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Box normalization of plane wave

  1. Oct 13, 2011 #1
    Hi everybody.
    I'm beginning my first course on quantum physics, and our professor introduced the box normalization for plane waves.
    My question is: why do we need to impose conditions on the borders? I haven't been able to find any explanation on the internet, as every text I found just introduce these conditions (which leads to a discretization of the spectrum) without explaining the physical motivation behind them.
    I asked my professor, he replied in quite elusive way: he said that such condition is there because it should stand for any kind of wave function, even the constant one (which would fail to satisfy any other kind of bonduary condition). This is rather a reply to "why this particular condition (ψ(x, y, z)=ψ(x+L, y, z)=ψ(x, y+L, z)=ψ(x, y, z+L))?" instead of my question, that is "why impose any condition at all?"

    Thanks in advance.
  2. jcsd
  3. Oct 13, 2011 #2
    I'm not sure I know precisely what "box normalization" entails, but if you are referring to a particle confined within boundaries, then we are able to take the boundary conditions for psi as zero...because the particle (wave) is CONFINED.... within the boundary.... so there is zero probability of it being at the boundary.

    The classic analogy of standing waves(stationary states) is the stretched string between rigid supports....it can't move at the fixed ends and this confinement (or bounding) forces the emergence of energy quantization....analogous to the quantized energy levels of an electron BOUND to a nucleus.
  4. Oct 13, 2011 #3
    no, that's not my case.
    This treatment is almost equal to my reference book: http://qfizik.upm.edu.my/phy3601/lecture05.pdf page 5.
    You see that the condition is not ψ(x, y, z)=ψ(x+L, y, z)=ψ(x, y+L, z)=ψ(x, y, z+L)=0, but the value at the boundary can be non-zero. So there can't be an analogy with stationary waves. But most importantly, the mere assumption that there's 0 probability of finding my particle out of the box doesn't imply any precise value at the limits of the box.
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  5. Oct 13, 2011 #4
    We require the wave function to be continuous, which means it shouldn't drop from some finite value instantly to zero at the boundary of the box. Recall that the momentum operator is proportional to d/dx, and the derivative of a discontinuous function is infinite at the discontinuity. Such an infinite concentration of momentum is unphysical. A wave function that was "close to discontinuous," i.e. dropped extremely sharply from some finite value to zero, would very quickly smooth itself out.
  6. Oct 14, 2011 #5
    The_Duck, you're assuming that this bonduary value is 0, but the lecture I referenced does not impose the equality to 0.
    Also, the equation of a plane wave, nor it squared absolute value (which has the physical significance of probability density of finding the represented particle in the specified (x, y, z, t) coordinates), can ever be 0: [itex]A e^{i(ωt-kr)}[/itex] is 0 only if A=0, which means, nothing interesting happens in my box.
  7. Oct 14, 2011 #6


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    It's useful to "put the system in a box" for several reasons.

    First, just having it in infinite space, without boundary conditions, leads to problems. For example, [itex]\exp(kx - iEt)[/itex] with [itex]k[/itex] real and [itex]E=-\hbar^2k^2/2m[/itex] solves the free-particle Schrodinger equation. This solution has negative energy, and is "unphysical".

    Second, we cannot normalize plane waves in infinite space. Working with normalizable wave packets is possible, but much more complicated.

    So, to put the system in a box, we have to specify boundary conditions on the walls of the box.

    A simple, physically motivated boundary condition is to require the wave function to vanish on the walls. This is what we would get by putting an arbitrarily large potential energy everywhere outside the box.

    Another, much less well motivated, set of boundary conditions is to assume that the box is actually a torus, so that moving across the box brings you back to the starting point. This is "periodic boundary conditions".

    It turns out that, for many purposes (and in particular for any question whose answer does not depend on the size of the box when the box is very large), it does not matter which boundary condition we use; we get the same results for these kinds of questions. In this case, it's often mathematically simpler to use periodic boundary conditions, and so this is a common choice.
  8. Oct 14, 2011 #7
    Why? I'm missing something big here. Putting something in a box from a mathematical standpoint means setting the wave function to zero elsewhere, but not touching anything within and at the boundary of the box.
    This cannot happen with a plane wave, again, from a mathematical standpoint.
    This, at last, makes some sense. Even though I'm a beginner in QM, could you give me a glimpse of why this happen?
  9. Oct 15, 2011 #8
    To see why we need boundary conditions, and why we need continuity of the wave function at the boundary, consider the classical mechanics problem of waves on a string. To solve this problem for a finite length of string you need to what is physically happening at the ends of the string. This knowledge of the physics enables you to write down boundary conditions, which are a prerequisite to solving the mathematical problem. You can tie the ends of the string down, corresponding to the boundary condition y(0) = y(L) = 0. You can tie the ends of the string together, making it a loop, corresponding to the boundary condition y(0) = y(L). You can attach the ends of the string to massless metal rings which slide freely on vertical rods, corresponding to something like y'(0) = y'(L) = 0. To answer the important question of what the normal modes of the string are, you need to know the physics of the boundaries, which when understood supply a mathematical prescription for the boundary conditions on y(x).

    For instance, let us choose to tie the ends of the string down. Now we need to solve the differential equation (d^2 y/dt^2) = 1/c^2 (d^2 y/dx^2). How about the following solution: y(x, t) = C, an arbitrary constant displacement, for 0 < x < L, and 0 outside that range. Over the string's length this solves the equation of motion, doesn't it? It even obeys the boundary condition y(0) = y(L) = 0. But this is clearly not a physically reasonable solution: the string's displacement jumps discontinuously from 0 to L as you go from x=0 to x=epsilon. Real strings don't get into such states. We should also impose continuity on the string displacement y(x). Mathematically it is NOT enough simply to set y(x) = 0 outside 0 < x < L.

