# How can we prove that the Hamiltonian for any quantum system is Hermitian? [closed]

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?

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

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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.

• Beat me to it; it’s a very simple argument :) – CR Drost Aug 2 at 16:34
• 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. – Sahand Tabatabaei Aug 2 at 17:41
• "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. – Emilio Pisanty Aug 2 at 19:13
• @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. – Sahand Tabatabaei Aug 2 at 19:16
• 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. – Emilio Pisanty Aug 2 at 19:28