Why is $\frac{d}{dt} \langle \psi | = (\frac{d}{dt} | \psi \rangle )^{\dagger} $?

Question

When we want to find the equation of motion of the density matrix, we use the property:

$$\frac{d}{dt} \langle \psi | = \left(\frac{d}{dt} | \psi \rangle \right)^{\dagger}. $$

Said differently, the time derivative and the hermitian conjugate "commute".

We use it in the proof about equation of motion for the density matrix:

$$ \frac{d}{dt} (| \psi \rangle \langle \psi |) = \frac{H}{ih} | \psi \rangle \langle \psi | - | \psi \rangle \langle \psi | \frac{H}{ih}.$$

But I don't understand why it is true from math point of view.

I see "intuitively" why it is true when I have a column vector that I transpose, to make the derivative before or after the transposition won't change the result.

But is there a more formal proof using some properties on Hermitian scalar product? I am looking for such a thing.

Answer 1

Formally we have $$ \left( \frac{d}{dt}| \psi(t) \rangle\right)^{\dagger} ~=~\left(\lim_{t^{\prime} \to t} \frac{ | \psi(t^{\prime}) \rangle - | \psi(t) \rangle}{t^{\prime}-t}\right)^{\dagger} ~=~\lim_{t^{\prime} \to t} \frac{ | \psi(t^{\prime}) \rangle^{\dagger} - | \psi(t) \rangle ^{\dagger}}{t^{\prime}-t}$$ $$~=~\lim_{t^{\prime} \to t} \frac{ \langle \psi(t^{\prime}) | - \langle \psi(t) |}{t^{\prime}-t} ~=~\frac{d}{dt}\langle \psi(t) |. $$

Answer 2

Let's see if this argument is good enough. Pick a basis of time-independent states $\left\{|\psi_i\rangle\right\}$. I think you can at least allow me to write \begin{equation} \left[ \frac{d}{dt} \left( \langle \phi| \psi_i \rangle \right)\right]^* = \frac{d}{dt} \left( \langle \phi| \psi_i \rangle^* \right) \, , \end{equation} being this just the derivative of a $\mathbb{C}$ number.

With that given, we have \begin{eqnarray} \left[ \left(\frac{d}{dt}\langle \phi| \right) | \psi_i \rangle \right]^* &=& \left[ \frac{d}{dt} \left( \langle \phi| \psi_i \rangle \right)\right]^* = \frac{d}{dt} \left( \langle \phi| \psi_i \rangle^* \right) = \frac{d}{dt} \left( \langle \psi_i | \phi \rangle \right) \\ &=& \langle \psi_i |\left( \frac{d}{dt} | \phi \rangle \right) = \left[ \left( \frac{d}{dt} | \phi \rangle \right)^{\dagger} | \psi_i \rangle\right]^* \, , \end{eqnarray} where in the last step we just used the definition of adjoint. This being valid for all $|\psi_i \rangle$, we can conclude we have found the adjoint we were looking for.

Answer 3

The bra-ket notation can muddy the issue somewhat. You want to know why, for some $\psi$,

$$\frac{d}{dt} \psi^\dagger = \big(\frac{d}{dt}\psi\big)^\dagger$$

Note that

$$\frac{d}{dt}\psi^\dagger := \lim_{\Delta t\rightarrow 0} \frac{\psi^\dagger(t+\Delta t) - \psi^\dagger(t)}{\Delta t}$$

But of course, since $\Delta t$ is a real number,

$$ \frac{\psi^\dagger(t+\Delta t) - \psi^\dagger(t)}{\Delta t} = \left(\frac{\psi(t+\Delta t) - \psi(t)}{\Delta t}\right)^\dagger $$

(ignoring subtleties like domain compatibility and such which is generally ignored in most physics contexts).

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To address the comments by OP:

Let $$ \lim_{\Delta t\rightarrow 0} \frac{A(t+\Delta t)-A(t)}{\Delta t} = A'(t)$$

This means that for all $\epsilon >0$, there exists some $\delta>0$ such that $$ |\Delta t|<\delta \implies \big\Vert\frac{A(t+\Delta t)-A(t)}{\Delta t} - A'(t) \big\Vert < \epsilon$$

where $\Vert \cdot \Vert$ is the operator norm. However, the operator norm of an operator $O$ is equal to the operator norm of $O^\dagger$, and since the adjoint operation distributes over addition and multiplication by real numbers, it immediately follows that

$$ |\Delta t|<\delta \implies \big\Vert\frac{A^\dagger(t+\Delta t)-A^\dagger(t)}{\Delta t} - [A'(t)]^\dagger \big\Vert < \epsilon$$

which is another way of saying that

$$ \frac{d}{dt}A^\dagger \equiv \lim_{\Delta t\rightarrow 0} \frac{A^\dagger(t+\Delta t)-A^\dagger(t)}{\Delta t} = [A'(t)]^\dagger \equiv \left[\frac{d}{dt}A\right]^\dagger$$