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I Why must the wave function be continuous in an infinite well?

  1. May 30, 2017 #1
    It is required to be continuous in the following text:
    Screen Shot 2017-05-31 at 4.06.31 AM.png

    The book's reason why wave functions are continuous (for finite V) is as follows. But for infinite V, ##\frac{\partial P}{\partial t}=\infty-\infty=## undefined, and so the reason that wave functions must be continuous is invalid.
    Screen Shot 2017-05-31 at 4.09.14 AM.png

    Source: Quantum Physics 3rd ed. by Stephen Gasiorowicz, p. 48
     
    Last edited: May 30, 2017
  2. jcsd
  3. May 30, 2017 #2

    PeterDonis

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    What source is this from?
     
  4. May 30, 2017 #3
    Quantum Physics 3rd ed. by Stephen Gasiorowicz, p. 48
     
  5. May 31, 2017 #4

    vanhees71

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    I don't understand the book's argument either, and the infinite-well example is everything else than simple. Ironically it is presented to students in QM 1 as if it were the most simple example.

    To answer your question we have to look at the Hamiltonian, defined on the Hilbert space ##L^2([0,a])##,
    $$\hat{H}=-\frac{1}{2m} \mathrm{d}_x^2.$$
    I use natural units with ##\hbar=1##.

    The Hamiltonian must be a self-adjoint operator, and this defines the domain, i.e., the function, where it is defined. Obviously the function should be differentiable twice. Now we check hermiticity. Integrating twice by parts you get
    $$\int_0^a \mathrm{d} x \psi_1(x)^* \psi_2''(x) = (\psi_1^* \psi_2'-\psi_1^{*\prime} \psi_2)|_0^a + \int_0^a \mathrm{d} x \psi_1'' \psi_2.$$
    For the Hamiltonian to be at least hermitean, the non-integral piece must vanish. This doesn't constrain the boundary conditions very much. Indeed possible conditions are (a) ##\psi(0)=\psi(a)=0##, (b) ##\psi(0)=\psi'(a)=0##, (c) ##\psi'(0)=\psi(a)=0##, (d) ##\psi## is a periodic function on entire ##\mathbb{R}##, restricted to the interval ##[0,a]##.

    So we need more constraints. One idea is to demand that ##\hat{x}##, defined as ##\hat{x} \psi(x)=x \psi(x)##, should be a self-adjoint operator. That it is Hermitean is trivial, but to be self-adjoint implies that the domain of ##\hat{x}## equals its co-domain. If we now check the above examples of conditions we see that (a) for sure fulfills this constraint, because if ##\psi(0)=\psi(a)=0## then also ##\hat{x} \psi(x)=x \psi(x)## fulfills this condition, and with ##\psi## also ##x \psi## is square integrable over the finite interval ##[0,a]##. So these boundary conditions make at least sense.

    Now we have to see, whether also ##\hat{H}## or for this matter ##\mathrm{d}_x^2## is indeed self-adjoint. To that end we evaluate the eigenfunctions of this operator
    $$u_{E}''(x)=-\epsilon u_E(x), \quad \epsilon=2 m E.$$
    The solutions are
    $$u_{E}(x)=A \sin(\sqrt{\epsilon} x)+B \cos(\sqrt{\epsilon} x).$$
    Since ##u_{E}(0)=0## we necessarily have ##B=0##. To also fulfill
    $$u_{E}(a)=A \sin(\sqrt{\epsilon} a)=0,$$
    we must have
    $$\sqrt{\epsilon} a=n \pi, \quad n \in \mathbb{Z}.$$
    Tha means the ##\epsilon \geq 0##. For ##n=0## the solution becomes identically 0. So it's not a solution either, because eigenfunctions must not be the null vector of the Hilbert space. For ##n>0## the solution with ##-n## leads to the same function up to a factor ##-1##, so it's no new solution. Thus the complete set of eigenvectors are
    $$u_{n}(a)=A \sin(\sqrt{\epsilon_n} a), \quad \epsilon_n=\left (\frac{n \pi}{a} \right)^2, \quad n \in \mathbb{N}=\{1,2,3,\ldots \}.$$
    Since
    $$u_n''(x)=-\epsilon u_n(x)$$
    fulfills the boundary conditions and since any function doing so can be written as a Fourier series on this interval,
    $$\psi(x)=\sum_{n=1}^{\infty} A_n \sin(n \pi x/a),$$
    The domain and the codomain of the Hermitean operator ##\hat{H}## are the same, and thus ##\hat{H}## is self-adjoint on the here considered space. Everything is a consistent description of a particle in the infinite-potential box.

