- #1

ja07019

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- Homework Statement
- Calculate <p> for an Infinite Potential Well for a classical particle

- Relevant Equations
- u(z,t) = Aexp{iu} where u=wt-kz

Okay so I begin first by mentioning the length of the well to be L, with upper bound, L/2 and lower bound, -L/2 and the conjugate u* = Aexp{-iz}

First I begin by writing out the expectation formula:

## \langle p \rangle = \int_{\frac{L}{2}}^{ \frac{L}{2} } Aexp(-iu) -i \hbar \frac{ \partial }{ \partial z} Aexp(iu) dz = \int_{\frac{L}{2}}^{ \frac{L}{2} } Aexp(-iu)Aexp(iu)-i \hbar i \frac{ \partial }{ \partial z} (wt-kz)dz = -A^2(i)^2\hbar \int_{\frac{L}{2}}^{ \frac{L}{2} } exp(-iu+iu) \frac{ \partial }{ \partial z} (wt-kz)dz = -A^2(-1) \hbar \int_{\frac{L}{2}}^{ \frac{L}{2} } exp(0) -k dz = -A^2(-1) \hbar (-k) \int_{\frac{L}{2}}^{ \frac{L}{2} } dz = -A^2 \hbar k (\frac{L}{2} - \frac{L}{2}) = -A^2 \hbar k (2L/2) = -A^2 \hbar k L ##

After normalizing the wavefunction, I ended up with ##A^2 = \frac{1}{L}##

Thus the entire thing reduces to ## -\frac{1}{L} \hbar k L = -\hbar k = -mv = p ##

The solution given by the book is <p> = 0 and the correct answer is also in agreement with the QM derived solution.

First I begin by writing out the expectation formula:

## \langle p \rangle = \int_{\frac{L}{2}}^{ \frac{L}{2} } Aexp(-iu) -i \hbar \frac{ \partial }{ \partial z} Aexp(iu) dz = \int_{\frac{L}{2}}^{ \frac{L}{2} } Aexp(-iu)Aexp(iu)-i \hbar i \frac{ \partial }{ \partial z} (wt-kz)dz = -A^2(i)^2\hbar \int_{\frac{L}{2}}^{ \frac{L}{2} } exp(-iu+iu) \frac{ \partial }{ \partial z} (wt-kz)dz = -A^2(-1) \hbar \int_{\frac{L}{2}}^{ \frac{L}{2} } exp(0) -k dz = -A^2(-1) \hbar (-k) \int_{\frac{L}{2}}^{ \frac{L}{2} } dz = -A^2 \hbar k (\frac{L}{2} - \frac{L}{2}) = -A^2 \hbar k (2L/2) = -A^2 \hbar k L ##

After normalizing the wavefunction, I ended up with ##A^2 = \frac{1}{L}##

Thus the entire thing reduces to ## -\frac{1}{L} \hbar k L = -\hbar k = -mv = p ##

The solution given by the book is <p> = 0 and the correct answer is also in agreement with the QM derived solution.