

    So the real killer argument is the evaluation of the eigenfunctions. It turns out that these are the parity even and parity odd functions
    $$u^{(+)}_n(x)=A \cos(p_n^{(+)} x), \quad p_n^{(+)}=\frac{(2 n+1) \pi}{2L}, \quad n \in \mathbb{N}_0=\{0,1,2,\ldots \},$$
    and
    $$u^{(-)}_n(x)=A \sin(p_n^{(-} x), \quad p_n^{(-)}=\frac{2 n \pi}{2L}, \quad n \in \mathbb{N}=\{1,2,\ldots \}.$$
    Here you see that it doesn't work out, because ##\hat{p} u^{(\pm)}## does not fulfill the rigid boundary conditions, and thus indeed ##\hat{p}## is not extendable from a Hermitean to a self-adjoint operator.

    However, the ##u_n^{(\pm})## are a complete set of orthogonal functions on this Hilbert space. The reason is also simple to see, because obviously ##\hat{H}=\hat{p}^2/(2m)## is extendable to a self-adjoint operator with the ##u_n^{(\pm)}## as a complete set of eigenfunctions. The eigenvalues of the Hamiltonian are ##E_n^{(\pm)}=(p_n^{(\pm)})^2/(2m)##.
     
  14. Nov 12, 2015 #13

    dyn

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    Thanks guys for your replies. Could someone please clarify the crux of my question - for momentum eigenstates in a box the walls cannot be infinite as we need periodic boundary conditions ?
     
  15. Nov 12, 2015 #14

    A. Neumaier

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    Yes, this is the crux. Is it clear now?
     
  16. Nov 13, 2015 #15

    dyn

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    Yes. Thank you.
     
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