1. Nov 24, 2015

Hi!!!

First of all I want apologize for my bad english!
Second, I'm doing a physical chemystry course about the main concepts of quantum mechanics !!

The Professor has given to me this definition of "the adjoint operator":

<φ|Aψ> = <Aφ|ψ>

My purpose is to verificate this equivalence so i gave some numeric values at <φ|, ψ> and at the matrix A (rappresentative of an operator).
Then i calculate the expression <φ|Aψ> multiplicating,at first ,the product |Aψ> = A|ψ> and then doing the scalar product <φ|Aψ>....The bra and ket are 1x3 and 3x1 matrix respectively ,while A is 3x3.

Now my problem is to calculate the "other" expression: <Aφ|ψ>

Because this expression says to calculate first <Aφ|.
BUT i dont' know how to calculate this because according to the linear algebra i can't do the product between A and <φ| (i.e. <Aφ|=A<φ|. In fact it would be a product between a 3x3 matrix and a 1x3 vector...I'm not able to do this but only the product 1x3 | 3X3 at most....

So how can i calculate the expression <Aφ|ψ> using the linear algebra?? I have to shift the matrix in order to do that product (where?)or what??

Thanks very much!! :)

2. Nov 24, 2015

blue_leaf77

A bra notation $\langle \ldots |$ is just the instruction to calculate the conjugate transpose of a vector it confines. To express something like $\langle A^\dagger \phi |$ in matrix notation, just first compute the matrix multiplication of the quantity inside the bra, and then take its conjugate transpose.

3. Nov 24, 2015

Hii!

I don't understand why i have to take its conjugate transpose. after having calculated Aφ|
To calculate <Aφ| ,assumed that i know the numerical values of the BRA <φ| and the matrix A what do i have to do?:

1) A<φ|

OR

2) <φ|A

Because i think that the relative order is important (the matric product usually is not commutative)...and the first expression according to linear algebra is not allowed ( matrix 3x3 * bra 1x3)

Thanks :)

4. Nov 24, 2015

Icaro Lorran

$\langle \psi | A$ means in traditional matrix notation $\psi ^\dagger A$. Similarly, if you try to put the A inside the bra like $\langle A^\dagger \psi |$, you'll have $\left({A^\dagger}\psi \right)^\dagger$, which is the same thing.

5. Nov 25, 2015

blue_leaf77

Lorran has explained the answer for me.

6. Nov 26, 2015

Thanks :)

The components of the vector ψ in the traditional matrix expression (Aψ) (corresponding to the bra <Aψ| ) are the same components of the ket vector |ψ> ??

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7. Nov 26, 2015

blue_leaf77

Yes, they are the same.

8. Nov 26, 2015