- #1

QuantumBunnii

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## Homework Statement

A particle of mass m in the infinite square well of width a at time t = 0 is in a linear superposition of the ground- and the first excited- eigenstates, specifically it has the wave function

$$| \Psi(x,t) > = A[ | \psi_1 > + e^{i \phi} | \psi_2 >$$

Find the expectation value <x(t)> of the position of the particle at time t.

## Homework Equations

[tex]< f(x,t) | g(x,t) > = \int_{-a/2}^{a/2} f(x,t)^{*}g(x,t)\ dx[/tex]

$$ <x(t)> = < \Psi(x,t) | x \Psi(x,t) > $$

## The Attempt at a Solution

Obtaining a solution for this problem isn't extraordinarily difficult-- given the stationary states [itex] \psi_1[/itex] and [itex] \psi_2 [/itex], we can simply plug them into the above equations and-- after having normalized for 'A'-- solve for <x(t)>. My question is a more methodological one, regarding the even-ness and odd-ness of the integrands. After going through a few steps to solve for <x>, we would obtain the following solution:

$$ <x> = A < \psi_1 | x \psi_1 > + B | \psi_1 | x \psi_2 > + C < \psi_2 | x \psi_1 > + D < \psi_2 | x \psi_2 > $$

where A, B, C, and D, are imaginary constants.

In the interest of time, I decided to consider the even-ness and odd-ness of the above integrands: ## \psi_{0,2,4,...} ## correspond to odd functions, and ## \psi_{1,3,5,...} ## correspond to even functions. 'x' should also be considered an odd function, since I have chosen the integral to run from -a/2 to a/2. Hence, we can determine whether each integrand is ultimately even or odd (even times even is even; odd times even is odd; etc.), and immediately set the odd integrands to zero (integrating an odd function centered at the origin yields a value of zero). This would mean: A = 0 and D = 0. However, this turns out to give me the wrong answer, as it seems I'm supposed to consider the value of *all* the integrals.

I tried to keep the mathematical calculations in this question to a minimum; I would just like to know whether the above methodology is mathematically intact (I almost hope it isn't...), and-- if so-- why it doesn't work out in this case.

Thanks for any help.