This might be a very simple question but I don't understand how to compute the complex conjugate of the Schrödinger equation: $$ i\partial_t \psi = H\psi $$ where $H$ is an hermitian operator. How to proceed now? All answers I found via the search function were not satisfactory for me.

First idea: Complex conjugate both sides: $$ (i\partial_t \psi)^* = (H\psi)^* $$ The problem is that I am not sure how to proceed here. What is the conjugate of an operator $A$ acting on a function $f$? So what is $(Af)^*$? Does the following hold: $(Af)^* = A^\dagger f^*$ ? If so, I get $$ -i\partial_t^\dagger \psi^* = H\psi^* $$ The problem is now, that I don't know what $\partial_t^\dagger$ is? If $\partial_t^\dagger = \partial_t$ holds then I have the common result!

Second idea: Say $C$ is the operator that conjugates a function: $Cf=f^*$. Then I can write $$ H\psi^* = HC\psi = (HC + CH - CH)\psi = [H,C]\psi + CH\psi = [H,C]\psi + Ci\partial_t\psi $$ where $[\cdot,\cdot]$ is the commutator. Moreover I find $$ Ci\partial_t\psi = Ci\partial_t\psi - i\partial_tC\psi + i\partial_tC\psi = \{i\partial_t,C\}\psi - i\partial_t\psi^* $$ where $\{\cdot,\cdot\}$ is the anticommutator. Thus I get $$ H\psi^* = - i\partial_t\psi^* + [H,C]\psi + \{i\partial_t,C\}\psi $$ If I could prove that $[H,C]= \{i\partial_t,C\} = 0$ then I would also get the common result.

But unfortunately I get stuck here... can someone resolve my problems?

  • 1
    $\begingroup$ $\partial_t$ is not an operator, which means $\partial_t^\dagger=\partial_t$. See physics.stackexchange.com/q/17477 $\endgroup$ – AccidentalFourierTransform Apr 1 '16 at 12:49
  • $\begingroup$ Why is $\partial_t$ not an operator? In the mathematical sense it is. And is my claim true that $(Af)^* = A^\dagger f^*$ holds? $\endgroup$ – thyme Apr 1 '16 at 12:58
  • $\begingroup$ @thyme: It depends on what you mean by "operator." $\endgroup$ – Buzz Apr 1 '16 at 13:03
  • $\begingroup$ I don't know what I mean with "operator". I guess I wouldn't have these problems if I knew it. So if I want to calculate $(i\partial_t\psi)^*$ is $\partial_t$ an operator or not? $\endgroup$ – thyme Apr 1 '16 at 13:08
  • 2
    $\begingroup$ Comments to the question (v3): 1. Time $t$ is not an operator but an external real parameter. 2. Complex conjugation acts trivially on the time derivative $\frac{\partial}{\partial t}$. 3. Complex conjugation $A^{\ast}$ and Hermitian conjugation $A^{\dagger}$ of a differential operator $A$ are in general different operations, cf. e.g. the example in footnote 1 in my Phys.SE answer here. So $(A\psi)^{\ast}= A^{\ast}\psi^{\ast}\neq A^{\dagger}\psi^{\ast}$ in general. $\endgroup$ – Qmechanic Apr 1 '16 at 13:35

Generally speaking the complex conjugate of an operator is not a standard notion of operator theory, though it can be defined after having introduced some general notions.

Definition. A conjugation $C$ in a Hilbert space $\cal H$ is an antilinear map $C : \cal H \to \cal H$ such that is isometric ($||Cx||=||x||$ if $x\in \cal H$) and involutive ($CC=I$).

There are infinitely many such maps, at least one for every Hilbert basis in $\cal H$ (the map which conjugates the components of any vector with respect to that basis). On $L^2$ spaces there is a standard conjugation $$C : L^2(\mathbb R^n, dx) \ni \psi \mapsto \overline{\psi}\in L^2(\mathbb R^n, dx)\:, $$ where $\overline{\psi}(x) := \overline{\psi(x)}$ for every $x \in \mathbb R^n$ and where $\overline{a+ib}:= a-ib$ for $a,b \in \mathbb R$.

Definition. An operator $H : D(H) \to \cal H$ (where henceforth $D(H)\subset \cal H$) is said to be real with respect to a conjugation $C$ if $$CHx=HCx \quad \forall x \in D(H)$$ (which implies $C(D(H)) \subset D(H)$ and thus $C(D(H)) = D(H)$ in view of $CC=I$, so that the written condition can equivalently be re-phrased $CH=HC$).

The complex conjugate $H_C$ of an operator $H$ with respect to a conjugation $C$, can be defined as $$H_C :=CHC\:.$$ This operator with domain $C(D(H))$ is symmetric, essentially self-adjoint, self-adjoint if $H$ respectively if is symmetric, essentially self-adjoint, self-adjoint. Obviously it coincides with $H$ if and only if $H$ is real with respect to $C$.

