Heisenberg Equations of Motion for Electron in EM-field

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?

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

Demystifier said:

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.

thatboi said:

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.

Demystifier said:

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')]##.

thatboi said:

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.

thatboi said:

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
$$[\psi^{\dagger},H]=-(h^*\psi^{\dagger})$$

Demystifier said:

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.

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

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

thatboi said:

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.)