
#1
May1112, 09:42 AM

P: 338

Hi, I am wondering what mistake i might have made in the following.
(v, Au) = [(v, Au)*]* = [(A†v, u)*]* = (u, A†v)* = ((A†)†u, v)* = (v, (A†)†u), where v and u are arbitrary vectors in a Hilbert space. That proves that A = (A†)†, doesn't it? My tutor says some of the steps are wrong. What might those be? 



#2
May1112, 12:44 PM

Mentor
P: 16,690

That looks correct. But I find it weird that you take the complex conjugate twice. Can't you just say
[tex]<v,Au>=<A^\dagger v,u>=<v,A^{\dagger\dagger} u>[/tex] Since this holds for all v, we conclude that [itex]Au=A^{\dagger\dagger}u[/itex] for all u. 



#3
May1112, 02:33 PM

P: 338

Wow! That is indeed a revelation! Thank you so much for accepting my reasoning.
Actually, the same question was in one of my problem sheets, and here was the solution: If we take a basis {e_{i}>}, then <e_{i}A†e_{j}> = <e_{j}Ae_{i}>*. Consider <e_{i}(B†)†e_{j}> = <e_{j}B†e_{i}>* = (<e_{i}Be_{j}>*)* = <e_{i}Be_{j}> for all i,j. Therefore, (B†)† = B. I am not sure why the guy has gone to all trouble of using the inconvenient Dirac notation and basis vectors. What are your thoughts? 



#4
May1112, 02:44 PM

Mentor
P: 16,690

adjoint of an operator
That's also correct. Perhaps that proof is easier for people used to braket notation, I don't consider it ver yeasy. It doesn't really matter really, there are many good proofs.




#5
May1212, 06:50 PM

P: 31

I've always wondered about this notation: <ejAei>.
I understand the pairing involved in <A ei,ej> but what does this <ejAei> mean? 



#7
May1212, 07:04 PM

P: 31

Thanks micromass. This notation is commonly found in books on quantum mechanics. So are physicists just complicating a notion that supposed to be simple?
I'm not attacking physicists but I feel that there must be more/other reasons why they use such notation... 



#8
May1212, 09:45 PM

Emeritus
Sci Advisor
PF Gold
P: 9,010

In braket notation, members of a Hilbert space ##\mathcal H## are written as ##\alpha\rangle,\ \beta\rangle##, or ##\psi\rangle,\ \phi\rangle##, instead of as x,y. And they're called "kets" instead of (or rather in addition to) "vectors". Members of the dual space ##\mathcal H^*## are called "bras". So far, it's just a weird notation and terminology. I will use the notation (x,y) for the inner product of x and y, and I will use the convention that the inner product is linear in the second variable. For each ##x\in\mathcal H##, I will denote the map ##y\mapsto (x,y)## from ##\mathcal H## into ##\mathbb C## by ##(x,\cdot)##. The Riesz representation theorem for Hilbert spaces says that ##x\mapsto (x,\cdot)## is an antilinear (=conjugate linear) bijection from ##\mathcal H## onto the dual space ##\mathcal H^*##. In braket notation, the member of ##\mathcal H^*## that corresponds to ##\alpha\rangle## via this antilinear bijection is denoted by ##\langle\alpha##. So ##\langle\alpha=\big(\alpha\rangle,\cdot\big)##, and the result of ##\langle\alpha## acting on ##\beta\rangle## is $$\langle\alpha\,\beta\rangle = \big(\alpha\rangle,\cdot\big)\beta\rangle =\big(\alpha\rangle,\beta\rangle\big).$$ The righthand side is the inner product of ##\alpha\rangle## and ##\beta\rangle##. It's conventional to simplify the notation for the thing on the lefthand side to ##\langle\alpha\beta\rangle##. This can be pretty confusing, since now it looks like the inner product of ##\alpha## and ##\beta##, but this would be nonsense, since ##\alpha## is undefined. The vector we've been talking about is denoted by ##\alpha\rangle##, not ##\alpha##. In order to make this notation easier to deal with, several other definitions are made. For example, the "product" of a bra and a linear operator is defined by $$\big(\langle\alphaA\big)\beta\rangle =\langle\alpha\big(A\beta\rangle\big).$$ This is why parentheses are unnecessary in the expression ##\langle\alphaA\beta\rangle##. What I just said should make it clear that the notation can be interpreted as two different things, both of which are equal to ##\big(\alpha\rangle,A\beta\rangle\big)##. It's either the bra ##\langle\alphaA## acting on the ket ##\beta\rangle## or the bra ##\langle\alpha## acting on the ket ##A\beta\rangle##. Another definition that's made out of convenience is to define the "product" of a ket and a bra by $$\alpha\rangle\langle\beta\, \gamma\rangle =\big(\langle\beta\gamma\rangle\big) \alpha\rangle.$$ We also define the product of a ket and a complex number (with the number on the right) by ##\alpha\rangle a =a\alpha\rangle##. This enables us to rewrite the previous equality as $$\big(\alpha\rangle\langle\beta\big) \gamma\rangle = \alpha\rangle\big(\langle\beta\, \gamma\rangle\big) = \alpha\rangle\big(\langle\beta \gamma\rangle\big).$$ This is why no parentheses are needed in the expression ##\alpha\rangle\langle\beta\gamma\rangle##. It can be interpreted as the operator ##\alpha\rangle\langle\beta## acting on the ket ##\gamma\rangle##, or as the product of the ket ##\alpha\rangle## and the number ##\langle\beta\gamma\rangle##. As you can see, the definitions are chosen so that all the "products" one might want to form are defined, and also associative. Now consider the equality that would be written as ##x=\sum_i (e_i,x)e_i## in the traditional (not braket) notation. In braket notation, we would write it as ##\alpha\rangle=\sum_ii\rangle\langle i\alpha\rangle##. This means that the identity operator can be written as ##1=\sum_ii\rangle\langle i##. This observation makes many problems significantly easier to solve. 



#9
May1312, 09:29 PM

P: 31

Thanks Fredrik! Your explanation clarifies a lot.




#10
May1312, 09:44 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

Fredrik explains that there is a very useful algebra involving scalars, vectors, covectors, and operators. The notation, I understand, was chosen specifically for the fact that the product of [itex]\langle x [/itex] and [itex] y \rangle[/itex] looks like [itex]\langle x, y \rangle[/itex]. Graphically suggestive notation is rather useful when what is suggested is actually true!
The algebra itself is rather useful in other contexts too, even if bras and kets aren't used. If the meaning of the notation is confusing, it may help to think in terms of ordinary matrix algebra  kets are column vectors, bras are row vectors, and operators are matrices. 


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