    Similarly if you want to do the quantum mechanical problem of a particle in a finite box, the physics of the box boundary implies certain boundary conditions on psi(x). If you are confining the particle to a box by making the potential large outside the allowed region, this corresponds to psi(boundary) = 0. We must also remember continuity, so a plane wave within the box that cuts off discontinuously at the boundary is unphysical. Instead in this case the energy eigenstates look like sin(kx) with k chosen so that this vanishes on the boundary.

    The periodic boundary conditions you are working with correspond to a particle moving on the boundary of a circle, so that x=0 and x=L are actually the same position; this circumstance then obviously requires psi(0) = psi(L), which are your boundary conditions.

    And indeed, when one's boundary conditions are psi(boundary) = 0, the plane waves exp[ikx] are not among the states one considers. You are using different boundary conditions, corresponding to a different physical situation.

    Avodyne obliquely referred to one reason. You can put a system in a box, of some sort, but if the system is much smaller than the box then it won't know the different between being in an extremely large box and being in free space. So you can consider it to be in a box or not and get the same result either way. For this reason the precise boundary conditions you use don't matter, since the system never sees the boundary. But imagining the system to be in a very large box can be convenient because then the energy and momentum eigenstates are discrete instead of continuous.
  10. Nov 5, 2011 #9
    No offense, but I didn't reply because I thought that after many clarifications of my doubt, all the replies that I got weren't satisfactory, and failed to get my point.

    But, finally, I found in my reference book something that explains it, and here it is, so you can understand what I was asking.

    The true reason for that boundary condition (which, let me say this again, is not about setting the wave function to equal to 0 at the bounds) is that without them you couldn't define the adjoint of a differential operator. Given the standard definition of an adjont F* of an operator F, that is F* satisfies

    1, F ψ2)=(F* ψ1, ψ2)

    consider the easiest differential operator, ∂/∂x, which operates on L2(a, b) (a and b can be finite or infinite) and let's find its adjoin using the integral form of the above scalar product:

    [itex]\int_a^b \psi _1(x){}^* \psi _2'(x) \, dx=\left[\psi _1(x){}^* \psi _2(x)\right]_a^b+\int_a^b \left(-\psi _1'(x){}^*\right) \psi _2(x) \, dx[/itex]

    Now, you can define the adjoin of ∂/∂x, that is (∂/∂x)*=-∂/∂x, if and only if the term out of integral is zero, and this happens if and only if ψ(a)=ψ(b), which is exactly the obscure boundary condition that we set when normalizing a plane function in a box. Moreover, if L2(a, b)=L2(-∞, ∞), then of course we request that ψ(-∞)=ψ(∞)=0.

    This whole problem now depends on the reason that we can't construct a quantum mechanics without differential operators. But that's another topic. I'm going to investigate what happens in momentum-space, where differential operators become simple multiplications, if I don't set these conditions.
  11. Nov 5, 2011 #10


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    I would say the boundary conditions follow from the standard Sturm-Liouville theory for differential equations. Remember that the Hamiltonian and the momentum operators in coordinate representation are actually differential operators, so that any spectral problem for energy or momentum asks to solve a Sturm-Liouville type ODE. Requirement of self-adjointness imposes the boundary conditions for the solutions of the ODE's. So post 9 by the OP contains the quoted boundary condition as being necessary for the adjoint of the momentum operator to exist. If the boundary condition is not applied, then there's no unique adjoint for p, thus p cannot be rendered self-adjoint.
  12. Nov 7, 2011 #11


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    I would say it is about completeness vs overcompleteness of a Fourier basis. A function defined on a finite interval 0<x<L may be expanded either in sin(2 pi n x/L) functions or cos(2 pi n x/L) functions. Each of the two bases is complete. But if you take both of them, then you deal with an overcomplete basis, which is something you don't want. So you must take ONLY ONE of them. It's not really important which one, but whatever it is, it corresponds to one boundary condition or another.
  13. Nov 8, 2011 #12


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    I think, the two bases belong to different boundary conditions, and in fact it makes a difference which basis you use. Of course, for the box problem with rigid boundary conditions there is trouble with the definition of a momentum operator.

    Thus, if you want to use the "box quantization" as a regulator for the full real line as position space, you should take periodic boundary conditions, and then you need both sets of functions or, equivalently, the more convenient basis

    [tex]u_{p}(x)=\frac{1}{L} \exp(\mathrm{i} p x), \quad p \in \frac{2 \pi}{L} \mathbb{Z}.[/tex]

    Here, the momentum operator is well defined as a self-adjoint operator, and the limit [tex]L \rightarrow \infty[/tex] for physical quantities is usually not too difficult to take.
  14. Nov 8, 2011 #13


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    I think it makes a difference at the boundary itself, but not at points arbitrarily close to the boundary. I mean, both bases are complete in the interval 0<x<L.

    Of course, if the box is a physical object, then the boundary condition has a physical root. But if the box is only a mathematical trick to make non-normalizable functions normalizable, then it really doesn't matter which boundary condition you take.

    By the way, pisto in #9 also has presented a very good argument in context of QM. But expansion in plane waves may be relevant even for classical waves, for which there is no need to introduce adjoints of operators.
    Last edited: Nov 8, 2011
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