# Antiunitary time reversal operator

1. Feb 3, 2010

### maverick280857

Hi,

I'm stuck with a seemingly simple question:

Let T denote the antiunitary time evolution operator, and A, B be two time dependent operators. Let A' and B' denote their time reversed versions. That is,

$$TAT^{-1} = A'$$
$$TBT^{-1} = B'$$

Then show that

$$TABT^{-1} = B'A'$$

I know I can't insert $T^{-1}T$ between A and B here. So how does one prove this? Hints would be appreciated.

2. Feb 3, 2010

### DrDu

$$\langle \psi | A B | \phi \rangle -> \langle \psi | A B | \phi \rangle^*=\langle \phi|B' A'|\psi \rangle$$

3. Feb 3, 2010

### maverick280857

Isn't that just a complex conjugation? How do you get a time reversal inside? You should just have a Hermitian adjoint of A and B, with the orders reversed. $A^{\dagger}$ and $A'$ are different.

4. Feb 3, 2010

### xepma

Why can't you insert $T^{-1}T$ in between? $T^{-1}T$ is the identity right? I don't really see the problem...

5. Feb 3, 2010

### DrDu

Ok, I was sloppy here, I'll try again: Any anti-unitary transformation can be decomposed into complex conjugation and a unitary transformation, U. Hence

$$\langle \psi | A B | \phi \rangle -> \langle \psi | U^*UAU^* UB U^* U| \phi \rangle^*=\langle \phi'|B' A'|\psi' \rangle$$ with $$|\psi' \rangle=U|\psi\rangle$$, $$A'=U A^+ U^*$$.

6. Feb 3, 2010

### maverick280857

Its interesting that you ask...if you do that, you'll get A'B' instead of B'A', which is what one should get as the order of these operations has to be reversed. How do you explain that?

T is parametrized by a variable t....

7. Feb 3, 2010

### maverick280857

Ah nice...yes, that makes sense. Thank you!

8. Feb 3, 2010

### Fredrik

Staff Emeritus
What is "the antiunitary time evolution operator"? Do you mean that it's the time reversal operator? Apparently not, since you later said that it's parametrized by a variable t.

I'm thinking that too. If $T^{-1}$ exists at all, then $T^{-1}T$ must be the identity. What else could we mean by $T^{-1}$?

Does it? I don't see it. For starters, what does the arrow mean?

9. Feb 3, 2010

### DrDu

Well, now that you intervene, Fredrik, I have doubts, too. This whole antilinear operator stuff is extremely nasty. I understand it for some days after having read Messiah but then I usually forget some important details. I'll have a look this evening.

However in the meanwhile I found the following article:
which claims to have proven just the statement under question using the same nomenclature.

10. Feb 3, 2010

### SpectraCat

Since T is anti unitary, then it can be written as the product of a unitary operator U and complex conjugation operator K. So, then we have $$T = UK$$ and $$T^{-1} = K^{-1}U^{-1}$$. Expanding the LHS of your last equation, we have:

$$UKABK^{-1}U^{-1} = U(AB)^{\dagger}U^{-1} = UB^{\dagger}A^{\dagger}U^{-1} = UB^{\dagger}U^{-1}UA^{\dagger}U^{-1} = B^{\dagger}A^{\dagger}$$

So, that works as long as your A' and B' are the adjoints of A and B, which they should be if T is the time-reversal operator.

EDIT: as pointed out below, this treatment is incorrect, because $$TAT^{-1} \neq A^{\dagger}$$, as I was assuming. I have fixed it in a subsequent post (I think).

Last edited: Feb 3, 2010
11. Feb 3, 2010

### maverick280857

I called it antiunitary because it is not unitary. I made the running comment about it being parametrized by 't' because writing T or T^{-1} imho makes no sense unless one also mentions the parameter it acts on. All I know is that A and B are matrix representations of operators, T is the 'time reversal operator' which is 'antiunitary', and A', B' represent 'time reversed versions' of A and B respectively. I was (and probably still am) required to show that $TABT^{-1} = B'A'$. If you look at the right hand side, this makes sense, because you've flipped the order of A and B and also flipped the 'direction of time' (this is implicit in the notation, which is why I made that rather irresponsible comment about the time parameter).

I remember my professor mentioning that one can't introduce $T^{-1}T$ in there, and we were supposed to find out why, and solve this problem after that. I spent all of last week worrying about it, forgot about it on Sunday, and now here I am. My only argument is that if one does make this introduction then, one gets A'B' instead of B'A', which does not make sense.

Furthermore, doesn't writing $T^{-1}$ like that imply that one can go back in time and come back again, as and when one pleases? I would like to believe that the prescription for time reversal is to sandwich the operator between $T$ and $T^{-1}$ and not regard either T or its inverse in isolation as time reversal operators.

