Antiunitary time reversal operator

[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

<br /> \langle{}a|a^{*}T^{\dagger}Tb|b\rangle{}<br /> =\langle{}Ta|ab^{*}|Tb\rangle{}<br /> =ab^{*}\langle{}Ta|Tb\rangle{}<br /> =ab^{*}\langle{}b|a\rangle{}<br />

whereas

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

?

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{}<br /> =\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&#039;=T^{-1}X^\dagger T, which implies

(AB)&#039;=T^{-1}(AB)^\dagger T=T^{-1}B^\dagger A^\dagger T=T^{-1}B^\dagger T^{-1}T A^\dagger T=B&#039;A&#039;

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

T[x,p]T^{-1} = T(xp-px)T^{-1} = p&#039;x&#039; - x&#039;p&#039; = (-p)(x)-(x)(-p) = xp-px = [x, p][/itex]<br />

<br /> Here you&#039;re using the result mentioned in post #1 in the second step, but you still haven&#039;t proved that (and you won&#039;t, because it&#039;s wrong, unless you define the primed operators with an extra dagger, like in the article that was mentioned).<br /> <br /> It doesn&#039;t make sense to say that [x,p]=i is &quot;invariant&quot;, since <i>i</i> isn&#039;t: TiT^{-1}=-i. If you&#039;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&#039;,p&#039;]=-i. In fact, all the following steps are valid:<br /> <br /> -i=TiT^{-1}=T[x,p]T^{-1}=TxpT^{-1}-TpxT^{-1}=TxTT^{-1}pT^{-1}-TpTT^{-1}xT^{-1}=x&amp;#039;p&amp;#039;-p&amp;#039;x&amp;#039;=[x&amp;#039;,p&amp;#039;]=[x,-p]=-[x,p]<br /> <br /> <blockquote data-attributes="" data-quote="maverick280857" data-source="post: 2563097" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> maverick280857 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Yes, I tend to think that T^{-1}T does not make sense as a stand-alone object either. In fact, in operator &#039;space&#039;, neither do T or T^{-1} by themselves right? </div> </div> </blockquote>They do. They are functions from a Hilbert space \mathcal H into the the same Hilbert space.<br /> <br /> <blockquote data-attributes="" data-quote="maverick280857" data-source="post: 2563097" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> maverick280857 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> One always finds them together, sandwiching an operator. In ket space however, T|\psi\rangle does have a significance, but appparently this means<br /> <br /> \langle\psi| = T|\psi\rangle<br /> <br /> rather than generating another ket. </div> </div> </blockquote>This is wrong. The bras are members of \mathcal H^* and the range of T is some subspace of \mathcal H.<br /> <br /> <blockquote data-attributes="" data-quote="maverick280857" data-source="post: 2563097" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> maverick280857 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> It seems this question has generated considerable [strike]interest and debate[/strike] confusion[/color] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":-p" title="Stick Out Tongue :-p" data-smilie="7"data-shortname=":-p" /><br /> ...<br /> What is the final verdict? </div> </div> </blockquote>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)&#039;=A&#039;B&#039;? Or is your professor just wrong about the properties of these operators?<br /> <br /> <blockquote data-attributes="" data-quote="turin" data-source="post: 2563129" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> turin said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> &lt;br /&gt; \langle{}a|a^{*}T^{\dagger}Tb|b\rangle{}&lt;br /&gt; =\langle{}Ta|ab^{*}|Tb\rangle{}&lt;br /&gt; =ab^{*}\langle{}Ta|Tb\rangle{}&lt;br /&gt; =ab^{*}\langle{}b|a\rangle{}&lt;br /&gt;<br /> <br /> whereas<br /> <br /> &lt;br /&gt; \langle{}a|a^{*}T^{-1}Tb|b\rangle{}&lt;br /&gt; =\langle{}a|a^{*}b|b\rangle{}&lt;br /&gt; =a^{*}b\langle{}a|b\rangle{}&lt;br /&gt; =\left(ab^{*}\langle{}b|a\rangle{}\right)^{*}&lt;br /&gt; \neq{}ab^{*}\langle{}b|a\rangle{}&lt;br /&gt; </div> </div> </blockquote>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&#039;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&#039;ll write the eigenvectors with an arrow on top.<br /> <br /> \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^*<br /> <br /> =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?

