Quick Braket Questions: A|B> & <B|A

[Domnu]

Let A be an operator while |B> be a state vector. Are all of the following correct?

A|B&gt; = (&lt;B|A^\dagger)
&lt;B|A = A&lt;B|
&lt;B|A = &lt;A^\dagger B|

I'm a bit shaky on the second one... importantly, |B> doesn't have to be an eigenstate of A.

 

[Fredrik]

The first one is correct. The right-hand sides of the other two don't make sense.

 

[Domnu]

Could you correct the other two? Why doesn't the third one make sense?

 

[Domnu]

Particularly, when A is an operator while |B> is a state ket (which is not necessarily an eigenket of A), is the following true:

<B|A = A<B|

My hunch is that this is incorrect, because the left is an operator while the right is a bra...

 

[Fredrik]

Keep these things in mind:

* A bra takes kets to numbers.
* An operator takes kets to kets when it acts to the right.
* An operator takes bras to bras when it acts to the left.

A<B| is undefined, as it is an operator with a bra on its right. \langle A^\dagger B| is even stranger, since it has a symbol representing an operator inside the area (between the < and the |) where you're suppose to put a symbol representing the vector. (When you're using the bra-ket notation, the vector is written as |B>, its dual as <B|, and B means nothing by itself).

 

[Domnu]

Could you give an example of the second and third statements you made just to clarify? Particularly, for the second statement, are you saying that A |B> = |C> ? For the third statement, are you saying that <A|B = <C| ?

 

[Fredrik]

There is at least one more combination of bras and kets that can be defined, for example |u\rangle\langle v| is defined as the operator that satisfies

|u\rangle\langle v|(|\alpha\rangle)=|u\rangle\langle v|\alpha\rangle

(You can put a number like \langle v|\alpha\rangle on either side of the ket).

One more thing: The reason why expressions like your A<B| are undefined is that we want all "products" involving bras and kets to be associative. For example, we have (<B|A)|C>=<B|(A|C>), but we can't write (A<B|)|C>=A(<B|C>) because the right-hand side would be an operator acting on a number (undefined), or alternatively as a number multiplying an operator (which yields another operator). So allowing operators to be on the left of a bra would destroy associativity.



Also, don't forget the things I talked about in the other thread:

A ket is just a vector in a Hilbert space H. A bra is a member of the dual space H*, i.e. it's a linear function from H into the set of complex numbers. The bra \langle\beta| is the member of H* that's defined by

\langle\beta|(|\alpha\rangle)=(|\beta\rangle,|\alpha\rangle)

There's a theorem that guarantees that there exists exactly one such bra for each ket. The left hand side isn't usually written like that. The convention is to drop the parentheses and one of the |s, and write it as \langle\beta|\alpha\rangle. So the conventional way to write the probability mentioned above is |\langle n|\alpha\rangle|^2.

We can define the sum of two bras in an obvious way, and the product of a complex number and a bra in an equally obvious way. These definitions turn H* into a vector space.

Note that we don't really need the dual space in quantum mechanics. We're just using it because it simplifies the notation sometimes.

Here's an example of the same equation written without bras and then with bra-ket notation. It's not meant as an example of why the notation is useful. It's just an example of how it's used.

(|\alpha\rangle,A|\beta\rangle)=(A^\dagger|\alpha\rangle,\beta\rangle)

\langle\alpha|(A|\beta\rangle)=(\langle\alpha|A)|\beta\rangle

I think it's easier to use the (,) notation for the scalar product (as I did here) when you're trying to explain the bra-ket notation to yourself.

 

[Domnu]

Hmm... okay. If you wouldn't mind, could you prove the following identity using braket notation?

<B|adj(X)|A> = <A|X|B>*

where adj(X) denotes the adjoint of X and the * denotes the conjugate.

 

[Fredrik]

Sure:

\langle B|X^\dagger|A\rangle = \Big(\langle B|X^\dagger\Big)|A\rangle =\Big(\langle A|\Big(X|B\rangle\Big)\Big)^*=\langle A|X|B\rangle^*

What I used here is a) that "products" are associative (this is something that we can prove once and for all using the notation I'm using below), b) that <u|v>=<v|u>* and c) the definition of the adjoint operator.

