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unscientific
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Homework Statement
A state at time t is given by:
[tex]|\psi\rangle = \frac{1}{\sqrt 2}\left[ e^{-\frac{i\omega t}{2}}|0\rangle + e^{-i\frac{3\omega t}{2}}|1\rangle \right] [/tex]
Where eigenfunctions are ##\phi_0 = \left(\frac{1}{a^2 \pi}\right)^{\frac{1}{4}}e^{-\frac{x^2}{2a^2}}## and ##\phi_1 = \left(\frac{4}{a^6 \pi}\right)^{\frac{1}{4}} x \space e^{-\frac{x^2}{2a^2}}##.I'm supposed to find ##\langle \psi|\hat p|\psi \rangle##.
Homework Equations
[tex]\hat p = -i\hbar \frac{\partial}{\partial x}[/tex]
The Attempt at a Solution
By symmetry, we know that ##\langle 0|\hat p|0 \rangle = \langle 1 | \hat p | 1 \rangle = 0##.
Starting:
[tex]\frac{1}{2} \langle 0 | e^{-i\omega t} \hat p |1\rangle [/tex]
[tex] = -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a^2}} \frac{\partial}{\partial x} \left(x e^{-\frac{x}{2a^2}}\right) dx[/tex]
[tex]= -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a^2}}\left(e^{-\frac{x^2}{2a^2}} - \frac{x^2}{a^2} e^{-\frac{x^2}{2a^2}}\right) dx [/tex]
[tex] = -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{a^2}} \left(1 - \frac{x^2}{a^2}\right) dx [/tex]
[tex] = -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \left[ \sqrt \pi a - \frac{1}{a^2} \frac{1}{2} \sqrt {\pi} a^3 \right] [/tex]
[tex] = -\frac{i\hbar}{2} \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}} \left(\frac{1}{2} \sqrt \pi a \right) e^{-i\omega t} [/tex]
[tex] = -\frac{i\hbar}{2\sqrt 2 a} e^{-i\omega t} [/tex]
Similarly, for ##\frac{1}{2}\langle 1 | e^{i\omega t} \hat p | 0 \rangle##, we take the complex conjugate of above.
Adding them both together, we get:
[tex]-\frac{\hbar}{\sqrt 2 a} sin (\omega t) [/tex]
Given ## \langle \psi |x|\psi\rangle = \frac{a}{\sqrt 2} cos (\omega t) ##, then I'm supposed to find:
[tex]\frac{\partial}{\partial t} \left( \langle \psi|p|\psi\rangle + m\omega^2 \langle \psi |x|\psi\rangle \right) [/tex]
They don't add up to zero, which is strange since the first term is simply the force, and the second term is the restoring force..
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