Let us come to your issue. Let us start form Schroedinger equation $$-i \frac{d}{dt} \psi_t = H\psi_t\:.$$ Here we have a vector valued map $$\mathbb R \ni t \mapsto \psi_t \in \cal H\:,$$ such that $\psi_t \in D(H)$ for every $t \in \mathbb R$ and the derivative is computed with respect to the topology of the Hilbert space whose norm is $||\cdot|| = \sqrt{\langle \cdot| \cdot \rangle}$: $$\frac{d}{dt} \psi_t = \dot{\psi}_t \in \cal H$$ means $$ \lim_{h\to 0} \left|\left| \frac{1}{h} (\psi_{t+h} -\psi_t) - \dot{\psi}_t \right|\right|=0\:.$$ (See the final remark)

If $C : \cal H \to \cal H$ is a conjugation, as it is isometric and involutive, in view of the definition above of derivative, we have $$C\frac{d}{dt} \psi_t = \frac{d}{dt} C\psi_t\tag{1}$$ where both sides exist or do not simultaneously.

Summing up, given a conjugation $C$, and the (self-adjoint) Hamiltonian operator $H$, both in the Hilbert space $\cal H$, the complex conjugate of the Schroedinger equation $$-i \frac{d}{dt} \psi_t = H\psi_t\:.\tag{2}$$ is a related equation satisfied by $C\psi_t$ and just obtained by applying $C$ to both sides of (2) and taking (1) and $CC=I$ into account, obtaining $$i \frac{d}{dt} C\psi_t = H_C\: C\psi_t\:.$$ If $H$ is real with respect to $C$ (this is the case for a particle without spin described in $L^2(\mathbb R^3)$, assuming the Hamiltonian of the form $P^2/2m + V$ and $C$ is the standard complex conjugation of wavefunctions), the equation reduces to $$i \frac{d}{dt} C\psi_t = H C\psi_t\:.$$

REMARK. It is worth stressing that $-i \frac{d}{dt}$ is not an operator in the Hilbert space $\cal H$ as, for instance, $H$ is. To compute $H\psi$, it is enough to know the vector $\psi \in D(H)$. To compute $\frac{d}{dt}\psi_t$ we must know a curve of vectors $$\gamma :\mathbb R \ni t \mapsto \psi_t \in \cal H\:.$$ $\frac{d}{dt}$ computes the derivative of such curve defining another curve $$\dot{\gamma} :\mathbb R \ni t \mapsto \frac{d}{dt}\psi_t \in \cal H\:.$$ More weakly one may view $\frac{d}{dt}|_{t_0}$ as a map associating vector-valued curves defined in aneighborhood of $t_0$ to vectors $\frac{d}{dt}|_{t_0}\psi_t$. In both cases it does not make sense to apply the derivative to a single vector $\psi$, contrarily to $H\psi$ is well defined.

  • $\begingroup$ Thank you for this detailed explanation. But for me the problem is now shifted to the question: Why is $H$ real with respect to $C$? You say that this is true for a Hamiltonian of the form $H=P^2/2m+V$. But why? $\endgroup$ – thyme Apr 1 '16 at 16:06
  • $\begingroup$ Assume $2m=1$ it does not matter. $P^2$ is $-\Delta$ the Laplacian operator, $V$ is a real function. Therefore, with the standard definition of conjugation in $L^2$ we have $(C (P^2+ V)\psi)(x) = (C(-\Delta + V)\psi)(x) = \overline{-\Delta_x \psi + V(x)\psi(x)} = -\Delta_x\overline{\psi(x) } + V(x)\overline{\psi(x)} = ((P^2+ V) C \psi)(x)$ so that $C (P^2+ V) = (P^2+ V)C$ and $C (P^2+ V)C = (P^2+ V)CC = P^2+ V$ $\endgroup$ – Valter Moretti Apr 1 '16 at 16:32
  • $\begingroup$ @thyme: Hamiltonians are always Self-adjoint, so they remain unchanged under the transpose conjugate operation. It is this property which ensures that the eigenvalues are real. $\endgroup$ – Peter Diehr Apr 1 '16 at 16:35

You actually want the transpose-conjugate, which applies to everything: $$ (i\partial_t |\psi\rangle )^\dagger = (H\|psi\rangle)^\dagger $$ This gives: $$ -i\partial_t |\psi\rangle^\dagger = |\psi\rangle^\dagger H^\dagger $$ which reduces to: $$ -i\partial_t \langle\psi| = \langle\psi|H $$

The differential operator remains unchanged.

Note if your units don't have $\hbar=1$, it belongs on the LHS.

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    $\begingroup$ I guess the $\dagger$'s in the last line should not be there? $\endgroup$ – thyme Apr 1 '16 at 13:56

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