Spreading of a Free-Particle Gaussian Wave Packet

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The discussion centers on the calculation of the expected position <x> for a free-particle Gaussian wave packet using Mathematica. The user attempted to integrate the position probability density P(x,t) but arrived at an unexpected result of Sqrt[2Pi]vt, which does not match the provided answer choices. It was pointed out that the issue lies in the normalization of P(x,t), which is not correctly normalized. The user also expressed confusion regarding the "in units of L" aspect of the problem. Proper normalization and understanding of the units are essential for accurate calculations in quantum mechanics.
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Homework Statement



It can be shown that the position probability density for the one-dimensional, free-particle Gaussian wave packet can be expressed as shown below, where m is the mass of the particle and L is the position uncertainty at time t=0, vgr = hbar *k0/m, k0 is the average wave number as discussed in class. Let x, t and P(x,t) be measured, respectively, in units of L, T=2mL^2/hbar, A=1/sqrt(2pi*L^2), then the position probability can be written as P(x,t)= sqrt(1/(1+t^2)) exp(-(x-vt)^2/(2 (1+t^2)), where v= 2 mLvgr/hbar. In order to demonstrate the spreading of the wave packet, use Mathematica to calculate <x> in units of L at time t./>

Homework Equations


<x> = integral from -infinity to infinity of x times psi squared

The Attempt at a Solution


I am both a mathematica and quantum mechanics novice, so I'm not sure where my mistake lies. Since psi[x,t]^2 = P[x,t], I integrated P(x,t)= sqrt(1/(1+t^2)) exp(-(x-vt)^2/(2 (1+t^2)) from -infinity to infinity with respect to x and got the answer Sqrt[2Pi]vt. That's not one of the answer choices. The answer choices are: 0, vt, sqrt(1+t^2 (1+v^2)), or sqrt(1+t^2 ). Where did I go wrong? [Note: I don't really understand all of that "in units of L" bit in the question.] Thanks in advance.
 
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P(x,t) is not properly normalized.
 

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