Same thing, without bras and with the scalar product written as (,):

\Big(|B\rangle,X^\dagger|A\rangle\Big) =\Big(X^\dagger|A\rangle, |B\rangle\Big)^* =\Big(|A\rangle,X| B\rangle\Big)^*

 

[Domnu]

Ah, I understand now :) Thank you very much, Fredrik, for your quick and helpful response!

 

[Seckin Sefi]

Just to remind

For example, we have (<B|A)|C>=<B|(A|C>)

must be (<B|A)|C>=<B|(dag(A)|C>) as Fredrik did as

<br /> \Big(|B\rangle,X^\dagger|A\rangle\Big) =\Big(X^\dagger|A\rangle, |B\rangle\Big)^* =\Big(|A\rangle,X| B\rangle\Big)^*<br />

 

[Fredrik]

must be (<B|A)|C>=<B|(dag(A)|C>)

Nope. The left-hand side is by definition (of a bra) equal to the scalar product of the ket that corresponds to <B|A and the ket |C>. The ket that corresponds to <B|A is in your notation dag(A)|B>. So the right-hand side of your equation should be (dag(A)|B>,|C>). Now you can continue: = (|B>,A|C>) = [again using the definition of a bra] = <B|(A|C>).

This is how you prove the associativity of the "product", if you have previously proved that the ket the corresponds to \langle B|A is A^\dagger|B\rangle.

 

[Fredrik]

Wait, that last comment was actually a mistake by me. You don't prove this associativity. The fact that (<B|A)|C>=<B|(A|C>) is the definition of the action of an operator on a bra. You can use that (and the standard definition of the adjoint) to prove that the ket that corresponds to \langle B|A is A^\dagger|B\rangle.

 

[tiny-tim]

Particularly, when A is an operator while |B> is a state ket (which is not necessarily an eigenket of A), is the following true:

<B|A = A<B|

My hunch is that this is incorrect, because the left is an operator while the right is a bra...

Keep these things in mind:

* A bra takes kets to numbers.
* An operator takes kets to kets when it acts to the right.
* An operator takes bras to bras when it acts to the left.

Hi Domnu!

The way I remember it is to think of an operator A as a matrix, with || round it: |A|.

And you can only fit one | next to another!

Then it's obvious that you can write |A||B> and <B||A|, but not |A|<B| or |B>|A|.

 

[Fredrik]

That's actually a pretty good mnemonic. The only things you didn't mention is that you can also put > next to < (pointy ends towards each other) and that whenever there "should" be two | next to each other, you only write one of them.

And every one of these "products" is associative, but I believe I already mentioned that.

 

[reilly]

Let A be an operator while |B> be a state vector. Are all of the following correct?

A|B&gt; = (&lt;B|A^\dagger)
&lt;B|A = A&lt;B|
&lt;B|A = &lt;A^\dagger B|

I'm a bit shaky on the second one... importantly, |B> doesn't have to be an eigenstate of A.

For practical purposes, none of these equations are correct. For the first equation,
the right hand side should be ((&lt;B|A^\dagger)) ^\dagger) -- it takes a dagger to turn a <B| into |B>, and vice versa.

<B| is, for all practical purposes, a row vector; A is an operator. In QM,
operator times row vector is not defined -- however, one could express the product in terms of matrix and vector components, and thus give sense to a very inelegant construction.

The last one is simply not defined in normal QM -- although, again, components could be used -- as in, if D is a three dimensional rotation matrix, then something like

R|x> = |Dx>, could hold, where R and D refer to the same rotation. R acts on states, D acts on coordinates. Why pull the operator inside the state vector in the first place?

The safest way to proceed with Dirac notation is simply to think of operators as matrices, which multiply column vectors, and can be multiplied by row vectors. Then, just follow the rules for matrix notation.
Regards,
Reilly Atkinson

 

[Fredrik]

That's funny...I didn't even notice that I called the first one "correct" when it clearly isn't. A ket can't be equal to a bra. I should have said something like "the right-hand side is the bra that corresponds to the ket on the left-hand side".