Antiunitary time reversal operator

[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.

Thanks in advance.

 

[DrDu]

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

 

[maverick280857]

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

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.

 

[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...

 

[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 $$|\psi' \rangle=U|\psi\rangle$$, $$A'=U A^+ U^*$$.

 

[maverick280857]

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

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...

 

[maverick280857]

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 $$|\psi' \rangle=U|\psi\rangle$$, $$A'=U A^+ U^*$$.

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

 

[Fredrik]

Let T denote the antiunitary time evolution operator,

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.

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

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 $T^{-1}$?

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 $$|\psi' \rangle=U|\psi\rangle$$, $$A'=U A^+ U^*$$.

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

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

 

[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:
http://www.google.de/url?sa=t&sourc...p5GvBg&usg=AFQjCNHNZOhNM4Yceala4o4U9EFvrkMlgw
which claims to have proven just the statement under question using the same nomenclature.

 

[SpectraCat]

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.

Thanks in advance.

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).

 

[maverick280857]

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

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.

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

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

Let A and B be two time dependent operators, with time reversed versions to be denoted by A' and B' respectively. If

$$TAT^{-1} = A'$$
and
[tex]TBT^{-1} = B'[/itex]

then show that

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

(Edit: Just noticed SpectraCat's post.)

 

[maverick280857]

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.

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? :-|

 

[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.

 

[maverick280857]

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.

What is "the standard treatment of T"?

 

[DrDu]

That in the book of Messiah or of Wigner himself.

 

[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.

 

[SpectraCat]

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.

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.

 

[maverick280857]

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.

So the time reversal operator does nothing?

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.

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.

 

[Physics Monkey]

So the time reversal operator does nothing?

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

According to what I've been taught

(b) $$T\sigma_{xz}T^{-1} = \sigma_{xz}$$ and $$T\sigma_{y}T^{-1} = -\sigma_{y}$$

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!

 

[maverick280857]

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 due to them). This isn't true, at least not in the usual conventions for time reversal.

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

Also, http://www.google.de/url?sa=t&sourc...p5GvBg&usg=AFQjCNHNZOhNM4Yceala4o4U9EFvrkMlgw

 

[SpectraCat]

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

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

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'.

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?

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

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? :-|

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.

 

[DrDu]

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.

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: $$(AB)^*=A^* B^*$$

 

[SpectraCat]

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.

Thanks in advance.

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.

 

[DrDu]

What does the ~ over alpha and beta stand for?

 

[SpectraCat]

What does the ~ over alpha and beta stand for?

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?

 

[Fredrik]

The operation $$(A^\dagger})^T$$ is usually abbreviated $$A^*$$.

I haven't seen that claim before. Are you sure you're not just confusing something with something else? These are my thoughts: Mathematicians use the notation $A^*$ for the operator that physicists write as $A^\dagger$. The definition is

$$\langle x,Ay\rangle=\langle A^* x,y\rangle$$ (mathematician's notation)

$$\langle x,Ay\rangle=\langle A^\dagger x,y\rangle$$ (physicist's notation)

$$|\psi\rangle=A|\phi\rangle \iff \langle\psi|=\langle\phi| A^\dagger$$ (bra-ket notation)

If the vector space is finite dimensional, the linear operator A corresponds to a matrix, let's call it A'. The matrix that $A^\dagger$ corresponds to in the same way is the complex conjugate of the transpose of A'. It would make sense to write that as $A'^*^T$, but linear algebra texts write it as $A'^*$.

 

[Fredrik]

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.

I think the key to understanding antilinear and antiunitary operators is this:

If $L$ is linear, the adjoint $L^\dagger$ is defined by

$$\langle x,Ly\rangle=\langle L^\dagger x,y\rangle$$

This definition wouldn't work for an antilinear operator, because suppose that $A$ is antilinear and that there exists an operator $A^\dagger$ such that

$$\langle x,Ay\rangle=\langle A^\dagger x,y\rangle$$

We already have a contradiction here, because the left-hand side is an antilinear function of y and the right-hand side is a linear function of y. (The physicist's convention is to have the inner product be antilinear in the first variable and linear in the second). So this $A^\dagger$ can't exist. Therefore, we define $A^\dagger$ by

$$\langle x,Ay\rangle=\langle A^\dagger x,y\rangle^*$$

A linear operator L is said to be unitary if

$$\langle Lx,Ly\rangle=\langle x,y\rangle$$

An antilinear operator A is said to be antiunitary if

$$\langle Ax,Ay\rangle=\langle x,y\rangle^*$$

However in the meanwhile I found the following article:
http://www.google.de/url?sa=t&sourc...p5GvBg&usg=AFQjCNHNZOhNM4Yceala4o4U9EFvrkMlgw
which claims to have proven just the statement under question using the same nomenclature.

