I want to show that for any $\mathcal{L}^2(\mathbb{C}^n\times\mathbb{R},\mathbb{C})$-function $\Psi(x,t)$ the differential operator $\partial_t$ is self-adjoint, i.e. $\langle\Phi|\partial_t\Psi\rangle=\langle\partial_t\Phi|\Psi\rangle$ for any $\Phi,\Psi\in\mathcal{L}^2(\mathbb{C}^n\times\mathbb{R},\mathbb{C})$, and therefore $\partial^\dagger_t=\partial_t$.

By definition the Hermitian scalar product of $\Psi,\Phi$ is $$\langle\Phi(t)|\partial_t\Psi(t)\rangle=\int_{\mathbb{C}^n}\Phi^\ast(x,t)\frac{\partial\Psi}{\partial t}(x,t)d^nx.$$

Because of $\frac{\partial\Phi^\ast\Psi}{\partial t}(x,t)=\Phi^\ast(x,t)\frac{\partial\Psi}{\partial t}(x,t)+\frac{\partial\Phi}{\partial t}(x,t)\Psi(x,t)$, we can rewrite this as

$$\int_{\mathbb{C}^n}\Phi^\ast(x,t)\frac{\partial\Psi}{\partial t}(x,t)d^nx = \int_{\mathbb{C}^n}\frac{\partial\Phi^\ast\Psi}{\partial t}(x,t)d^nx-\int_{\mathbb{C}^n}\frac{\partial\Phi^\ast}{\partial t}(x,t)\Psi(x,t)d^nx.$$

Since $\Phi(x,t),\Psi(x,t)$ are $\mathcal{L}^2$-integrable over $\mathbb{C}^n$ for each constant $t\in\mathbb{R}$ by default we can exchange the integral and the differential for the first summand, which gives us:

$$\int_{\mathbb{C}^n}\frac{\partial\Phi^\ast\Psi}{\partial t}(x,t)d^nx = \frac{\partial}{\partial t}\int_{\mathbb{C}^n}\Phi^\ast\Psi(x,t)d^nx = \frac{\partial}{\partial t}|\langle\Phi(t)|\Psi(t)\rangle|^2.$$

This is equal to zero, which we can show by using the unitary time evolution operator $U(t,t_0)$. Let $|\Psi(t)\rangle = U(t,t_0)|\Psi(t_0)\rangle$. Then $\langle\Phi(t)| = \left(U(t,t_0)|\Phi(t_0)\rangle\right)^\dagger=\langle\Phi(t_0)|U^\dagger(t,t_0)$. Therefore, the scalar product of both vectors is time-independent:

$$ \frac{\partial}{\partial t}|\langle\Phi(t)|\Psi(t)\rangle|^2 = \frac{\partial}{\partial t}\langle\Phi(t_0)|U^\dagger(t,t_0)U(t,t_0)|\Psi(t_0)\rangle = \frac{\partial}{\partial t}\langle\Phi(t_0)|\Psi(t_0)\rangle = 0.$$

Hence, we now have the following equation: $$\int_{\mathbb{C}^n}\Phi^\ast(x,t)\frac{\partial\Psi}{\partial t}(x,t)d^nx = -\int_{\mathbb{C}^n}\frac{\partial\Phi^\ast}{\partial t}(x,t)\Psi(x,t)d^nx.$$

Doesn't this rather mean that $\partial^\dagger_t=-\partial_t$? This is where I am stuck.

  • $\begingroup$ It is a pity that this question moved here form PS, since the physical nature of the question is prominent. $\endgroup$
    – V. Moretti
    Jan 10, 2017 at 9:35

2 Answers 2


What you wrote is wrong, sorry. If the Hilbert space is $L^2(\mathbb C^n \times \mathbb R, \mathbb C)$, then the relevant integral is $$\int_{\mathbb C^n \times \mathbb R} \overline{\psi(x,t)} \phi(x,t) d^nx dt\tag{1}$$ where (I am interpreting your notations) the functions are complex-valued and $d^nx$ is the Lebesgue measure over $\mathbb C^n$ viewed as $\mathbb R^{2n}$ and $dt$ the one over $\mathbb R$. Presumably your $\mathbb C^n$ should be $\mathbb R^n$.

