Why does XY|a> = <a|Y*X* in bra-ket notation?

[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*

 

[nrqed]

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*

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

 

[indigojoker]

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

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

 

[antonantal]

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

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

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*

 

[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 don't know how that is possible

 

[antonantal]

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

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.

 

[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 doesn't really help

 

[antonantal]

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 doesn't really help

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

 

[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| = ...

 

[indigojoker]

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

 

[antonantal]

using your hint of <a|XY = <X*a|Y=<Y*X*a| and <a|(XY) = <(XY)*a| to prove it

That's also a solution.

 

[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*

 

[antonantal]

im still not sure how my book gets from X(Y|a>) to (<a|Y*)X*

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.

 

[indigojoker]

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

 

[antonantal]

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

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

Check the wikipedia page on bra-ket notation.

 

[indigojoker]

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

 

[antonantal]

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

Exactly.

 

[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:
$$\langle a|b \rangle = \delta _{a,b} = \delta(a-b)$$

 

[antonantal]

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:
$$\langle a|b \rangle = \delta _{a,b} = \delta(a-b)$$

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.

 

[indigojoker]

so if a and b were orthonormal sets, would:

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

or

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

 

[antonantal]

<a| and <b| are vectors. If they are orthogonal then <a|b> = 0, and indeed, in the general case $$<a|b> = \delta(a-b)$$ but only if <a| and <b| are of infinite dimension. If they are of finite dimension then $$<a|b> = \delta_{a,b}$$

 

[Gokul43201]

The original question can be answered trivially if you recall that the Hermitian adjoint of an operator is defined using the mapping onto a dual space of the ket space.

$A ^{\dagger}$ is defined as: $$A|a \rangle \leftrightarrow \langle a|A^{\dagger}$$

So, $$X (Y |a \rangle ) \leftrightarrow ( \langle a | Y ^{\dagger} ) X ^{\dagger}$$

 

[indigojoker]

why isn't it $$X (Y |a \rangle ) \leftrightarrow X ^{\dagger} ( \langle a | Y ^{\dagger} )$$

 

[dextercioby]

Why would it be ? Y|a> is a ket, call it |b>. The dual of |b> is <b| which is <a|Y^{\dagger}. Now <b| stays to the left of X^{\dagger} by the duality correspondence.