Actually, it proves a different statement, because it defines the primed operators in a different way. (See (12)). This must be the correct definition. (Edit: I changed my mind about that. See my more recent posts). When you guys use that one instead of the one in #1, I think you'll find the proof quite easy, and also that it does involve inserting $T^{-1}T$ in one location.

Maverick, if you had posted a reference to where you found the claim you were trying to prove in #1, we could have figured this out a lot sooner. Let that be a lesson to you.

 

[Physics Monkey]

Hi maverick,

Just a friendly heads up, the linked paper which you are studying from appears a bit dubious in my opinion. I've not checked the author's argument carefully for internal consistency, but I think you should at least be aware that the author is explicitly going against the widely accepted formulation of time reversal (in my experience and in many textbooks). I would be curious to know if there is someone actively telling you something different?

PM

 

[turin]

You don't have to insert T^-1T; it is already there by definition on the RHS of the equation.

TABT^-1 = B'A' ≡ TBT^-1TAT^-1

Apparently, either A, B, and T have special properties such that TABT^-1=TBAT^-1, or else T and T^-1 violate the associative property. This bothered me for a bit, until I realized, "Of course! T and T^-1 do violate the associative property!" In fact, I don't think that T^-1T even makes sense as a stand-alone object. After all, T is not even a linear operator. That is, if x and y are vectors and a is a member of the complex field of the vector space to which x and y belong, then:

T(ax+y) ≠ aTx+Ty

I don't even think it is possible to decompose a general vector into eigenstates of T, is it? Anyway, try the inner product idea again, but this time expanding the vectors in terms of some arbitrary basis (with complex coefficients!).

 

[maverick280857]

Maverick, if you had posted a reference to where you found the claim you were trying to prove in #1, we could have figured this out a lot sooner. Let that be a lesson to you.

Hi maverick,

Just a friendly heads up, the linked paper which you are studying from appears a bit dubious in my opinion. I've not checked the author's argument carefully for internal consistency, but I think you should at least be aware that the author is explicitly going against the widely accepted formulation of time reversal (in my experience and in many textbooks). I would be curious to know if there is someone actively telling you something different?

PM

I would have posted a reference if I had read it from a reference . This problem was posed in the final particle physics class before my exam (which is scheduled for Friday). We were also told (as I pointed out in my last post) that there are 3 different formulations of time reversal -- due to Wigner, Racah and the third I don't remember now. I forgot to mention another thing my professor told us right before he asked us to figure it out: he said there are two choices we have for defining T, based on the requirement that $|\langle\psi|\varphi\rangle|$ be invariant. They are

Choice 1: Linear Unitary Operation

$$\langle T\psi|T\varphi\rangle = \langle \psi|\varphi\rangle$$

Choice 2: Antilinear Anti-Unitary Operation

$$\langle T\varphi|T\psi\rangle = \langle \psi|\varphi\rangle$$

We're using the second 'choice'. The invariance of [x, p] = i under this operation is not difficult to prove, since

[tex]T[x,p]T^{-1} = T(xp-px)T^{-1} = p'x' - x'p' = (-p)(x)-(x)(-p) = xp-px = [x, p][/itex]

This is as much as we were told. The paper I gave you folks a link to, was actually posted first by someone else on this thread.

However in the meanwhile I found the following article:
http://www.google.de/url?sa=t&source...4o4U9EFvrkMlgw
which claims to have proven just the statement under question using the same nomenclature.

I came across this paper while I was typing my question here, but I decided to post the question here first.

 

[maverick280857]

This bothered me for a bit, until I realized, "Of course! T and T^-1 do violate the associative property!" In fact, I don't think that T^-1T even makes sense as a stand-alone object. After all, T is not even a linear operator. That is, if x and y are vectors and a is a member of the complex field of the vector space to which x and y belong, then:

T(ax+y) ≠ aTx+Ty

The general "rule" for an anti-linear operator is

$$T(aX + bY) = a^{*}T(X) + b^{*}T(Y)$$

Yes, I tend to think that $T^{-1}T$ does not make sense as a stand-alone object either. In fact, in operator 'space', neither do $T$ or $T^{-1}$ by themselves right? One always finds them together, sandwiching an operator. In ket space however, $T|\psi\rangle$ does have a significance, but appparently this means

$$\langle\psi| = T|\psi\rangle$$

rather than generating another ket.