 

[Fredrik]

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

Because T is antilinear. T\big(i|\psi\rangle\big)=i^*T|\psi\rangle=-iT|\psi\rangle

What are the 3 different formulations of time reversal?

I don't know.

Anyhow, the question isn't wrong.

It is.

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

I agree. They should be defined the way you defined them. But if we define them that way, we get (AB)'=A'B' from the absolutely trivial fact that T^{-1}T is the identity.

 

[DrDu]

Ok, seems I missed a lot of funny stuff while sleeping. I had a look at Wigner's book on symmetry, where he defines time reversal. To add some confusion on notation, Wigner uses the prime (') to denote transposition. Wigner doesn't use Diracs notation, which is a good way to remain sane.
I found the formula Spectracat was giving in #23 also in the book of Schwabl, "advanced quantum mechanics", and I think it states what I tried to write down in #5.
I think all conclusions by Fredrik in #34 are correct and can be derived from that formula. Hence the original statement by marverick is wrong and I also suppose that some daggers are missing somewhere.
As to any alternative definitions of the time reversal operator, I only know about the Racah definition which is t-> -t, but no complex conjugation. While eventually possible in relativistic QM, it can be excluded in non-relativistic QM as it would lead to the conclusion that Hamiltonians have to be unbound from below. Anyhow, being a unitary operation, it cannot bring about a change of order of the transformed operators.

 

[SpectraCat]

Ok, seems I missed a lot of funny stuff while sleeping. I had a look at Wigner's book on symmetry, where he defines time reversal. To add some confusion on notation, Wigner uses the prime (') to denote transposition. Wigner doesn't use Diracs notation, which is a good way to remain sane.
I found the formula Spectracat was giving in #23 also in the book of Schwabl, "advanced quantum mechanics", and I think it states what I tried to write down in #5.
I think all conclusions by Fredrik in #34 are correct and can be derived from that formula. Hence the original statement by maverick is wrong and I also suppose that some daggers are missing somewhere.
As to any alternative definitions of the time reversal operator, I only know about the Racah definition which is t-> -t, but no complex conjugation. While eventually possible in relativistic QM, it can be excluded in non-relativistic QM as it would lead to the conclusion that Hamiltonians have to be unbound from below. Anyhow, being a unitary operation, it cannot bring about a change of order of the transformed operators.

I agree with this, and with Fredrik's post #34. There is something missing/wrong in the definition given in the OP, which is highlighted by the "derivation of invariance of [x,p]" in post #30 by maverick. IOW, the (AB)'=B'A' leads to the *wrong* conclusion in that case, as already pointed out by maverick, since if T is anti-unitary, then TiT^{-1}=-i must be true by definition.

I was thinking about this in bed last night, and I came to the conclusion that the relation in #1 must be wrong as given, because T[x,p]T^{-1}=[x,-p]=-[x,p] is the way things must work, both in the intuitive physical picture of what time-reversal must mean, as well as from the mathematical treatment using T properly as an anti-unitary operator.

Also, while the equation from Sakurai I gave in post #23 is correct and leads to the correct conclusion with regard to the time-reversal of [x,p], the subsequent interpretation I wrote looks nonsensical to me now. I should have realized then that it pointed to an error in post #1 ... I was taking on faith that it must be correct .

 

[turin]

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.