    Some puzzle to think about: What about the operator ##\hat{p}=-\mathrm{i} \mathrm{d}_x## (a candidate for a momentum operator)? Is it Hermitean? Is it self-adjoint?
     
  6. Jun 1, 2017 #5
    1. How does case (d) make the non-integral piece vanish?

    If ##\psi(x)## is periodic, then ##\psi(0)=\psi(a)##. Then ##(\psi_1^* \psi_2'-\psi_1^{*\prime} \psi_2)|_0^a=\psi_1^*(0)\big(\psi_2'(a)-\psi_2'(0)\big)+\psi_2(0)\big(\psi_1^{*'}(0)-\psi_1^{*'}(a)\big)##, which may not vanish.

    2. I think you missed out one case: ##\psi'(0)=\psi'(a)=0##. Is that right?
     
  7. Jun 1, 2017 #6

    vanhees71

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    In case 1 both ##\psi## and ##\psi'## are periodic. Concerning 2. you are right. It would be interesting to check out these cases too. I think this example is far from being as trivial as stated in the textbooks. There they simply give the boundary conditions ##\psi(0)=\psi(a)=0##, but it's not clear, why these are the correct ones.

    Everything is clear for finite potential wells. Maybe one should investigate which boundary conditions come out when investigating the finite potential well and then taking the limit to the infinite potential.
     
  8. Jun 1, 2017 #7
    For an infinite potential well the wave function is zero at the boundaries. If the potential is infinite there is not probability of it extending any way into the region beyond the wall.
     
  9. Jun 1, 2017 #8
    That only explains why ##u(x)=0## for ##x<0## and for ##x>a##, but not for ##x=0## and for ##x=a##, since the book's reason for requiring the wave function to be continuous fails at the boundaries, where the potential is infinite.
     
    Last edited: Jun 1, 2017
  10. Jun 1, 2017 #9
    Why does it need to be continuous at the boundary if it cannot be found there?
     
  11. Jun 1, 2017 #10
    So are you saying that a zero probability of finding the particle at ##x=0##, ##P(x=0)=0##, implies that ##\psi(0)=0##? I don't see how this is implied. Even if ##\psi(0)\neq0##, ##P(x\leq0)=\int_{-\infty}^0\psi^*(x)\psi(x)dx## still vanishes. So a non-vanishing wave function at ##x=0## can still be consistent with the fact that it cannot be found there.
     
  12. Jun 2, 2017 #11

    jtbell

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    I agree. The infinite well is an idealization that cannot be attained in practice, but is useful for calculations. Years ago, I made graphs of ##\psi## for finite wells of various depths, which illustrated that ##\psi(0)## and ##\psi(a)## become smaller as the well becomes "deeper." However, I did not try to set up a formal mathematical limit to show that they both ##\rightarrow 0## as the depth ##\rightarrow \infty##.
     
  13. Jun 2, 2017 #12
    Why take minus infinity as the bottom of the integral? Could you consider a bigger number and a smaller range for the integral?
     
  14. Jun 2, 2017 #13

    bhobba

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    As others have alluded to its simply an exercise in QM that you find in beginner and intermediate texts. One makes 'reasonable' assumptions in analysing it at that level, but it cant actually occcur in practice.

    Continuity is basically an assumption you make because of Schrodinger's equation - if you can take derivatives it must be continuous.

    However the whole thing is clouded by much more advanced treatments where the concept of Rigged Hilbert space is introduced - but I wont go into that that here except to mention its a further generalization of distribution theory which you should know about.

