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Solutions of schrodinger equation contradicts HUP

  1. Oct 14, 2014 #1
    By solving the schrodinger equation, we get atmost two solutions for wavefunctions with definite wavenumber and definite wavelength. Thus, we know specifically the momentum of the particle. But this is contradicted by HUP. Please explain.
    I would appreciate an explanation in the context of a particle in a box.
  2. jcsd
  3. Oct 14, 2014 #2


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    The HUP does not preclude states with definite momentum (so-called momentum eigenstates). It simply states that the particle then has equal probability of being anywhere within the well.
  4. Oct 14, 2014 #3


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    The particle-in-a-box energy eigenfunctions do not have definite momentum. To see this clearly, calculate the momentum-space wave function for e.g. the ground state, using $$\phi(p) = \int_a^b {e^{-ipx/\hbar} \psi(x)\,dx}$$ where a and b are the "walls" of the box; and then the momentum probability distribution ##\phi^*(p)\phi(p)##.

    Going further, you can calculate ##\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}## and ##\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}## and then their product. If you do this correctly, the width of the box should cancel out, and you will get a multiple of ##\hbar## that is greater than ##0.5\hbar## in keeping with the HUP.
    Last edited: Oct 14, 2014
  5. Oct 14, 2014 #4
    For particle in a box, we have a specific wavenumber = n*(pi)/L. This corresponds to a momentum. Actually, what is this momentum? It is not of course the average momentum of all the contributing sinusoids (fourier transform components), because the expectation value of momentum in the particle in a box case is 0. So, what is the explanation of this momentum?
  6. Oct 15, 2014 #5


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    No! For a particle in a box with rigid boundary conditions (i.e., the box with infinitely high potential walls, which already shows that this is an artificial example to begin with) there doesn't even exist a momentum observable.

    This can be easily shown as follows: The momentum operator is, by definition via Noether's theorem, the operator that generates spatial translations. Let's look at the 1-dimensional box for simplicity. From this definition of momentum it follows that
    [tex]\hat{p}=-\mathrm{i} \partial_x.[/tex]
    The eigenvalue problem reads
    [tex]\hat{p} u_p(x)=-\mathrm{i} \partial_x u_p(x)=p u_p(x).[/tex]
    It follows that
    [tex]u_p(x)=N(p) \exp(\mathrm{i} p x).[/tex]
    However, the Hilbert space of wave functions of the finite box is given by the functions that vanish at the boundaries of the box, e.g., the interval [itex](-L/2,L/2)[/itex].

    This boundary condition is not fulfilled by any of the putative eigenfunctions and thus there is no essentially self-adjoint momentum operator on this Hilbert space, and thus there is no momentum observable in the usual sense.

    The whole thing is different for the Hamiltonian, which you formally define as
    [tex]\hat{H}=\frac{\hat{p}^2}{2m}=-\frac{1}{2m} \partial_x^2.[/tex]
    The eigenfunctions are
    [tex]\hat{H} u_E(x)=-\frac{1}{2m} \partial_x^2 u_E(x)=E u_E(x).[/tex]
    This leads to the solutions
    [tex]u_E(x)=N(E) [\exp(\mathrm{i} k x)+B \exp(-\mathrm{i} k x)],[/tex]
    where [itex]k=\sqrt{2m E}[/itex].

    Now we can fulfill the boundary conditions:
    [tex]\exp(\mathrm{i} k L/2)+B \exp(-\mathrm{i} k L/2)=0,[/tex]
    [tex]\exp(-\mathrm{i} k L/2) + B \exp(\mathrm{i} k L/2)=0.[/tex]
    From the first equation you get
    [tex]B=\exp(2 \mathrm{i} k L/2)=\exp(\mathrm{i} k L)[/tex]
    and from the second
    [tex]B=\exp(-\mathrm{i} k L)[/tex]
    This means
    [tex]\exp(2\mathrm{i} k L)=1.[/tex]
    This determines the possible values of [itex]k[/itex] by the condition
    [tex]2 k L=2 \pi n \; \Rightarrow \; k=\pi n/L,[/tex]
    where [itex]n \in \mathbb{Z}[/itex].

    Now for each [itex]n \in \mathbb{Z}[/itex] you can solve for [itex]B[/itex] using one of the equations. This gives
    Thus you have
    [tex]u_E(x)=N(E) [\exp(\mathrm{i} \pi n x/L)+(-1)^n \exp(-\mathrm{i} \pi n x/L)].[/tex]
    Now switching from a given [itex]n \in \mathbb{N}[/itex] to [itex]-n[/itex] leads to the same solution (up to a sign, which is insignificant, because it's just a phase factor). Thus you have to take only [itex]n \in \{0,1,2,3,\ldots\}[/itex].

    These solutions are a complete orthogonal set in [itex]L^2[(-L/2,L/2)][/itex], because it's just
    N_n \cos(\pi n x/L) & \text{for} \quad n \quad \text{even},\\
    N_n \sin(\pi n x/L) & \text{for} \quad n \quad \text{odd}.
    The normalization constant is given by
    [tex]\int_{-L/2}^{L/2} |u_n(x)|^2=1 \; \Rightarrow \;N_0=1/\sqrt{L}, \quad N_n=\sqrt{2}{L} \quad \text{for} \quad n \in \{1,2,\ldots\}.[/tex]
    The well-known theorems about Fourier series thus tell you that indeed this is a complete set of orthonormal function, because obviously
    [tex]\langle u_n|u_{n'} \rangle=\delta_{nn'}[/tex]
    with the so chosen normlization constants.
    Last edited: Oct 18, 2014
  7. Oct 15, 2014 #6


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    However, this is the unique and definite (magnitude of) momentum of the particle only if this is a monochromatic wave that extends to infinity in both directions. Two infinitely-long monochromatic waves with equal wavelength and frequency, traveling in opposite directions, would produce a standing wave with momentum +p and -p, which we might represent as the sum of two Dirac delta functions in momentum space. This standing wave also extends to infinity in both directions.

    To get a particle-in-a-box wavefunction, we have to "chop off" this infinitely-long standing wave at two of the nodes. In terms of a superposition of infinitely-long waves, we have to add more waves with a variety of wavelengths, chosen such as to cancel the original waves outside the box. This is where the "extra" momentum values come from, for the particle-in-a-box.
    Last edited: Oct 15, 2014
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