We know that a time-reversal operator $T$ can be represented as $$T=UK$$ where $U$ is some unitary operator and $K$ is the complex conjugation operator.

Then under time-reversal operation, a quantum state $|\psi\rangle$ will transform as the following: $$|\psi^R\rangle=T|\psi\rangle=UK|\psi\rangle=U|\psi^*\rangle$$ If we require time-reversal symmetry to the system, then we need to have $$\langle\psi^R|O^R|\phi^R\rangle=\langle\psi|O|\phi\rangle$$ where $|\psi\rangle$ and $|\phi\rangle$ are some arbitrary quantum states and $O$ is some operator. From the above equation, we would have $$\langle\psi^*|U^{\dagger}O^RU|\phi^*\rangle=\langle\psi|O|\phi\rangle$$ So based on this equation, how do we obtain the result (given in the book "random matrices" by Mehta) that $$O^R=UO^TU^{\dagger}$$ where $O^T$ means the transpose of $O$.

My second question is that, what if we do NOT assume time reversal symmetry?

  • $\begingroup$ I am reading your question, but I have a doubt. Time reversal is not only complex conjugate, what it has to do is also to transpose the items on which it acts (vectors, matrices). $\endgroup$
    – Sofia
    Commented Dec 3, 2014 at 22:33
  • 2
    $\begingroup$ Also, for reference \langle and \rangle produce nice looking brackets, $\langle \varphi | \psi \rangle$ instead of $< \varphi | \psi >$, if you care about that sort of thing. $\endgroup$
    – user12029
    Commented Dec 4, 2014 at 0:17
  • $\begingroup$ @Sofia also I don't think time reversal operator needs to transpose the operand $\endgroup$
    – M. Zeng
    Commented Dec 4, 2014 at 7:00

3 Answers 3


One problem with your formula that $T$ factorizes as the product of a unitary operator $U$ and complex conjugation $K$ is that 'complex conjugation' is meaningless in a Hilbert space a priori.

Let me be more formal about this; consider a vector $\psi \in H$, with $H$ a $n$-dimensional Hilbert space, that is, isomorphic to $\mathbb{C}^n$. How is an (abstract) vector space isomorphic to the canonical Hilbert space $\mathbb{C}^n$? In the following way: pick a basis $\{e_i\}_{i=1,...,n}$ of $H$. Then decompose $\psi$ in this basis:

$$ \psi = \sum_i \psi_i e_i \ .$$

Finally map $\psi$ on the $n$-tupel $(\psi_1,...,\psi_n)$.

As you can see, this isomorphism, lets denote it by $E$, very strongly depends on the chosen basis.

Let's ignore this anyway. On $\mathbb{C}^n$ we can define complex conjugation $K$, it is simply the map

$$ K(\psi_1,...,\psi_n) = (\overline{\psi_1},...,\overline{\psi_n}) \ . $$

Hence we can define a complex conjugation $K_E$ on $H$ simply by

$$ K_E = E^{-1} \circ K \circ E \ .$$

Now let's see what happens if we change a basis; consider another basis $E' = \{e'_i\}_{i=1,...,n}$ with $e_i = \sum_j M_{ij} e'_j$. Then considering $M_{ij}$ as the matrix of an operation on $\mathbb{C}^n$:

$$ K_{E} = E^{-1} \circ K \circ E = E'^{-1} \circ M^{-1} \circ K \circ M \circ E' = E'^{-1} \circ \overline{M}M^{-1} K \circ E' \ ,$$

i.e. unless $\overline{M} = M$, $K_E \neq K_{E'}$. Hence there is no invariant notion of complex conjugation in a complex Hilbert space.

The statement is true that if we fix a basis $E$, we may write complex conjugation w.r.t. any other basis as

$$ K_{E'} = U_{E,E'} K_E \ ,$$

where $U_{E,E'}$ is a unitary. Note that complex conjugation depends on $E$ only through the equivalence class $[E]$ of basis connected to $E$ through real transformations.

In this language one could say that a time-reversal operation is a choice of preferred equivalence class of basis. That is, at least, if $T^2 = 1$. If $T^2 = -1$, we should map $H$ to a $\mathbb{H}^{n/2}$, where $\mathbb{H}$ are the quaternions.

This leaves you with a recipe for computing the time-reversed operator, for $T^2=1$: simply represent $O$ in a real basis (a basis invariant under $T$), then take the complex conjugate of this matrix.

Notice that neither of these manipulations depend on $T$ being a symmetry.


Time reversal is not only complex conjugate, what it does is also to transpose the items on which it acts (vectors, matrices).

$$T\langle \phi|\hat{O}|\psi\rangle = \langle \psi T|\hat{O}|T \phi\rangle.$$

Notice the change of places of the functions in the right wing with respect to the left wing. Also, I used the fact that $\hat{O}$ is unchanged at time-reversal.

