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I'm having a problem with an expectation value that doesn't seem to add up for me.

What I know is, that $\psi(\vec{r})$ is a wavefunction for a particle in three dimensions. The Hamiltonian is given by:

$\hat{H} = \frac{\hat{p}^{2}}{2m}+\hat{V}(\vec{r})$

The expectation value of the momentum in the state $\psi$ is given by:

$\langle \psi,\hat{\vec{p}}\psi\rangle = \vec{p}_{0}$

Another state is given by:

$\phi(\vec{r})=\psi(\vec{r}) \cdot e^{i\vec{k} \cdot \vec{r}} $,

where $\vec{k}$ is a constant vector.

Now I have to calculate the expectation value of the momentum in the state $\phi$. Then I just thought I had to do this:

$\langle\vec{p}\rangle=\langle\phi|\hat{\vec{p}}|\phi\rangle = e^{-i\vec{k} \cdot \vec{r}}e^{i\vec{k} \cdot \vec{r}} \langle\psi|\hat{\vec{p}}|\psi\rangle$,

which would end up being $\vec{p}_{0}$.

But according to the solution I have to get:

$\langle\vec{p}\rangle = \vec{p}_{0}+\hbar\vec{k}$,

which I can't see how is done ? I know that the momentum is given by $p = \hbar k$ in general, but I can't see how it comes out in the solution ?

So anyone who might be able to give me a hint ?

Thanks in advance.

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Write the momentum operator in position space and think how this acts on the wavefunction. –  Will Jun 9 '13 at 23:51
    
Hmmm, I see what you mean. Then I get the $\hbar \vec{k}$ actually. But why is that not multiplied with $\vec{p}_{0}$ insead of adding it ? When the operator works on the exponential function, I just get the $\hbar \vec{k}$, but where does the plus sign come from ? :/ –  Denver Dang Jun 10 '13 at 0:09
3  
The momentum operator in position space is a derivative. So you're getting the chain rule. e.g. $\frac{\partial}{\partial x}\left(e^{ikx}\phi(x)\right) = \left(\frac{\partial}{\partial x}e^{ikx}\right)\phi(x) + e^{ikx}\left(\frac{\partial}{\partial x}\phi(x)\right)$. Hence addition. –  Will Jun 10 '13 at 0:13
    
Ahhh, didn't think of it as a product. Thank you Will. –  Denver Dang Jun 10 '13 at 0:21
    
*product rule. You're welcome. –  Will Jun 10 '13 at 0:28
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1 Answer

up vote 1 down vote accepted

This is effectively an application of the product rule.


In terms of your wavefunction $\psi(\vec{r})$ the expectation of $\hat{p}$ can be written \begin{eqnarray} \langle\psi|\hat{p}|\psi\rangle &=& \int d^3rd^3r^\prime \langle\psi|\vec{r}^\prime\rangle\langle\vec{r}^\prime|\hat{p}|\vec{r}\rangle\langle\vec{r}|\psi\rangle\\ &=& \int d^3rd^3r^\prime \langle\psi|\vec{r}^\prime\rangle\delta^3(\vec{r}-\vec{r}^\prime)\frac{\hbar}{i}\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\langle\vec{r}|\psi\rangle\\ &=& \int d^3r~ \psi^*(r)\frac{\hbar}{i}\frac{\partial}{\partial \vec{r}}\psi(\vec{r}) \end{eqnarray} where I have defined $\frac{\partial}{\partial \vec{r}} = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$ for notational simplicity. We have been given that \begin{equation} ~~~~~~~~~~~\boxed{\langle\psi|\hat{p}|\psi\rangle = \vec{p}_0 \implies \int d^3r~ \psi^*(r)\frac{\hbar}{i}\frac{\partial}{\partial \vec{r}}\psi(\vec{r}) = \vec{p}_0}~~~~~~~~~~~~~~~~~~~~~~~~(1) \end{equation} If we now define $\phi(\vec{r}) = e^{i\vec{k}\cdot\vec{r}}\psi(\vec{r}) = \langle \vec{r}|\phi\rangle$ we have expectation of momentum for this state as \begin{eqnarray} ~~~~~~~~~~~~~~~~\langle\phi|\hat{p}|\phi\rangle &=& \int d^3 r ~\phi^*(\vec{r})\frac{\hbar}{i}\frac{\partial}{\partial \vec{r}}\phi(\vec{r})\\ &=& \int d^3 r~ \psi^*(\vec{r})e^{-i\vec{k}\cdot\vec{r}}\frac{\hbar}{i}\frac{\partial}{\partial \vec{r}}\left(e^{i\vec{k}\cdot\vec{r}}\psi(\vec{r})\right)~~~~~~~~~~~~~~~~~~~~~~~(2) \end{eqnarray} Now applying the product rule $$\frac{\partial}{\partial \vec{r}}\left(e^{i\vec{k}\cdot\vec{r}}\psi(\vec{r})\right) = i\vec{k}~e^{i\vec{k}\cdot\vec{r}}\psi(\vec{r}) + e^{i\vec{k}\cdot\vec{r}}\frac{\partial}{\partial \vec{r}}\psi(\vec{r}) = e^{i\vec{k}\cdot\vec{r}}\left(i\vec{k}~\psi(\vec{r}) + \frac{\partial}{\partial \vec{r}}\psi(\vec{r})\right)$$ putting this into Eq. (2) we have \begin{eqnarray} \langle\phi|\hat{p}|\phi\rangle &=& \int d^3 r~ \psi^*(\vec{r})e^{-i\vec{k}\cdot\vec{r}}\frac{\hbar}{i}e^{i\vec{k}\cdot\vec{r}}\left(i\vec{k}~\psi(\vec{r}) + \frac{\partial}{\partial \vec{r}}\psi(\vec{r})\right)\\ &=& \hbar\vec{k}\int d^3 r~ \psi^*(\vec{r})\psi(\vec{r}) + \int d^3 r~ \psi^*(\vec{r})\frac{\hbar}{i} \frac{\partial}{\partial \vec{r}}\psi(\vec{r})\\ &=& \hbar\vec{k} + \vec{p}_0 \end{eqnarray} where in the last line we have used Eq. (1) and normalization of the wavefunction ($\langle\psi|\psi\rangle = 1$). And so we see $$\boxed{\langle\phi| \hat{p}|\phi\rangle = \hbar\vec{k} + \vec{p}_0}$$ as required.

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