Yes, between your post and my acceptance of DrDu's response, I thought about this too. My first thought was: Decompose T as UK where U is a unitary operator and K is the complex conjugation operation. He is looking at a scalar -- inner product, and he operates T on the entire scalar, and then uses antilinearity of T to assert that there's a complex conjugation sitting outside, and the unitary part U acts inside in the manner that he has shown it.

(It is implicit here that $|\psi_{R}\rangle = T|\psi\rangle$ implies $\langle\psi_{R}| = \langle \psi|T^{\dagger}$)

DrDu, I came across the article you just linked to. I'll check it out again. To be precise, the statement under question should not require invoking a matrix element or an inner product. One should be able to show it using the matrix representation too, shouldn't one?

The original question is

(Edit: Just noticed SpectraCat's post.)

Last edited: Feb 3, 2010
12. Feb 3, 2010

### maverick280857

Ok, this is probably trivial but what do you mean by

$$KAK^{-1} = A^{\dagger}$$

How did this happen?

If K is the complex conjugation operation, then shouldn't it be equal to its own inverse? After all, $K^2 = 1$.

And continuing with your reasoning, one could still argue that

$$TABT^{-1} = TAT^{-1}TBT^{-1} = (UKAK^{-1}U^{-1})(UKBK^{-1}U^{-1}) = A^{\dagger}B^{\dagger}$$

What is wrong with that?

EDIT: K is not a unitary operation, but two successive operations by K should yield the same result. That prompts me to write $K^2 = 1$. Now for the parity operator (which is unitary), Perkins reasons that $P^2 = 1$ implies that P is a unitary operator. So if that is universally true, $K^2 = 1$ would imply that K is a unitary operator too. Is the fact that an operator whose square is an identity, a unitary operator a tautology?? Now that I think about it, I think this has to do with your writing $KAK^{-1}$. What does this mean? Also, what is $K^{-1}$? If one can write $K^{-1}K = 1$, then one should certainly be allowed to assert that $TT^{-1} = 1$!

What is seriously haywire with this convoluted reasoning? :-|

Last edited: Feb 3, 2010
13. Feb 3, 2010

### DrDu

Spectracat, I remember vaguely that in ordinary systems without spin, the time inversion operator is that of complex conjugation but not the one that would transform an operator into its hermitian adjoint. This can be seen most easily for p as $$T p T^{-1}=-p$$ but $$p^\dagger=p$$. In fact, in the standard treatment of T, $$TABT^{-1}$$ does not change the order of (the transformed) A and B.

14. Feb 3, 2010

### maverick280857

What is "the standard treatment of T"?

15. Feb 3, 2010

### DrDu

That in the book of Messiah or of Wigner himself.

16. Feb 3, 2010

### Physics Monkey

Isn't the original statement that maverick280857 set out to prove just not true? In equations: $$T A B T^{-1} = T A (T^{-1} T ) B T^{-1} = (T A T^{-1} ) ( T B T^{-1})$$ with no order change. In particular, one is free to insert $$1 = T^{-1} T$$ in the equation.

For example, consider a spin half system. What is $$T S_x S_z T^{-1}$$? According to what I said above, we should just get $$S_x S_z$$ since $$T S_i T^{-1} = - S_i$$. Note that this differs from the reversed form by a minus sign.
A way to convince yourself of the truth of this statement is to first write $$S_x S_z = -i S_y /2$$ and compute $$T (- i S_y ) T^{-1} = i T S_y T^{-1} = i (- S_y) = -i S_y$$. Thus we obtain exactly what we started with and no extra minus sign.

17. Feb 3, 2010

### SpectraCat

Yes, you are correct. It is not as simple as I made it out to be in my post. The complex conjugation operator K is different from taking the Hermetian adjoint. I will annotate my earlier post, and post the correction below.

18. Feb 3, 2010

### maverick280857

So the time reversal operator does nothing?

According to what I've been taught

(a) there are (I think) 3 different formulations of the Time Reversal operator, due to Wigner, Racah and someone else
(b) $$T\sigma_{xz}T^{-1} = \sigma_{xz}$$ and $$T\sigma_{y}T^{-1} = -\sigma_{y}$$

Part (b) follows from the following

(1) T = UK where U is unitary and K denotes complex conjugation.
(2) $\sigma_{x,z}$ are real, whereas $\sigma_{y}$ is imaginary.

19. Feb 3, 2010

### Physics Monkey

Haha, well it does something, it just doesn't reverse the order of operators.