Now we do the following change which is allowed under the integral if the two functions vanish at infinity:

$$\langle UK\psi|\hat{O}|UK\phi\rangle = \langle \phi|U\hat{O}U^\dagger|\psi\rangle.$$

So, we got the time-reversed of $\hat{O}$.

  • $\begingroup$ I don't think we can assume that the operator is unchanged under time-reversal $\endgroup$
    – M. Zeng
    Commented Dec 4, 2014 at 6:55
  • $\begingroup$ @Timo: you may think or not, but this is the assumption in the text of the question, see the 3rd equation. $\endgroup$
    – Sofia
    Commented Dec 4, 2014 at 21:32
  • $\begingroup$ I'm sorry but your argument seems contradictory. We are supposed to find out how the operator transform under time-reversal, but you assumed that $O$ is unchanged under time-reversal. My assumption is that that the inner product is invariant instead of the operator. $\endgroup$
    – M. Zeng
    Commented Dec 5, 2014 at 2:19
  • $\begingroup$ @Timo: when reversing the time, we see that the particles reverse their movement. One starts from the final state, (we introduce the change ⟨ϕ|†), we apply on it Ô†, and as the exercise hints, the vector obtained should be now projected on the initial state (which therefore should appear as |ψ⟩†. (I am not sure which are the initial and final states, but if I trust the LHS of the 3rd equality, i.e. that he took the transposed of the product of components, it seems that |ψ⟩ is the initial state. But, let me return to you some later. $\endgroup$
    – Sofia
    Commented Dec 5, 2014 at 10:39
  • $\begingroup$ @Timo: please see why I thought that Ô is a symmetrical operator. Before the 3rd equality in the question, is written "if we require time-reversal symmetry, ..." $\endgroup$
    – Sofia
    Commented Dec 5, 2014 at 14:17

I am going to use the notation used by Mehta instead of using Dirac brackets because I think it makes understanding the equations easier.

\begin{align} (\Phi, A\Psi) =& (\Psi^{R}, A^{R}\Phi^{R})\\ =& (T\Psi, A^{R}T\Phi)\\ =& ((A^{R})^{\dagger}T\Psi, T\Phi)\\ =& (T^{\dagger}(A^{R})^{\dagger}T\Psi, \Phi)^{*}\\ =& (\Phi, T^{\dagger}(A^{R})^{\dagger}T\Psi) \end{align}

Third and fourth lines are considering that $A^{R}$ is not an antiunitary operator, so its adjoint is taken normally, while since $T$ is an antiunitary operator so it follows: $(x, Ty)=(T^{\dagger}x,y)^{*}$.

Now the last equation implies that: $ A=T^{\dagger}(A^{R})^{\dagger}T \Rightarrow A^{\dagger}=T^{\dagger}A^{R}T$ which implies: \begin{align} A^{R} =& (T^{\dagger})^{-1}A^{\dagger}T^{-1}\\ =& (T^{-1})^{\dagger}A^{\dagger}T^{-1}\\ =& (KU^{\dagger})^{\dagger}A^{\dagger}T^{-1}\\ =& UK^{\dagger}A^{\dagger}KU^{-1} \end{align}

where I have used the relations: $T=UK\Rightarrow$ \begin{align} T^{-1}=& K^{-1}U^{-1}\\ =& KU^{-1}\\ =& KU^{\dagger} \end{align}

Now I need to prove that $K^{\dagger}A^{\dagger}K=A^{T}$.

For complex conjugation operator $K$ we have the relation: $O^{*}=KOK^{-1}$ for any operator $O$, for example here, https://www.its.caltech.edu/~xcchen/img/Ph129b2020/lecture/lecture0312.pdf . This implies: \begin{align} O^{*}=& K^{-1}OK^{-1}\\ =& K^{-1}KK^{\dagger}OK^{-1}\\ =& K^{\dagger}OK^{-1}. \end{align}

Now if we replace $O=A^{*}\Rightarrow O^{*}=A$ which implies: \begin{align} &A=K^{\dagger}A^{*}K^{-1}\\ \Rightarrow &AK=K^{\dagger}A^{*}\\ \Rightarrow &(AK)^{\dagger}=(K^{\dagger}A^{*})^{\dagger}=(A^{*})^{\dagger}K=A^{T}K\\ \Rightarrow & K^{\dagger}A^{\dagger}=A^{T}K\\ \Rightarrow & A^{T}=K^{\dagger}A^{\dagger}K^{-1}\\ \Rightarrow & A^{T}=K^{\dagger}A^{\dagger}K. \end{align}

Edit: You can find a much better explanation in Appendix A of the book written by Kota https://books.google.co.in/books/about/Embedded_Random_Matrix_Ensembles_in_Quan.html?id=BLK5BQAAQBAJ&redir_esc=y.


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