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Orthogonality of wave functions to negative momentum states

  1. Feb 25, 2014 #1
    This is a question I have about the textbook discussion, so I'll do away with the standard format.

    The author of my QM book (Shankar, Principles of Quantum Mechanics) used the term "negative momentum states," all of a sudden, and I've never heard of it before. He has a little note saying that it is "left-going."

    I'll quickly run through the discussion. Its the single step potential scattering problem.
    Consider the (normalized) Gaussian
    $$\psi_I(x)=(\pi\Delta^2)^{-1/4}e^{ik_0(x+a)}e^{-(x+a)^2/2\Delta^2}$$
    This packet has mean momentum ##p_0=\hbar k_0## and mean position ##\langle X\rangle=-a##. We will assume that ##\Delta X## is large and that the wave function in k space is very sharply peaked near ##k_0##.
    He then goes on to talk about uncertainty calculations and reflection and transmission coefficients, but I think they are irrelevant for my problem. He presents the solution to the Schrödinger equation as
    $$\psi_E(x)=A\left[\left(e^{ik_1x}+\frac{B}{A}e^{-ik_1x}\right)\Theta(-x)+\frac{C}{A}e^{ik_2x}\Theta(x)\right]$$
    Where A, B and C are arbitrary, ##k_1=\sqrt{\frac{2mE}{\hbar^2}}## and ##k_2=\sqrt{\frac{2m(E-V_0)}{\hbar^2}}## and ##V_0## is the potential of the step.
    Since to each E there is a unique ##k_1=+\sqrt{\frac{2mE}{\hbar^2}}##, we can label the eigenstates by ##k_1## instead of E. Eliminating ##k_2## in favor of ##k_1##, we get
    $$\psi_{k_1}(x)=A\left[\left(e^{ik_1x}+\frac{B}{A}e^{-ik_1x}\right)\Theta(-x)+\frac{C}{A}e^{ix\sqrt{k_1^2-2mV_0/\hbar^2}}\Theta(x)\right]$$
    Great foresight tells us that ##A=\frac{1}{\sqrt{2\pi}}##, so that our wave function is normalized. Now consider
    $$a(k_1)=\langle\psi_{k_1}\vert\psi_I\rangle=\frac{1}{\sqrt{2\pi}}\left(\int_{-\infty}^{\infty}\left[e^{-ik_1x}+\frac{B^*}{A^*} e^{ik_1x}\right]\Theta(-x)\psi_I(x)dx+\int_{-\infty}^{\infty}\frac{C^*}{A^*}e^{-ik_2x}\Theta(x)\psi_I(x)dx\right)$$
    This mess ought to reduce to the Fourier transform
    $$a(k_1)=\frac{1}{\sqrt{2\pi}}\int e^{-ik_1x}\psi_I(x)dx$$
    I get that the second integral vanishes because of the Heavyside function. My question is why the second term in the first integral vanishes. To quote Shankar, "Similarly, the second piece of the first integral also vanishes since ##\psi_I## in k space is peaked around ##k=+k_0## and is orthogonal to (left-going) negative momentum states." Does this mean that ##e^{ik_1x}## is a negative momentum state? I'm very confused by this.

    Any help would be greatly appreciated.
     
  2. jcsd
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