Mixture of States Proof

A particle of mass m sits in a one-dimensional square well with infinitely high walls and width L . The particle is in a 50–50 mixture of states, half in the ground state and half in the first excited state. I want to show how to derive a formula for the complete, time-dependent wave-function of the particle.

I begin by the using the normalizing condition, that \int|\Psi|^{2}dx=1 . This is because the probability of finding the particle somewhere in space must equal 1 at all times. Because the particle is in a mixture of states, my wave function will take the form \Psi = A(\Psi_{1}+\Psi_{2}) . Combining this with the normalizing condition, I get the equation A^{2}\int(\Psi_{1}+\Psi_{2})(\Psi_{1}^{*}+\Psi_{2}^{*})dx = 1 . The individual wave-functions will take the forms \Psi_{1} = \sqrt{\frac{2}{L}}\sin(\frac{\pi x}{L})e^{-i\omega_{1}t} and \Psi_{2} = \sqrt{\frac{2}{L}}\sin(\frac{\pi x}{L})e^{-i\omega_{2}t} . I can then plug these two equations in the the above normalizing condition for a particle in mixed states, convert all sine functions to cosine functions and cancel out like terms until I get down to the simple expression that 2A^{2}=1 which implies that A = \frac{1}{\sqrt{2}} . Now I can use this constant to write down my mixed wave-function which will look as follows \Psi = \frac{1}{\sqrt{L}}(\sin(\frac{\pi x}{L})e^{-\omega_{1}t} + \sin(\frac{2\pi s}{L})e^{-i\omega_{2}t}) . Now, I want to show that the probability of finding the particle between positions \frac{1}{4}L and \frac{1}{4}L+dx , as measured from the left-hand side of the well, as a function of time, is [\frac{3}{2}+\sqrt{2}\cos(\frac{3E_{1}t}{\hbar})]\frac{dx}{L} . This is done by finding the probability amplitude which is the square modulus of the wave function. This will look as follows |\Psi_{1}+\Psi_{2}|^{2} = (\Psi_{1}+\Psi_{2})(\Psi_{1}^{*}+\Psi_{2}^{*}) . After filling in each wave function and multiplying out each term I obtain \frac{1}{L}(\sin(\frac{\pi x}{L}))^{2} + \frac{1}{L}(\sin(\frac{2\pi x}{L}))^{2} + \frac{1}{L}\sin(\frac{2\pi x}{L}) \sin(\frac{\pi x}{L}) ( e^{-i\omega_{1} t}e^{i\omega_{2} t}+e^{-i\omega_{2} t}e^{i\omega_{1} t}) . After converting the sine terms to cosines and converting the exponentials to trigonometric functions, I can plug in x for L , which will give me\frac{1}{4}L and \frac{1}{4}L+dx which is what I wanted to show.

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