It seems this question has generated considerable interest and debate ..I just logged in this morning. Thanks for all the replies folks.

What is the final verdict?

 

[turin]

What is the final verdict?

I wonder if

$$\mathcal{O}\stackrel{T}{\rightarrow}T\mathcal{O}T^{-1}$$

is correct. Shouldn't it be

$$\mathcal{O}\stackrel{T}{\rightarrow}T\mathcal{O}T^\dagger$$

with $T^\dagger$ obviously NOT equal to $T^{-1}$ because

$$\langle{}a|a^{*}T^{\dagger}Tb|b\rangle{} =\langle{}Ta|ab^{*}|Tb\rangle{} =ab^{*}\langle{}Ta|Tb\rangle{} =ab^{*}\langle{}b|a\rangle{}$$

whereas

$$\langle{}a|a^{*}T^{-1}Tb|b\rangle{} =\langle{}a|a^{*}b|b\rangle{} =a^{*}b\langle{}a|b\rangle{} =\left(ab^{*}\langle{}b|a\rangle{}\right)^{*} \neq{}ab^{*}\langle{}b|a\rangle{}$$

?

Actually, I am now utterly confused by the meaning of such things as $T\mathcal{O}T^\dagger$.

 

[maverick280857]

$$\mathcal{O}\stackrel{T}{\rightarrow}T\mathcal{O}T^\dagger$$

Actually, I am now utterly confused by the meaning of such things as $T\mathcal{O}T^\dagger$.

What do you even mean here by $\langle{}a|a^{*}T^{\dagger}Tb|b\rangle{} =\langle{}Ta|ab^{*}|Tb\rangle{}$? What are 'a' and 'b'? Shouldn't you be looking at $\langle Ta|Tb\rangle$ instead?

 

[Fredrik]

Just a friendly heads up, the linked paper which you are studying from appears a bit dubious in my opinion. I've not checked the author's argument carefully for internal consistency, but I think you should at least be aware that the author is explicitly going against the widely accepted formulation of time reversal (in my experience and in many textbooks).

I admit that I didn't look very closely at that paper. I just saw that it defines the primed operators by $X'=T^{-1}X^\dagger T$, which implies

$$(AB)'=T^{-1}(AB)^\dagger T=T^{-1}B^\dagger A^\dagger T=T^{-1}B^\dagger T^{-1}T A^\dagger T=B'A'$$

so I just assumed that the extra dagger is supposed to be there. I clearly didn't think that through. My calculation here is definitely 100% correct, but the time-reversed X is supposed to be $TXT^{-1}$, not $T^{-1}X^\dagger T$.

You don't have to insert T^-1T; it is already there by definition on the RHS of the equation.

TABT^-1 = B'A' ≡ TBT^-1TAT^-1

I don't know what you mean by this equation. The second equality is just the definition of the primed operators, and the first is what Maverick says he's trying to prove.

Apparently, either A, B, and T have special properties such that TABT^-1=TBAT^-1, or else T and T^-1 violate the associative property. This bothered me for a bit, until I realized, "Of course! T and T^-1 do violate the associative property!" In fact, I don't think that T^-1T even makes sense as a stand-alone object.

This is all wrong. All of these symbols represent functions (some of them linear, and some of them antilinear), and the "multiplication" is just the composition operation, $XY=X\circ Y$, which is definitely associative. And we definitely have $T^{-1}T=I$, where I is the identity map of the Hilbert space on which T is defined. This holds for all types of (invertible) functions, not just for the very special cases of linear and antilinear functions.

We're using the second 'choice'. The invariance of [x, p] = i under this operation is not difficult to prove, since

[tex]T[x,p]T^{-1} = T(xp-px)T^{-1} = p'x' - x'p' = (-p)(x)-(x)(-p) = xp-px = [x, p][/itex]

Here you're using the result mentioned in post #1 in the second step, but you still haven't proved that (and you won't, because it's wrong, unless you define the primed operators with an extra dagger, like in the article that was mentioned).