I think that y'all were talking about inserting a resolution of the identity in there on the very LHS, and then identifying the result in terms of the definition:

B&#039;A&#039;\doteq{}TABT^{-1}=TATT^{-1}BT^{-1}\equiv{}A&#039;B&#039;

whereas I was talking about "going backwards":

TABT^{-1}\doteq{}B&#039;A&#039;\equiv{}TBT^{-1}TAT^{-1}=TBAT^{-1}

So, y'all were getting the contradiction of B&#039;A&#039;=A&#039;B&#039; whereas I get the contradiction of TABT^{-1}=TBAT^{-1}, without having to "insert" anything "extra". I now realize that both ways of reaching this same basic contradiction rely on exactly the same two assumptions (namely associativity and definition of time reversed operator) just in reverse order, so I don't think my approach adds anything useful to the conversation.

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.

Thank you for affirming this. I don't understand this. I will have to think about and study the composition of functions. (I know what the composition of functions is, I think, but my mind is stuck on the notion that associativity refers to matrices, i.e. only applies to linear functions.)

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.

OK, yes, that makes sense.

... something is wrong in the question you asked in #1, and you need to find out what that is.

Crap! I committed one of the fundamental sins in education, and one that I admonish other students about: I just took for granted that the excercise was meaningful. It confused me when maverick said that we were not allowed to insert TT^{-1}, and this recalled those exciting and uncomfortable moments during my education when I was confronted with deep, underlying assumptions that don't always apply.

See #27 for the correct definition of the adjoint of an antilinear operator (and the reasons for it).

I do not understand the reason for it. How do you know that \langle{}A^{\dagger}x,y\rangle is linear (as opposed to antilinear) in y? What is wrong with:

\langle{}A^{\dagger}x,cy_{1}+y_{2}\rangle{}=c^{*}\langle{}A^{\dagger}x,y_{1}\rangle{}+\langle{}A^{\dagger}x,y_{2}\rangle{}

?

You use the same symbol for an eigenvalue and an eigenvector corresponding to that eigenvalue. That would be confusing ...

You are correct; it is confusing. I have already confused maverick by it. Actually, I didn't even mean eigenvector and eigenvalue, just some vector with some complex coefficient. That was really sloppy notation on my part.

\ldots{}\langle\vec a, T^\dagger ab^* T\vec b\rangle=\langle T\vec a,ab^*T\vec b\rangle^*\ldots

Yeah, I missed that conjugation. But I still don't understand why it must be defined this way. This is one of the reasons that I can't grasp the associativity of these antilinear operators.

Anyway, I just got distracted by the adjoint, and now I'm flip-flopping. I think that it should be \mathcal{O}\stackrel{T}{\rightarrow}T\mathcal{O}T^{-1} as originally described.

 

[Fredrik]

I will have to think about and study the composition of functions. (I know what the composition of functions is, I think, but my mind is stuck on the notion that associativity refers to matrices, i.e. only applies to linear functions.)

f\circ g is defined by f\circ g(x)=f(g(x)) for all x in the domain of g such that g(x) is in the domain of f. You know that part. Linear and antilinear operators are both functions, and a "product" XY is just a composition written without the \circ symbol. An easy excercise that you really should do is to use the definition to prove that (f\circ g)\circ h=f\circ(g\circ h).

I do not understand the reason for it. How do you know that \langle{}A^{\dagger}x,y\rangle is linear (as opposed to antilinear) in y?

This is part of the definition of a Hilbert space. The inner product is required to be linear in the second variable and antilinear in the first. (This is the physicist's convention. Mathematicians define inner products on complex vector spaces to be linear in the first variable and antilinear in the second).

 

[turin]

... for all x in the domain of g such that g(x) is in the domain of f.

This provision makes the associativity trivial, right? So, I'm considering x as some (any) vector (say |ψ>) in the Hilbert space, f as some observable (either A or B), and g as the inverse of the time reversal operator (T^-1). Is it always true that T^-1|ψ> is in the Hilbert space (acted on by A and B)? I suppose you are going to tell me that it is? Can you help me to prove this?