I am studying for my test in Quantum Mechanics, and there is something I don't quite understand about the Heisenberg picture and Heisenberg's equation of motion. In the lecture, we derived Heisenberg's equation: $$\frac{\mathrm{d}\hat{A}_H}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{A}_H, \hat{\mathscr{H}}_H\right] + \left(\frac{\partial\hat{A}_S}{\partial t}\right)_H$$ where $$\hat{\mathscr{H}}_H = \hat{\mathscr{U}}^{\dagger}\hat{\mathscr{H}}\hat{\mathscr{U}}$$ is the Hamiltonian operator in the Heisenberg picture ($\hat{\mathscr{U}}$ is the time evolution operator), which is equal to the Hamiltonian in the Schrodinger picture for most systems (including the one discussed below), and: $$\left(\frac{\partial\hat{A}_S}{\partial t}\right)_H = \hat{\mathscr{U}}^{\dagger}\frac{\partial\hat{A}_S}{\partial t}\hat{\mathscr{U}}$$ is the explicit time dependence of $\hat{A}_S$ if there is one.

This last term of the explicit time dependence is the one that bothers me. In many texts, I see it written and used without the time evolution operator: $$\frac{\mathrm{d}\hat{A}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{A}, \hat{\mathscr{H}}\right] + \frac{\partial\hat{A}}{\partial t}$$ This form of Heisenberg's equation makes more sense in terms of its relation with the classical Hamilton's equations, and it certainly makes sense if we only work in the Heisenberg picture and we treat $\hat{A}_S$ as $\hat{A}(t = 0)$ (So we don't need to know about the existence of the Schrodinger picture at all). However, I can't understand why we can ignore the time evolution operator like that.

To give a more concrete example, in a follow-up lecture, we learned about a quantum particle in electromagnetic fields with the Hamiltonian: $$\hat{\mathscr{H}} = \frac{1}{2m}\left(\hat{\mathbf{P}} - q\mathbf{A}\right)^2 + q\phi$$ where $\phi(\mathbf{x}, t)$ and $\mathbf{A}(\mathbf{x}, t)$ are the electric and magnetic potentials which depend on the position operator and possibly on time explicitly. We defined the mechanical momentum operator: $$\hat{\mathbf{\Pi}} = \hat{\mathbf{P}} - q\mathbf{A}$$ so the Hamiltonian becomes: $$\hat{\mathscr{H}} = \frac{1}{2m}\hat{\mathbf{\Pi}}^2 + q\phi$$ The lecturer then went on to use Heisenberg's equation on $\hat{\mathbf{\Pi}}$: $$\frac{\mathrm{d}\hat{\mathbf{\Pi}}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{\mathbf{\Pi}}, \frac{1}{2m}\hat{\mathbf{\Pi}}^2 + q\phi\right] + \frac{\partial \hat{\mathbf{\Pi}}}{\partial t}$$ He then said that $$\frac{\partial \hat{\mathbf{\Pi}}}{\partial t} = -q\frac{\partial \mathbf{A}}{\partial t}$$ since $\mathbf{A}$ can depend on time explicitly and continued solving to derive the quantum version of the Lorentz force: $$m\frac{\mathrm{d}^2\hat{\mathbf{X}}}{\mathrm{d}t^2} = q\left(\mathbf{E} + \frac{1}{2}\left(\frac{\mathrm{d}\hat{\mathbf{X}}}{\mathrm{d}t} \times \mathbf{B} - \mathbf{B} \times \frac{\mathrm{d}\hat{\mathbf{X}}}{\mathrm{d}t}\right)\right)$$ In this derivation, there was no use of the time evolution operator and this is confusing me. Weren't we supposed to write: $$\frac{\mathrm{d}\hat{\mathbf{\Pi}}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{\mathbf{\Pi}}, \frac{1}{2m}\hat{\mathbf{\Pi}}^2 + q\phi\right] + \hat{\mathscr{U}}^{\dagger}\frac{\partial \hat{\mathbf{\Pi}}}{\partial t}\hat{\mathscr{U}}$$ instead? I don't think it's equivalent to the equation he wrote since these operators don't commute: $$\left[\hat{\mathscr{U}}, \frac{\partial \hat{\mathbf{\Pi}}}{\partial t}\right] = \left[e^{-\frac{it}{\hslash}\hat{\mathscr{H}}}, -q\mathbf{A}\right] \ne 0$$ As I stated before, it feels right to write the equation as: $$\frac{\mathrm{d}\hat{A}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{A}, \hat{\mathscr{H}}\right] + \frac{\partial\hat{A}}{\partial t}$$ and use it like the lecturer did for $\hat{\mathbf{\Pi}}$, but the derivation clearly shows that the last term with the partial time derivative should be sandwiched between the time evolution operator and its adjoint. Why are we able to use Heisenberg's equation like that without worrying about the time evolution operator at all?