Instead you use the measure $d^nx$ over the factor factor only. $$\int_{\mathbb C^n} \overline{\psi(x)} \phi(x) d^nx$$ In this case it does make any sense to wonder whether or not $\partial_t$ is Hermitian or not, since it is not an operator over the linear space $L^2(\mathbb C^n, \mathbb C)$, that is on the functions $\psi=\psi(x)$

(Even dealing with the whole Hilbert space $L^2(\mathbb C^n \times \mathbb R, \mathbb C)$ and the scalar product (1) there would be other issues regarding regularity of functions but the problem arising from your approach is much more relevant.)

In elementary Quantum Mechanics the Hilbert space is $L^2(\mathbb R^n)$ referred to Lebesgue's measure $d^nx$ over $\mathbb R^n$.

Operators are linear maps $A: D(A) \to L^2(\mathbb R^n)$ where $D(A) \subset L^2(\mathbb R^n)$ is a (usually dense) linear subspace.

You are instead considering a function from $\mathbb R$ to $L^2(\mathbb R^n)$ physically representing the temporal evolution of a vector $\psi_0$ at $t=0$ $$\mathbb R \ni t \mapsto \psi_t \in L^2(\mathbb R^n) $$ The derivative you are considering acts on such a family $$\partial_t \psi_t := \lim_{h \to 0} \frac{\psi_{t+h} -\psi_t}{h} \tag{2}$$ and, if exists, is computed with respect to the topology of $L^2(\mathbb R^n)$. You see that to compute (2) you need a $t$-parametrized family of vectors.

With a single vector $\psi \in L^2(\mathbb R^n)$, $\partial_t \psi$ would not have any meaning, differently from, say, $A\psi$ -- where $A$ is a true linear operator in $L^2(\mathbb R^n)$ and $\psi \in D(A)$ -- which is meaningful.

  • $\begingroup$ Ah, I see! This is also what bothered me since adjoint operators are only defined with respect to the scalar product of the Hilbert space they operate on and I was not quite sure the integral I used was indeed a scalar product or the $\mathcal{L}^2$-space I used a Hilbert space at all. $\endgroup$ Jan 10, 2017 at 10:00
  • $\begingroup$ I came across this problem while trying to derive the von Neumann equation from the Schroedinger equation, where one adjoints the equation. I was sceptical, if that is even formally possible, i.e. $$i\hbar\partial_t\Psi(t) = H\Psi(t) \Leftrightarrow -i\hbar\left(\partial_t\Psi(t)\right)^\ast = \Psi^\ast(t)H$$. $\endgroup$ Jan 10, 2017 at 10:07
  • $\begingroup$ I am not sure to understand well. However the adjoint equation has here a formal meaning, it does not mean that $\partial_t$ is or is not self-adjoint with respect to the Hilbert space structure... $\endgroup$
    – V. Moretti
    Jan 10, 2017 at 11:04
  • $\begingroup$ As far as I see, it is a (notationally confused since the horrible left-action of operators is used) way to write the Schroedinger equation in the dual Hilbert space. $\endgroup$
    – V. Moretti
    Jan 10, 2017 at 11:06

So, it seems that I had a "Brett vorm Kopf" as they say in German.

Since $\partial_t\Phi^\ast(x,t)$ is a real derivative, complex conjugation and derivation commute. Thus, indeed we have $-\int_{\mathbb{C}^n}\frac{\partial\Phi^\ast}{\partial t}(x,t)\Psi(x,t)d^nx = \int_{\mathbb{C}^n}\left(-\frac{\partial\Phi}{\partial t}(x,t)\right)^\ast\Psi(x,t)d^nx$, which concludes

$$\langle\Phi(t)|\partial_t\Psi(t)\rangle = \langle-\partial_t\Phi(t)|\Psi(t)\rangle \Leftrightarrow \partial^\dagger_t=-\partial_t.$$

Hence, my claim in the very beginning of my post was wrong. This is the answer I was looking for. The post can be closed.


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