    For any area of applied math, physics included you should really study distribution theory:
    https://www.amazon.com/Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    The answer is the test space is continuous, infinitely differentiable etc and that is what is used - however the well problem containing a discontinuity does not fit into the test space so you imagine that it does and is so close to it, its physically indistinguishable from it.

    Thats the intuitive way of doing it - the rigorous way (using RHS's) has been discussed here before:
    https://www.physicsforums.com/threads/dimension-of-hilbert-space-in-quantum-mechanics.258277/

    Thanks
    Bill
     
  15. Jun 2, 2017 #14

    bhobba

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    Again the answer is in RHS's and distribution theory. Distribution theory is one of those areas outside physics that uses Dirac notation. In Dirac notation |f> is some set of functions called test functions. It's dual (the set of all linear functions defined on the test functions) is another space whose elements are written as <f| and the linear function is written of f applied to g ie f(g) is written as <f|g>. A simple way to get a linear function if f is some kind of ordinary function is ∫fg where the bounds are -∞ to ∞. That ∞ is used is just convention. The test spaces, be they functions of compact support or the slightly more complex good and fairly good functions (as an exercise look them up) all have interesting properties in distribution theory. For example it can be shown the Fourier transform of a good function is also a good function. This makes defining the Fourier transform of much more complex functions, and even things like the Dirac Delta function that isn't even a function in the regular sense a snap.

    Its defined this way. Let g be any such function, even weird ones, then F(g) where F is the Fourier transform is defined by <F(g)|f> = <g|F(f)>. By definition in distribution theory <f|g> is written as ∫ fg where the integral is, again by definition, -∞ to ∞. Just out of interest the space of good functions is called a Schwartz space and fairly good functions have the nice property of when multiplied by a good function it is also a good function.

    Thanks
    Bill
     
    Last edited: Jun 4, 2017
  16. Jun 4, 2017 #15

    hilbert2

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    If we could have a discontinuous wave function in an infinite potential well, it would be possible to just say that ##\psi (x)## is constant inside the well and zero outside it. Then both the total energy and the expectation value ##\left< p^2 \right>## would be zero, and this just isn't how a quantum particle behaves, it must always have a nonzero kinetic energy when confined in a finite space. A similar problem arises if we let the derivative of the wave function be discontinuous at points other than the bounding walls of the well, for instance if the well is located at ##-L \leq x \leq L## and ##\psi (x) = 1 - |x|/L## inside it and 0 outside.
     
  17. Jun 5, 2017 #16

    dextercioby

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    This is - for me - by far the most interesting problem asked on PF in a whole lot of time. Because it seems simple, but it really isn't. Let's rephrase the original problem. For a Hamiltonian operator ## H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} ## of a particle confined in a box of width L symmetric to the origin (thus, it makes sense to treat complex wavefunctions only for ##x\in I = \left[-\frac{L}{2},\frac{L}{2}\right]##), discuss all the possible boundary conditions imposed to the wavefunctions, so that the Hamiltonian is correctly defined as a linear, self-adjoint operator on the Hilbert space ##L^2 (I) ## (dense everywhere subset of).

    Introductory physics textbooks mention that - as a necessity - ##\psi \left(-\frac{L}{2}\right) = \psi \left(\frac{L}{2}\right) = 0## as the "physical boundary conditions" which are said to be the right ones to use to find the acceptable wavefunctions and the possible energy values. But this is clearly wrong, because, just as vanhees71 showed, simply requiring that the Hamiltonian be formally symmetric as a linear differential operator provides us with a larger class of boundary value solutions, namely:

    H is formally symmetric in ##L^2 (I) ## iff ## \mbox{for} ~~\phi (x), \psi (x) \in L^2 (I) ~, ~~ \phi^{*} \left(\frac{L}{2}\right)\psi ' \left(\frac{L}{2}\right) + \phi ' ^{*} \left(-\frac{L}{2}\right) \psi \left(-\frac{L}{2}\right) = \phi^{*} \left(-\frac{L}{2}\right) \psi' \left(-\frac{L}{2}\right) + \phi ' ^{*} \left(\frac{L}{2}\right) \psi \left(\frac{L}{2}\right) ## (1)

    As one can see, (1) is not satisfied only by "the physical boundary conditions" ## \phi \left(-\frac{L}{2}\right) = \phi \left(\frac{L}{2}\right) = 0 ## (2), but also by a larger set, namely all functions ##\phi (x)## for which ## \phi ' \left(-\frac{L}{2}\right) = \lambda \phi \left(-\frac{L}{2}\right)## and ## \phi ' \left(\frac{L}{2}\right) = \lambda \phi \left(\frac{L}{2}\right)## with ##\lambda \in \mathbb{R}## (3). So the Hamiltonian is then formally symmetric under a larger class of boundary conditions.

