Revisions to Mahan's derivation of energy current of a free-particle system

Post Reopened by John Rennie, Vincent Thacker, Roger V.

occurred Jul 11 at 11:19

Left closed in review as "Original close reason(s) were not resolved" by Jon Custer, Miyase, Michael Seifert

occurred Jul 10 at 15:56

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edited Jul 8 at 20:31

Redcrazyguy

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On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}\textbf{r}\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger \textbf{r}\nabla^2\psi+2\psi^\dagger(\nabla \textbf{r}\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-\textbf{r}\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$$$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(\textbf{r}\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger \textbf{r}\nabla^2\psi+2\psi^\dagger(\nabla \textbf{r}\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-\textbf{r}\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}\textbf{r}\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger \textbf{r}\nabla^2\psi+2\psi^\dagger(\nabla \textbf{r}\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-\textbf{r}\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}\textbf{r}\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(\textbf{r}\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger \textbf{r}\nabla^2\psi+2\psi^\dagger(\nabla \textbf{r}\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-\textbf{r}\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

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edited Jul 8 at 14:02

Redcrazyguy

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On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}r\psi^\dagger\psi$$$$\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}\textbf{r}\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger r\nabla^2\psi+2\psi^\dagger(\nabla r\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-r\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$$$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger \textbf{r}\nabla^2\psi+2\psi^\dagger(\nabla \textbf{r}\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-\textbf{r}\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}r\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger r\nabla^2\psi+2\psi^\dagger(\nabla r\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-r\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger \textbf{r}\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}\textbf{r}\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger \textbf{r}\nabla^2\psi+2\psi^\dagger(\nabla \textbf{r}\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-\textbf{r}\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

Post Closed as "Needs details or clarity" by Roger V., Matt Hanson, Miyase

occurred Jul 7 at 18:08

added 2 characters in body

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edited Jul 5 at 15:24

Redcrazyguy

383
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On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}r\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger r\nabla^2\psi+\psi^\dagger(\nabla r\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-r\psi^\dagger\psi+\psi^\dagger\nabla\psi\right]$$$$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger r\nabla^2\psi+2\psi^\dagger(\nabla r\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-r\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}r\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger r\nabla^2\psi+\psi^\dagger(\nabla r\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-r\psi^\dagger\psi+\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?

On p.25 of the 3rd Ed. analogously to the polarisation operator $\textbf{P}$ for particle currents $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he defines an operator $$\textbf{R}_E=\frac{1}{2}\int\textbf{r}\mathcal{H}(\textbf{r})+\mathcal{H}(\textbf{r})\textbf{r}d^3r$$ He defines the energy current, $\textbf{j}_E=\partial_t\textbf{R}_E$, then claims that for a free-particle system, the energy current is

$$ \textbf{j}_E=\sum_{\textbf{p}\sigma}\textbf{v}_{\textbf{p}}\epsilon_{\textbf{p}}c_{\textbf{p}\sigma}^\dagger c_{\textbf{p}\sigma} $$

My question is how does he get to this? I think since $\partial_t$ doesn't make sense for a free particle system, he's using the commutator relation to get the time evolution, but if you plug $\psi=a_ke^{ikr}$ into the definition of $\textbf{R}_E$, using $H=\nabla^2$ the first term gives $$\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2\psi=\frac{\hbar^2}{2m}\psi^\dagger r\nabla^2e^{ikr}=-\frac{\hbar^2k^2}{2m}r\psi^\dagger\psi$$ for the second term, we use the product rule (and reusing an old short-hand he's used previously) to find $$\frac{\hbar^2}{2m}\psi^\dagger\nabla^2(r\psi)=\frac{\hbar^2}{2m}\left[\psi^\dagger r\nabla^2\psi+2\psi^\dagger(\nabla r\cdot\nabla\psi)\right]=\frac{\hbar^2}{2m}\left[-r\psi^\dagger\psi+2\psi^\dagger\nabla\psi\right]$$ where the only difference is an additional $\frac{i\hbar^2k}{2m}\psi^\dagger\psi$ term. However, once we carry out the integral terms with $\textbf{r}$ dominate and the integral diverges. Using the alternative definition of $$\frac{\partial}{\partial t}\textbf{R}_E=\frac{1}{2}\int\textbf{r}\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})+\frac{\partial}{\partial t}\mathcal{H}(\textbf{r})\textbf{r}d^3r=\textbf{j}_E$$ doesn't seem to change anything either. Even if we got to a stage where we had something to put into the commutator $\partial_t\textbf{R}_E=i[H,\textbf{R}_E]$ the number operators would commute, meaning the current would be zero. How does he get the energy current for a free particle system?