If by $$\sigma_{xz}$$ you mean $$\sigma_x \sigma_z$$ then I agree with you here. My point is that in your first post you set out to prove that $$T \sigma_x \sigma_z T^{-1} = \sigma_z \sigma_x = - \sigma_x \sigma_z$$ i.e. that operators switch position under time reversal (in addition to whatever else time reversal may do to them). This isn't true, at least not in the usual conventions for time reversal.

Note added: I now suspect you may mean that you've been taught that $$T\sigma_x T^{-1} = \sigma_x$$ and similarly for the z component. I don't agree with this statement. This is the action of complex conjugation alone in this basis since $$\sigma_x$$ is real, but time reversal involves also the unitary transformation you mentioned. Time reversal, however you implement it, should flip all the components of the electron by rotational invariance. For example, you can represent $$T = \exp{(-i \pi S_y )} K$$ for a spin half system and obtain all the formulas I wrote. Sorry if this note is irrelevant!

Last edited: Feb 3, 2010
20. Feb 3, 2010

### maverick280857

Sorry, by $\sigma_{xz}$ I really meant $\sigma_{x}$ or $\sigma_{z}$.

21. Feb 3, 2010

### SpectraCat

Yeah, that was a mistake on my part. I think the following is the correct description:

$$KAK^{-1} = [A^{\dagger}]^{T}$$

IOW, the K operator acts as a transformation of the elements of the matrix A, converting them to their complex conjugates in place (i.e. without taking the transpose). I guess there is probably a concise notation for this, but I don't remember it, and I couldn't find it described online with a quick search.

As DrDu pointed out, the easiest way to see this is with a pure imaginary operator, like the momentum operator, where time-reversal produces the following:

$$T\hat{p}T^{-1} = -\hat{p}$$

So, assuming the above result is correct, using the identity $$(AB)^{T}=B^{T}A^{T}$$, we can modify my earlier derivation as follows:

$$TABT^{-1} = UKABK^{-1}U^{-1} = U[(AB)^{\dagger}]^{T}U^{-1} = U[B^{\dagger}]^{T}[A^{\dagger}]^{T}U^{-1}$$

which simplifies to the desired result, where your A' and B' matrices are now equivalent to the $$[A^{\dagger}]^{T}$$ and $$[B^{\dagger}]^{T}$$ above.

I think this is now consistent, and it is also different from what one would get by inserting $$T^{-1}T$$ in the initial expression.

EDIT: Nope ... turns out this is wrong too ... it ends up being identical to assuming that $$T^{-1}T$$ can be used as the identity operator, IOW, it would predict that the correct result on the RHS should be A'B', instead of B'A'.

Nothing. You are correct AFAICS, based on my initial (incorrect) analysis.

I don't know ... it does seem like it can't all be true, and still produce the result from your initial question. I dimly recall that there are several logical traps like this when dealing with anti-unitary operators, but I can't recall how they work precisely. I will have to look it up in Messiah.

Last edited: Feb 3, 2010
22. Feb 3, 2010

### DrDu

The operation $$(A^\dagger})^T$$ is usually abbreviated $$A^*$$. But note that both taking the hermitian adjoint (the "dagger") and taking the transpose each lead to a change of order of the operators on their own. So the combined operation of complex conjugation (*) does not change the order of the operators: $$(AB)^*=A^* B^*$$

23. Feb 3, 2010

### SpectraCat

Ok .. starting over ... I found the solution in Sakurai, eq. 4.4.36 on p. 273. Basically, the anti-unitarity of $$T$$ yields the following identity for the linear operator $$\hat{A}$$:

$$\left\langle\beta\right|\hat{A}\left|\alpha\right\rangle = \left\langle\tilde{\alpha}\right|T\hat{A}^{\dagger}T^{-1}\left|\tilde{\beta}\right\rangle$$

where $$\left|\tilde{\alpha}\right\rangle = T\left|\alpha\right\rangle ,\ \left|\tilde{\beta}\right\rangle = T\left|\beta\right\rangle$$.

So, the I think the issue here is that the operators A and A' in your initial example are really operators on different sets of kets. A operates on "normal" kets and A' operates on their time-reversed analogs. At least that is the only way I can make sense of your initial example.

Anyway, assuming that is the correct interpretation, the proof of your original expression is trivial with the identity above, since $$\left(AB\right)^{\dagger} = B^{\dagger}A^{\dagger}$$, again keeping in mind that the RHS and LHS describe operators on different sets of kets.

I think this is really correct now, and I'm sorry for any confusion my earlier incorrect responses may have caused.

Last edited: Feb 3, 2010
24. Feb 3, 2010

### DrDu

What does the ~ over alpha and beta stand for?

25. Feb 3, 2010

### SpectraCat

Sorry about that ... I accidentally clicked submit instead of preview (to check my TeX code), and it looks like you replied to my earlier incomplete version. Is it clear now?