Solving the Quantum State Problem: Expressing Psi with Orthogonal States

[Lindsayyyy]

Hi everyone

Homework Statement

I have a quantum state

[tex] \mid \Psi \rangle= a_1 \mid \Psi_1 \rangle + a_2 \mid \Psi_2 \rangle[/tex]

wheres as psi1 and psi are normalized orthognal states.

Not I want to express the psi with the following two states

[tex] \mid \Psi_3 \rangle = \frac {1}{\sqrt{2}} ( \mid \Psi_1 \rangle +\mid \Psi_2 \rangle)[/tex]

and

[tex] \mid \Psi_3 \rangle = \frac {1}{\sqrt{2}} ( \mid \Psi_1 \rangle -\mid \Psi_2 \rangle)[/tex]

Homework Equations

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The Attempt at a Solution

Well, I don't have much of an idea actually. I know how to calculate coefficients if I have different basis vectors, but that doesn't seem to help here. Can anyone give me a little hint on how to approach this?

Thanks for your help

 

[Fightfish]

The problem is essentially this:

You want to express [itex]\mid \Psi \rangle[/itex] in terms of the two orthogonal basis vectors [itex]\mid \Psi_{3} \rangle[/itex] and [itex]\mid \Psi_{4} \rangle[/itex].

This means that
[itex]\mid \Psi \rangle = c_{3}\mid \Psi_{3} \rangle + c_{4}\mid \Psi_{4} \rangle[/itex]

and your goal is to determine these two coefficients.

I'm not sure why your usual methods don't work here; they should: Clearly,
[itex]\langle \Psi_{3}\mid \Psi \rangle = c_{3}[/itex]
and so on in the usual fashion. Give it another shot.

 

[Lindsayyyy]

If I do this I get to the following expression

[tex] \langle \Psi_3 \mid \Psi \rangle = \langle \frac {1}{\sqrt 2} (\langle \Psi_1 \mid + \langle \Psi_2 \mid) \mid(a_1 \mid \Psi_1 \rangle +a_2 \mid \Psi_2 \rangle) \rangle [/tex]

But I don't know how to ease this expression up a bit.

 

[andrien]

hey,don't you think you should just solve it like an algebraic eqn.and get psi1 and psi2 in terms of psi3 and psi4(there is an error with third eqn) and just substitute back.

 

[Lindsayyyy]

I think I should do it the way Fightfish told me to. I tried it also your way before that was my idea aswell, but that didn't work out for me.

 

[andrien]

they are rather same.

 

[Fightfish]

If I do this I get to the following expression

[tex] \langle \Psi_3 \mid \Psi \rangle = \langle \frac {1}{\sqrt 2} (\langle \Psi_1 \mid + \langle \Psi_2 \mid) \mid(a_1 \mid \Psi_1 \rangle +a_2 \mid \Psi_2 \rangle) \rangle [/tex]

But I don't know how to ease this expression up a bit.

The inner product is associative. So,
[tex](\langle \Psi_1 \mid + \langle \Psi_2 \mid) (a_1 \mid \Psi_1 \rangle +a_2 \mid \Psi_2 \rangle) = a_1 \langle \Psi_1\mid \Psi_1\rangle + a_2\langle \Psi_1 \mid \Psi_2 \rangle + a_1\langle \Psi_2\mid\Psi_1\rangle + a_2 \langle \Psi_2\mid\Psi_2\rangle[/tex]

You could do it Andrien's way as well, but the method above is a general one that is easily applicable to most cases, especially when working with an infinite Hilbert space.

 

[Lindsayyyy]

Alright. Thank you very much. I understood it now. I was actually confused that the bra and kets are sums. I know how to deal with it when I have an operator as a sum but that was new to me.