- #1
Cogswell
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Homework Statement
Consider an electron confined by a 1 dimensional harmonic potential given by ## V(x) = \dfrac{1}{2} m \omega^2 x^2##. At time t=0 the electron is prepared in the state
[tex]\Psi (x,0) = \dfrac{1}{\sqrt{2}} \psi_0 (x) + \dfrac{1}{\sqrt{2}} \psi_4 (x)[/tex]
with ## \psi_n (x) = \left( \dfrac{m \omega}{\pi \hbar}\right)^{0.25} \dfrac{1}{\sqrt{2^n n!}} H_n \left( \sqrt{\dfrac{m \omega}{\hbar}}x \right) exp \left( - \dfrac{m \omega}{2 \hbar} x^2 \right) ##
the normalised solutions to the time independent Schrodinger equation, with ##H_n## being the Hermite Polynomials.
(a) Express the square of the momentum operator as ladder operators.
(b) Give the ratio of the expectation values of the kinetic and potential energies as a function of time.
Homework Equations
## \hat{x} = \sqrt{\dfrac{\hbar}{2m \omega}} (a_+ + a_-) ##
## \hat{p} = i \sqrt{\dfrac{\hbar m \omega}{2}} (a_+ - a_-) ##
The Attempt at a Solution
For the first one I got:
[tex]\hat{p}^2 = \left[ i \sqrt{\dfrac{\hbar m \omega}{2}} (a_+ - a_-) \right]^2 = -\dfrac{\hbar m \omega}{2} (a_+a_+ - a_-a_+ - a_+ a_- + a_-a_-) [/tex]For the second one, I attempted it using the operators
[T = kinetic energy, V = potential energy]
[tex]\displaystyle <T> = < \dfrac{\hat{p}}{2m} > = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (a_+a_+ - a_-a_+ - a_+ a_- + a_-a_-) \right] \Psi dx [/tex]
[tex]\displaystyle <V> = <0.5 \omega ^2 \hat{x}^2> = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (a_+a_+ + a_-a_+ + a_+ a_- + a_-a_-) \right] \Psi dx[/tex]Now each of them will split into 4 integrals. The a+a+ and a-a- operator integrals will evaluate to zero because of its orthogonality properties. That leaves us with:
[tex]\displaystyle <T> = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (- a_-a_+ - a_+ a_-) \right] \Psi dx [/tex]
[tex]\displaystyle <V> = \int \Psi^* \left[ \dfrac{\hbar \omega}{4} (a_-a_+ + a_+ a_-) \right] \Psi dx[/tex]That'll mean those terms disappear, and then... we end up with -1 as a ratio?? What did I do wrong?