It doesn't make sense to say that [x,p]=i is "invariant", since i isn't: $TiT^{-1}=-i$. If you're allowed to use the transformation properties of x and p, $TxT^{-1}=x$ and $TpT^{-1}=-p$, then you can immediately see that [x',p']=-i. In fact, all the following steps are valid:

$$-i=TiT^{-1}=T[x,p]T^{-1}=TxpT^{-1}-TpxT^{-1}=TxTT^{-1}pT^{-1}-TpTT^{-1}xT^{-1}=x'p'-p'x'=[x',p']=[x,-p]=-[x,p]$$

Yes, I tend to think that $T^{-1}T$ does not make sense as a stand-alone object either. In fact, in operator 'space', neither do $T$ or $T^{-1}$ by themselves right?

They do. They are functions from a Hilbert space $\mathcal H$ into the the same Hilbert space.

One always finds them together, sandwiching an operator. In ket space however, $T|\psi\rangle$ does have a significance, but appparently this means

$$\langle\psi| = T|\psi\rangle$$

rather than generating another ket.

This is wrong. The bras are members of $\mathcal H^*$ and the range of T is some subspace of $\mathcal H$.

It seems this question has generated considerable [strike]interest and debate[/strike] confusion
...
What is the final verdict?

That something is wrong in the question you asked in #1, and you need to find out what that is. Are the primed operators supposed to be defined with an extra dagger or something? Is the equation supposed to say (AB)'=A'B'? Or is your professor just wrong about the properties of these operators?

$$\langle{}a|a^{*}T^{\dagger}Tb|b\rangle{} =\langle{}Ta|ab^{*}|Tb\rangle{} =ab^{*}\langle{}Ta|Tb\rangle{} =ab^{*}\langle{}b|a\rangle{}$$

whereas

$$\langle{}a|a^{*}T^{-1}Tb|b\rangle{} =\langle{}a|a^{*}b|b\rangle{} =a^{*}b\langle{}a|b\rangle{} =\left(ab^{*}\langle{}b|a\rangle{}\right)^{*} \neq{}ab^{*}\langle{}b|a\rangle{}$$

See #27 for the correct definition of the adjoint of an antilinear operator (and the reasons for it). Since I absolutely hate the notation that mixes bra-ket notation with simply writing the inner product of x and y as $\langle x|y\rangle$, I'm going to have to reply in a different notation. You use the same symbol for an eigenvalue and an eigenvector corresponding to that eigenvalue. That would be confusing if we stick to the same notation throughout the calculation, so I'll write the eigenvectors with an arrow on top.

$$\langle\vec a, a^*T^\dagger Tb\vec b\rangle=\langle\vec a, T^\dagger ab^* T\vec b\rangle=\langle T\vec a,ab^*T\vec b\rangle^*=\big(ab^*\langle T\vec a,T\vec b\rangle\big)^*=a^*b\langle T\vec a,T\vec b\rangle^*$$

$$=a^*b\langle\vec a,\vec b\rangle=\langle\vec a,a^*b\vec b\rangle=\langle\vec a,a^*T^{-1}Tb\vec b\rangle$$

 

[maverick280857]

It doesn't make sense to say that [x,p]=i is "invariant", since i isn't: $TiT^{-1}=-i$. If you're allowed to use the transformation properties of x and p, $TxT^{-1}=x$ and $TpT^{-1}=-p$, then you can immediately see that [x',p']=-i. In fact, all the following steps are valid:

Why is $TiT^{-1} = -i$?

This is wrong. The bras are members of $\mathcal H^*$ and the range of T is some subspace of $\mathcal H$.

Yeah, I just wrote that out of the hat. I couldn't see why a bra could become a ket, although that is what seems to be happening in the inner product transformation I wrote in my very last post.

That something is wrong in the question you asked in #1, and you need to find out what that is. Are the primed operators supposed to be defined with an extra dagger or something? Is the equation supposed to say (AB)'=A'B'? Or is your professor just wrong about the properties of these operators?

What are the 3 different formulations of time reversal? I think one is by Schwinger, the other by Racah, and the other by Wigner? Anyhow, the question isn't wrong. I think we need to find an authoritative source now.

Also, I don't understand why the primes should be equal to the dagger or "defined with an extra dagger or something". The primes represent the time reversed versions of the operator, not the transpose or the Hermitian adjoint or the conjugate...unless you're telling me otherwise?