I Heisenberg Equations of Motion for Electron in EM-field

[thatboi]

Consider the Heisenberg picture Hamiltonian $$H(t) = \int_{\textbf{r}}\psi^{\dagger}(\textbf{r},t)\frac{(-i\hbar\nabla+e\textbf{A})^{2}}{2m}\psi(\textbf{r},t)$$ where $\psi(\textbf{r},t)$ is a fermion field operator. To find the equations of motion that $\psi,\psi^{\dagger}$ obey. I would invoke the Heisenberg equations of motion $i\hbar\partial_{t}\psi = [\psi(t),H(t)]$ and $i\hbar\partial_{t}\psi^{\dagger} = [\psi(t)^{\dagger},H(t)]$. I know the equations of motion should be $$i\hbar\partial_{t}\psi = \frac{1}{2m}((-i\hbar\nabla+e\textbf{A})^{2})\psi$$ and $$i\hbar\partial_{t}\psi^{\dagger} = -\frac{1}{2m}((i\hbar\nabla+e\textbf{A})^{2})\psi^{\dagger}$$
But I have redone this calculation so many times for $\psi^{\dagger}$ and do not understand how come the $i$ flips signs for the $\psi^{\dagger}$ case inside $((i\hbar\nabla+e\textbf{A})^{2})$. In both instances, the calculation requires us taking the commutator with the same Hamiltonian $H(t)$ so why is there suddenly a sign flip now?

 

[Demystifier]

Hint: Can you explain why the momentum operator $p=-i\hbar\nabla$ is hermitian?

 

[thatboi]

Hint: Can you explain why the momentum operator $p=-i\hbar\nabla$ is hermitian?

Do you mean to say there is some kind of integration by parts argument I am overlooking here? I don't think I see how it plays into the calculation of the 2 commutators.

 

[Demystifier]

Do you mean to say there is some kind of integration by parts argument I am overlooking here? I don't think I see how it plays into the calculation of the 2 commutators.

Yes, there's a partial integration implicitly involved. Let me first write the Hamiltonian as
$$H=\int d^3r \psi^{\dagger} (h \psi)$$
where
$$h=\frac{(-i\hbar\nabla+eA)^2}{2m}$$
The notation $(h\psi)$ reminds us that $h$ contains a derivative operator acting on $\psi$. Then
$$H^{\dagger}=\int d^3r (h \psi)^{\dagger} \psi =\int d^3r (h^* \psi^{\dagger}) \psi$$
In the last expression, $h^*$ acts only on $\psi^{\dagger}$, not on $\psi$. Now, if you want that the derivative operator acts on $\psi$, and not on $\psi^{\dagger}$, you must perform a partial integration. This partial integration flips the sign, so
$$H^{\dagger}=\int d^3r \psi^{\dagger} (h\psi)=H$$
confirming that $H$ is hermitian. The moral is, when you use the operator $h$ containing the derivative operator, you must always know on which object this derivative acts. When you keep track of this, you can see how the sign flips.

 

[thatboi]

Yes, there's a partial integration implicitly involved. Let me first write the Hamiltonian as
$$H=\int d^3r \psi^{\dagger} (h \psi)$$
where
$$h=\frac{(-i\hbar\nabla+eA)^2}{2m}$$
The notation $(h\psi)$ reminds us that $h$ contains a derivative operator acting on $\psi$. Then
$$H^{\dagger}=\int d^3r (h \psi)^{\dagger} \psi =\int d^3r (h^* \psi^{\dagger}) \psi$$
In the last expression, $h^*$ acts only on $\psi^{\dagger}$, not on $\psi$. Now, if you want that the derivative operator acts on $\psi$, and not on $\psi^{\dagger}$, you must perform a partial integration. This partial integration flips the sign, so
$$H^{\dagger}=\int d^3r \psi^{\dagger} (h\psi)=H$$
confirming that $H$ is hermitian. The moral is, when you use the operator $h$ containing the derivative operator, you must always know on which object this derivative acts. When you keep track of this, you can see how the sign flips.

Right, I can understand that argument but what I need to calculate is $[\psi,H]$ and $[\psi^{\dagger},H]$. I don't see why I cannot use the same H for both commutators. The $\psi$ are just field operators so what matters most is their commutation relations right. In both cases we would end up with $[\psi(r), \psi^{\dagger}(r')\psi(r')],[\psi(r)^{\dagger}, \psi^{\dagger}(r')\psi(r')]$.

 

[Demystifier]

Right, I can understand that argument but what I need to calculate is $[\psi,H]$ and $[\psi^{\dagger},H]$. I don't see why I cannot use the same H for both commutators.

You can use the same H, as I will explain.

The $\psi$ are just field operators so what matters most is their commutation relations right. In both cases we would end up with $[\psi(r), \psi^{\dagger}(r')\psi(r')],[\psi(r)^{\dagger}, \psi^{\dagger}(r')\psi(r')]$.

Using a short-hand notation $\psi(x)=\psi$, $\psi(x')=\psi'$, etc., I take
$$H=\int dx' \psi'^{\dagger}(h'\psi')$$
First I compute $[\psi,H]$ and after a straightforward calculus I obtain
$$[\psi,H]=(h\psi)$$
without using partial integration. Similarly, with the same expression for $H$, after a straightforward calculus I obtain
$$[\psi^{\dagger},H]=-\int dx' \psi'^{\dagger} (h'\delta(x-x'))$$
Now the point is that $h'$ contains a derivative acting on the $\delta$-function, which is ill-defined. Hence I want that $h'$ acts on $\psi'^{\dagger}$, and to achieve this I need a partial integration. The partial integration flips the sign, i.e. $h'\to h'^*$, giving
$$[\psi^{\dagger},H]=-(h^*\psi^{\dagger})$$

 

[thatboi]

You can use the same H, as I will explain.


Using a short-hand notation $\psi(x)=\psi$, $\psi(x')=\psi'$, etc., I take
$$H=\int dx' \psi'^{\dagger}(h'\psi')$$
First I compute $[\psi,H]$ and after a straightforward calculus I obtain
$$[\psi,H]=(h\psi)$$
without using partial integration. Similarly, with the same expression for $H$, after a straightforward calculus I obtain
$$[\psi^{\dagger},H]=-\int dx' \psi'^{\dagger} (h'\delta(x-x'))$$
Now the point is that $h'$ contains a derivative acting on the $\delta$-function, which is ill-defined. Hence I want that $h'$ acts on $\psi'^{\dagger}$, and to achieve this I need a partial integration. The partial integration flips the sign, i.e. $h'\to h'^*$, giving
$$[\psi^{\dagger},H]=-(h^*\psi^{\dagger})$$

Ahh, there was the missing minus sign I was looking for! Thanks so much, I was not careful enough with my treatment of the derivative of $\delta(x-x')$ and lost more time than I'd like to admit on this.

 

[Demystifier]

By the way, I have just checked, the same results are obtained also if $\psi$ is quantized as a bosonic field, rather than fermionic.

 

[thatboi]

Yes I agree, though I suppose if we were to think about bosonic fields, perhaps the sign of $e$ should change as well?

 

[PeterDonis]

Yes I agree, though I suppose if we were to think about bosonic fields, perhaps the sign of $e$ should change as well?

Why would the sign of $e$ need to change? Bosonic fields can have charge of either sign. (Consider the $W^+$ and $W^-$ fields in the Standard Model, for example, or charged pions.)