# Bra-ket algebra

1. Sep 22, 2007

### indigojoker

I need to show that (XY)^(dagger)=(Y)^(dagger)(X)^(dagger) using bra-ket algebra
where X and Y are operators

say we started out with: if we dagger it (using *):

XY|a> = X(Y|a>)=(X(Y|a>))=((Y|a>)X*)=<a|Y*X*

we also know that XY|a>=<a|(XY)* by definition, so (XY)^(dagger)=(Y)^(dagger)(X)^(dagger)

My question is on how valid this statement is:
XY|a> = X(Y|a>)=(X(Y|a>))=((Y|a>)X*)=<a|Y*X*

2. Sep 22, 2007

### nrqed

I don't understand at all X(Y|a>)=(Y|a>)X*

3. Sep 22, 2007

### indigojoker

that's what i'm not sure about, actually, i get here and I dont know what to do:

XY|a> = X(Y|a>) = (X<a|Y*)

4. Sep 22, 2007

### antonantal

That's not correct. When you say X<a|Y* you are multiplying a matrix by a row vector, which is impossible.

Basically you have two properties of the adjoint which are useful in this problem:

|Xa> = X|a>

and

<Xa| = <a|X*

And also don't forget that the product of two operators can be treated as another operator.

EDIT: In fact only one property: <Xa| = <a|X*

Last edited: Sep 22, 2007
5. Sep 22, 2007

### indigojoker

i'm not sure what the difference between |Xa> and X|a> is

also, my books says you can go from X(Y|a>) to (<a|Y*)X*

i dont know how that is possible

6. Sep 22, 2007

### antonantal

If you translate |Xa> into words it is the column vector resulting from the multiplication of the matrix X by the a row vector having the components a0, a1, a2, ...
X|a> would translate as the result of multiplying the matrix X by the column vector |a>
Because |a> has the components a0, a1, a2, ... it results that |Xa> = X|a>

On the other hand <Xa| is the conjugate transpose of the column vector resulting from the multiplication of the matrix X by the a column vector having the components a0, a1, a2, ...
<a|X would translate as the result of multiplying the conjugate transpose of |a> by the matrix X. It is clear that they are not necessarily equal.

Try to use the property <Xa| = <a|X*.
Start from <XYa|. Considering that XY is also an operator, apply the property above. Then considering that Ya is a vector and X an operator, apply the property above.

Last edited: Sep 22, 2007
7. Sep 22, 2007

### indigojoker

so I'm trying to get from X(Y|a>) to (<a|Y*)X*

X(Y|a>)=X(<a|Y*)

the rules that has been listed does not say what to do in this situation: X(<a|Y*)

i could do X<Ya| but that still doesnt really help

8. Sep 22, 2007

### antonantal

I already told you that X(<a|Y*) can't be calculated. Use the hints that I gave you.

9. Sep 22, 2007

### antonantal

Look, I'll do the second part and you do the first part.
I start from <XYa| and considering that Ya is a vector and X an operator, I apply the property <Xa| = <a|X* and get:

<XYa| = <X(Ya)| = <Ya|X* then, I apply the property for <Ya| i.e. <Ya| = <a|Y* and get:
<XYa| = <X(Ya)| = <Ya|X* = <a|Y*X*

Now all you have to do is to start from <XYa| , consider that XY is an operator, and apply the same property <Xa| = <a|X*

You will get <XYa| = <(XY)a| = ...

10. Sep 22, 2007

### indigojoker

using your hint of <a|XY = <X*a|Y=<Y*X*a| and <a|(XY) = <(XY)*a| to prove it, but im still not sure how my book gets from X(Y|a>) to (<a|Y*)X*

11. Sep 22, 2007

### antonantal

That's also a solution.

12. Sep 22, 2007

### indigojoker

i know that is the solution, my question is trying to understand what my book did.

how it went from X(Y|a>) to (<a|Y*)X*

13. Sep 22, 2007

### antonantal

I don't think they are sayng that X(Y|a>) = (<a|Y*)X* but rather X(Y|a>) <-> (<a|Y*)X* which means that (<a|Y*)X* is the corresponding bra of the ket X(Y|a>) which is true.

14. Sep 22, 2007

### indigojoker

what does "the corresponding bra of the ket" mean?

15. Sep 22, 2007

### antonantal

<a| is the corresponding bra of the ket |a>

Last edited: Sep 22, 2007
16. Sep 22, 2007

### indigojoker

so, the transpose conjugate of X(Y|a>) is (<a|Y*)X* ?

17. Sep 22, 2007

### antonantal

Exactly.

18. Sep 22, 2007

### indigojoker

thanks!

also, i was wondering, is the kronecker delta the same thing as the dirac delta in the form delta(a-b)

that is, $$\delta _{a,b} = \delta(a-b)$$

I'm really wondering if a and b were orthonomal sets, then:
$$\langle a|b \rangle = \delta _{a,b} = \delta(a-b)$$

Last edited: Sep 22, 2007
19. Sep 22, 2007

### antonantal

They are not the same but the Dirac delta has a similar role as the Kronecker delta, when the dimension of the space is infinite.

20. Sep 22, 2007

### indigojoker

so if a and b were orthonormal sets, would:

$$\langle a|b \rangle = \delta(a-b)$$

or

$$\langle a|b \rangle = \delta _{a,b}$$