Average value of components of angular momentum for a wave packet

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The discussion centers on the calculation of angular momentum components for a wave packet, specifically questioning why integrals yield zero when classical mechanics suggests a free particle should have nonzero angular momentum. Participants noted that the y and z components were not shown, and one contributor pointed out an error in the derivative calculation that led to the incorrect zero result. After addressing the mistake, the correct calculation produced the expected nonzero value for the y component, specifically ##bp_0##. The conversation highlights the importance of careful mathematical derivation in quantum mechanics.
Nelsonc
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Homework Statement
Given a wave packet as shown (see below), find the mean value of angular momentum components L_x, L_y, L_z with regard to point (a,0,-b) where a and b (the impact parameter) are nonzero
Relevant Equations
\frac{1}{\pi^{3/4} \sqrt{\sigma_x\sigma_y\sigma_z}}e^{-(x^2/2{\sigma_x}^2+y^2/2{\sigma_y}^2+z^2/2{\sigma_z}^2)}e^{i(p_0/\hbar)x}
I have typed up the main problem in latex (see photo below)
problem.png

It seems all such integrals evaluates to 0, but that is apparantly unreasonable for in classical mechanics such a free particle is with nonzero angular momentum with respect to y axis.
 
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Nelsonc said:
It seems all such integrals evaluates to 0, but that is apparantly unreasonable for in classical mechanics such a free particle is with nonzero angular momentum with respect to y axis.
You didn't show your work for the y and z components of angular momentum. You shouldn't get zero for the y component.
 
Thanks for the reminder, but I have already done so and it turns out they all goes to 0, so there must be something awry with my method. In particular, there seems always to be one spacial part of the integrand to be antisymmetric so that the whole integral goes to 0 (please refer to the image attached). Moreover, I know that classically a free particle moving in such fashion would have y angular momentum component being ##bp_0##

problem_2.png
 
In deriving equation (7), check your result for ##\large \frac{\partial \psi'}{\partial x}##. Did you use the product rule when taking the derivative of the product of the Gaussian function and the function ##e^{i(p_0/\hbar)x}##?
 
I see, thanks so much for catching that error! Now the calculation generates ##bp_0## as a result.
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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