    But the Hamiltonian needn't be only symmetric, it must be self-adjoint. One shows that the Hamiltonian - in absence of boundary values for the wavefunctions, as the theory requires - is closed, symmetric and has deficiency indices (2,2), thus possesses a 2 parameter family of self-adjoint extensions. Now you are prepared to read the article by Bonneau et al. I have attached below (no worries about copyright, it comes from the ArXiv preprint server).
     
  18. Jun 6, 2017 #17

    vanhees71

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    In the mean-time I also thought of the problem from a more physical perspective, i.e., I first looked at the finitely high potential well. Indeed, it's much simpler to make the potential symmetric around ##x=0##, because then parity can be easily defined and is a good quantum number (i.e., it commutes with ##\hat{H}##, and you can separately look for energy eigenfunctions for even and odd wave functions, i.e., you need to bother only with the boundary conditions at ##x=+1/2##). For the finite potential well the wave function and its first derivative should be continuous, and this gives enough constraints. You find (a finite number of) bound states and an continuously infinite number of scattering states (with twofold degeneracy for each energy value and each parity, because in the asymptotic right you can have left- and right-moving solutions and all superpositions thereof for any given energy larger than the potential). Now one can take the limit ##V_0 \rightarrow \infty##, and you get the expected homogeneous boundary conditions, leading to a jump in the first derivative. Of course in this limit there are no more scattering states and you get the well-known countable set of bound states. The Hamiltonian is self-adjoint, but not the "would-be momentum operator", ##- \mathrm{i} \hbar \partial_x##, which is only Hermitean, but not self-adjoint.
     
  19. Jun 7, 2017 #18
    Yes you can. It doesn't matter as long as the bottom limit is not positive: the integral still vanishes. And so a non-vanishing wave function at ##x=0## can still be consistent with the fact that the particle cannot be found there and anywhere outside the well.
     
  20. Jun 9, 2017 #19
    From a physical point of view, could we think of it as, an infinite amount of energy would be required in order for a wave configuration to have a discontinuity? Or even a kink for that matter. In the Schrodinger equation form, this can be seen from the expectation value of the kinetic energy operator ##\hat{K}##: ##\left<\frac{\hbar^2}{2m}\mathbf{\nabla}^2\right>##. In the hydrodynamical/De-Broglie-Bohm formalism, where one writes the wavefunction in the polar form ##\psi=\sqrt{\rho}e^{i\frac{\theta}{\hbar}}##, the Schrodinger equation $$i\hbar\partial_t\psi=\left[\hat{K}+\hat{V}\right]\psi$$, takes the form of a Bernoulli/Hamilton-Jacobi equation $$\partial_t\theta+\frac{1}{2}mv^2+V+Q=0$$ where ##\textbf{v}=\frac{\mathbf{\nabla}\theta}{m}## is the flow of the fluid/probability density i.e. ##\textbf{J}=\rho\textbf{v}## and ##Q=-\frac{\hbar^2}{2m}\frac{\nabla^2\sqrt{\rho}}{\sqrt{\rho}}## is a quantum energy potential/pressure. The point is that ##\hat{K}\psi## leads to both a kinetic energy of flow ##\frac{1}{2}mv^2## determined by the phase twist/winding, and this additional ##Q## determined by the "curvature" of the probability density. In other words this separates out the energies associated with spatial variation of the phase and the density. So a discontinuity in the phase means infinite flow (which is maybe a more specific take on what the Gasiorowicz textbook was implying?) and a discontinuity in the probability density means infinite quantum pressure.
     
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