1 Answer 1


The lecturer then went on to use Heisenberg's equation on $\hat{\mathbf{\Pi}}$: $$\frac{\mathrm{d}\hat{\mathbf{\Pi}}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{\mathbf{\Pi}}, \frac{1}{2m}\hat{\mathbf{\Pi}}^2 + q\phi\right] + \frac{\partial \hat{\mathbf{\Pi}}}{\partial t}$$ He then said that $$\frac{\partial \hat{\mathbf{\Pi}}}{\partial t} = -q\frac{\partial \mathbf{A}}{\partial t}$$

... Weren't we supposed to write: $$\frac{\mathrm{d}\hat{\mathbf{\Pi}}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{\mathbf{\Pi}}, \frac{1}{2m}\hat{\mathbf{\Pi}}^2 + q\phi\right] + \hat{\mathscr{U}}^{\dagger}\frac{\partial \hat{\mathbf{\Pi}}}{\partial t}\hat{\mathscr{U}}$$ instead?

I think the problem you are having might be remedied by a specific example. Suppose, for example, that we can write: $$ \mathbf{A}_S = \alpha \mathbf{X}_S \sin(\omega t)\;, $$ where $\alpha$ and $\omega$ are just real-number parameters, and $\mathbf{X}_S$ is the position operator. Such a form of $\mathbf{A}_S$ looks like a time-dependent dipole term.

We have $$ \frac{\partial \mathbf{A}_S}{\partial t} = \alpha\omega \mathbf{X}_S \cos(\omega t)\tag{1} $$ and $$ U^\dagger \frac{\partial \mathbf{A}_S}{\partial t} U = \alpha\omega \mathbf{X}_H(t) \cos(\omega t)\tag{2}\;. $$

See how Eq. (2) looks like Eq. (1) except we have made the replacement $\mathbf{X}_S \to \mathbf{X}_H$? This is the notational shorthand that your professor is seemingly exploiting (perhaps too liberally).

I don't think it's equivalent...

Right, they aren't necessarily equivalent if you take the $\frac{\partial}{\partial t}$ symbol to have its previous literal meaning.

The way to make sense of the equation, and to see how they can be equivalent, is to no longer think of $\partial/\partial t$ as meaning the derivative where $X_S$ and $P_S$ are fixed, but rather think of it as the derivative where $X_H$ and $P_H$ are fixed. Yes, that does not immediately seem to make sense if you are thinking about the Schrodinger picture operators, but it turns out to work, and it is not so strange after a little thought. Operationally, you can just think, "I'm making the replacements $X_S \to X_H$ and now I'm holding $X_H$ fixed for my partial derivatives with respect to time."

In my example, we have: $$ \mathbf{A}_S = \alpha \mathbf{X}_S \sin(\omega t) $$ and $$ \mathbf{A}_H = \alpha \mathbf{X}_H \sin(\omega t)\;, $$ and so, if I think of my partial derivative with respect to time as taking $\mathbf{X}_H$ as fixed, I have: $$ \left(\frac{\partial\mathbf{A}_H}{\partial t}\right)_{X_H,P_H} = \alpha\omega \mathbf{X}_H \cos(\omega t)\;, $$ just like Eq. (2) above.

Anyways, if you ever get worried about the notational/symbolic shorthand, you can always just return to your previous, more explicit definition: $$\frac{\mathrm{d}\hat{\mathbf{\Pi}_H}}{\mathrm{d}t} = \frac{1}{i\hslash}\left[\hat{\mathbf{\Pi}_H}, \frac{1}{2m}\hat{\mathbf{\Pi}_H}^2 + q\phi\right] + U^\dagger\frac{\partial \hat{\mathbf{\Pi}_S}}{\partial t}U\;,$$ where the partial dervitivative with respect to time now means we are holding $X_S$ and $P_S$ fixed.

Finally, to see more formally why $$ U^\dagger \left(\frac{\partial A_S}{\partial t}\right)_{X_S,P_S}U = \left(\frac{\partial A_H}{\partial t}\right)_{X_H,P_H} $$ you can look at a power series. The result follows simply by inserting factors of $U U^\dagger =1$ where needed in the power series to convert Schrodinger operators to Heisenberg operators where needed. E.g.,: $$ A_S(t) = \sum_n c_n(t) (X_S)^n\;; $$ $$ A_H(t) = \sum_n c_n(t) (X_H)^n\;; $$ $$ \left(\frac{\partial A_S}{\partial t}\right)_{X_S} = \sum_n \frac{dc_n}{dt} (X_S)^n\;; $$ $$ U^\dagger\left(\frac{\partial A_S}{\partial t}\right)_{X_S} U= \sum_n \frac{dc_n}{dt} (X_H)^n\tag{3}\;; $$ $$ \left(\frac{\partial A_H}{\partial t}\right)_{X_H} = \sum_n \frac{dc_n}{dt} (X_H)^n\tag{4}\;. $$

The right hand sides of Eq. (3) and Eq. (4) are the same.


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