1
$\begingroup$

By applying partial time derivative to $$\psi_t \rightarrow U \psi_{t_0}$$ we end up with an expression for the Hamiltonian $$H = i\hbar\frac{\partial U_{t}}{\partial t}U^{\dagger}_{t}$$ where $U$ is the unitary time evolution operator. How can we show that $(\partial U_{t} / \partial t) U^\dagger_t$ is anti-Hermitian? If $(\partial U_t / \partial t)U^\dagger_t$ is anti-Hermtian surely we can multiply $i$ to $(\partial U_t / \partial t)U^\dagger_t$ which would make $i(\partial U_t/\partial t)U^\dagger_t$ Hermitian. Right?

$\endgroup$

closed as off-topic by Aaron Stevens, Jon Custer, Cosmas Zachos, ZeroTheHero, tpg2114 Aug 5 at 3:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, Cosmas Zachos, ZeroTheHero, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

3
$\begingroup$

Since $U_t$ is unitary, we have:

$$U_t U_t^\dagger = \mathbb I$$

Taking a derivative from both sides:

$$\frac{\partial U_t}{\partial t} U_t^\dagger +U_t (\frac{\partial U_t}{\partial t})^\dagger =0$$

so:

$$\frac{\partial U_t}{\partial t} U_t^\dagger =-U_t (\frac{\partial U_t}{\partial t})^\dagger$$

or:

$$\frac{\partial U_t}{\partial t} U_t^\dagger = -\Big(\frac{\partial U_t}{\partial t} U_t^\dagger \Big)^\dagger$$

Meaning that $\frac{\partial U_t}{\partial t} U_t^\dagger$ is anti-Hermitian.

EDIT:

The OP brings up an interesting question in the comments: Can any anti-Hermitian operator-valued function $A(t)$ be written in the form $\dot{U}(t) U^\dagger(t)$ for some unitary operator $U(t)$?

The answer is yes, as long as $A(t)$ is differentiable. To see this, let's define the operator $U(t)$ through the following initial value problem:

$$\dot{U}(t) = A(t) U(t) \qquad \text{with} \quad U(0) = \mathbb I \qquad (*)$$

First of all, taking the Hermitian conjugate of $(*)$ we get:

$$\dot{U}^\dagger(t) = -U^\dagger(t)A(t) \qquad \text{with} \quad U^\dagger(0) = \mathbb I \qquad (**)$$

Using $(*),(**)$ we can show that $U(t)$ is indeed unitary. Let's calculate the derivative of $U^\dagger(t) U(t)$:

$$\frac{\partial }{\partial t} \Big(U^\dagger(t) U(t) \Big) = \dot{U}^\dagger(t) U(t) +U^\dagger(t) \dot{U}(t) = -U^\dagger (t) A(t) U(t)+U^\dagger(t) A(t) U(t) = 0$$

So $U^\dagger(t) U(t)$ is constant in time meaning that:

$$U^\dagger(t) U(t)= U^\dagger(0) U(0) = \mathbb I$$

Therefore, $U(t)$ is indeed unitary. Now from $(*)$ we can easily write:

$$A(t) = \dot{U}(t) U^\dagger(t)$$

meaning that we've proved our claim.

Second EDIT:

Emilio Pisanty points out in the comments that our defining initial value problem for $U$ may not have a solution in the general case of an infinite dimensional Hilbert space. The above argument obviously breaks down for those settings.

$\endgroup$
  • 1
    $\begingroup$ Beat me to it; it’s a very simple argument :) $\endgroup$ – CR Drost Aug 2 at 16:34
  • 1
    $\begingroup$ I meant the definition of an anti-Hermitian operator. An operator $A$ is called anti-Hermitian when it's equal to to minus its own adjoint, so $A = - A^\dagger$. Now take $A = \partial U_t / \partial t \times U_t^\dagger$ in this definition, that's the same as the final equation in my answer. $\endgroup$ – Sahand Tabatabaei Aug 2 at 17:41
  • 2
    $\begingroup$ "let's define the operator $U(t)$ through the following initial value problem" - so long as the IVP has a solution, which need not be guaranteed; this is where all the complicated functional analysis lives. $\endgroup$ – Emilio Pisanty Aug 2 at 19:13
  • 2
    $\begingroup$ @EmilioPisanty That's true, I guess I swept those technicalities under the rug. This would at least work for finite-dimensional Hilbert spaces. But the solution is not guarantied to exist in the general case. $\endgroup$ – Sahand Tabatabaei Aug 2 at 19:16
  • 3
    $\begingroup$ I don't think sweeping technicalities under the rug is a problem, particularly if your audience wouldn't cope, but it's always best to mark those points explicitly in the text. $\endgroup$ – Emilio Pisanty Aug 2 at 19:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.