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

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.

• To assemble such a proof, you might start by producing an eigenbasis of the time derivative operator. – probably_someone Feb 20 '18 at 18:09

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

• Nice, do we need any conditions on the second equality (existence of the derivative?) or does conjugation always commute with limiting? – Alec Rhea Feb 20 '18 at 22:35
• Conjugation $\dagger :{\cal H}\to {\cal H}^{\ast}\cong {\cal H}$ is an anti-linear isometry between Hilbert spaces, and hence continuous, so that it commutes with limiting. – Qmechanic Feb 21 '18 at 9:03
• But $\mathcal{H}\cong\mathcal{H}^*$ only holds in the finite dimensional case or the bounded infinite dimensional case; are those requirements necessary? – Alec Rhea Feb 21 '18 at 12:12

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 $$\left[ \frac{d}{dt} \left( \langle \phi| \psi_i \rangle \right)\right]^* = \frac{d}{dt} \left( \langle \phi| \psi_i \rangle^* \right) \, ,$$ 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.

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).

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

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

• I tried to do this but here you use the fact you can intervet the limit and the dagger. How do we know we can do this ? – StarBucK Feb 20 '18 at 19:55
• That's a reasonable question, and the answer depends on how deeply you want to go. The operator norm of some operator $A$ is equal to the operator norm of $A^\dagger$ - you can use this fact to demonstrate that if you have a sequence of operators $\{a_n\}$ which converges to $A$, then the sequence $\{a_n^\dagger\}$ converges to $A^\dagger$, and take the limit that way. – J. Murray Feb 20 '18 at 20:14
• @StarBucK I've added to my answer to justify the exchange of the limit and the adjoint operation. – J. Murray Feb 20 '18 at 20:35