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A Solving the Schrödinger equation for free electrons

  1. Jan 2, 2018 #1

    SeM

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    Dear all, sorry I made a new post similar to the previous post "Initial conditions..", however, a critical point was missed in the previous discussion:

    The initial conditions y(0)=1 and y'(0)=0 are fine and help in solving the Schrödinger equation, however, studying free electrons, the equation cannot be solved neither for the particle in the box , or for a finite linear motion harmonic oscillator as far as I can see from the literature.

    When I solve the eqn, HY=EY , I can set E= p/2m or E= ground state energy. In either cases, I get a result where some unknown constant appears.

    This constant can be solved using the Born Sommerfeld condition, but because the solution is not square integrable, it gives complex values.

    How can I solve the unknown constant for a solution of the Schrödinger eqn. where the above given initial conditions are used?

    Thanks!
     
  2. jcsd
  3. Jan 2, 2018 #2

    vanhees71

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    I'm a bit insecure what you mean with your "initial conditions" and most of the rest of your posting, but let's briefly discuss free non-relativistic electrons. The Hamiltonian reads
    $$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}.$$
    Working in the position representation ("wave mechanics") we have
    $$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}.$$
    For simplicity also let's neglect the spin of the electron. Then to get a complete set of energy eigenvectors it's clear that we can as well look for a complete set of common eigenvectors of ##\hat{\vec{p}}## (note that ##[\hat{p}_j,\hat{p}_k]=0##, and that thus there's a set of common eigenvectors of all three momentum components). To find these eigenvectors we have to solve the eigenvalue equation
    $$\hat{\vec{p}}u_{\vec{p}}(\vec{x})=-\mathrm{i} \hbar \vec{\nabla} u_{\vec{p}}(\vec{x}) =\vec{p} u_{\vec{p}}(\vec{x}).$$
    It's easy to see that the solution is
    $$u_{\vec{p}}(\vec{x})=N(\vec{p}) \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right), \quad \vec{p} \in \mathbb{R}^3.$$
    It's convenient to normalize the eigenvectors "to a ##\delta## distribution":
    $$\langle u_{\vec{p}} |u_{\vec{p}'} \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} N^*(\vec{p}) N(\vec{p}') \exp \left [\frac{\mathrm{i} \vec{x} \cdot (\vec{p}'-\vec{p})}{\hbar} \right] =|N(\vec{p})|^2 (2 \pi)^3 \hbar^3 \delta(\vec{p}-\vec{p}') \; \Rightarrow \; N(\vec{p})=\frac{1}{(2 \pi \hbar)^{3/2}}.$$
    So finally we have
    $$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right).$$
    Of course, each ##u_{\vec{p}}## is an energy eigenstate,
    $$\hat{H} u_{\vec{p}}(\vec{x})=\frac{\vec{p}^2}{2m} u_{\vec{p}}(\vec{x})=E_{\vec{p}} u_{\vec{p}}(\vec{x}).$$
    The initial value is the wave function at time ##t_0## (not some strange other "initial values" you quote in the OP). The general solution of the initial value problem is now easily given in terms of the just found energy eigenfunctions. Defining
    $$\tilde{\psi}_0(\vec{p})=\langle u_{\vec{p}}|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle u_{\vec{p}}|\vec{x} \rangle \langle \vec{x}|\psi_0 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}}^*(\vec{x}) \psi_0(\vec{x}),$$
    you get
    $$\psi(t,\vec{x})=\langle \vec{x} | \exp[-\mathrm{i} \hat{H}(t-t_0)/\hbar]|\psi_0 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \langle \vec{x}|\exp[-\mathrm{i} \hat{H}(t-t_0)/\hbar] u_{\vec{p}} \rangle \langle u_{\vec{p}}|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}(\vec{x}) \exp[-\mathrm{i} E_{\vec{p}}(t-t_0)/\hbar] \tilde{\psi}_0(\vec{p}).$$
     
  4. Jan 2, 2018 #3

    SeM

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    Thanks for the fantastic answer!

    I will need some time to go through this. Thanks a lot!
     
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