Find an orthogonal quantum state: introduction to dirac notation.

[knowlewj01]

Homework Statement

Suppose we have a spin 1/2 Particle in a prepared state:

\left|\Psi\right\rangle = \alpha \left|\uparrow\right\rangle + \beta\left|\downarrow\right\rangle

where

\left|\uparrow\right\rangle \left|\downarrow\right\rangle

are orthonormal staes representing spin up and spin down respectively.

also: \left|\alpha\right|^2 + \left|\beta\right|^2 = 1

\alpha & \beta are complex numbers.

find a state which is orthogonal to \left|\Psi\right\rangle

Homework Equations

The Attempt at a Solution

I went about this first by saying that the inner product of two states which are orthogonal is 0, so propose that:

\left\langle\Phi\right|\left|\Psi\right\rangle = 0

where
\left|\Phi\right\rangle = \gamma\left|\uparrow\right\rangle + \delta\left|\downarrow\right\rangle

where \gamma & \delta are complex numbers:

\therefore

\left\langle\Phi\right|\left|\Psi\right\rangle = \alpha \gamma^* + \beta \delta^* = 0

Not sure where to go from here, i must be missing something. anyone know what it is?

 

[fzero]

Note that if |\Phi\rangle is orthogonal to |\Psi\rangle, then so is c|\Phi\rangle for any complex number c. You can use this to scale away one of the coefficients in your expression for |\Phi\rangle and then solve for the other.

 

[knowlewj01]

Not sure I'm following. I could just say that:

\alpha \gamma^* = -\beta \delta^*

then the following relationship would satisfy.

\gamma^* = \beta , \delta^* = -\alpha

so:
\left|\Phi\right\rangle = \beta\left|\uparrow\right\rangle - \alpha\left|\downarrow\right\rangle

 

[grey_earl]

Of course, that's one particular solution. fzero only said that for any complex c
|\Theta\rangle = c |\Phi\rangle also is orthogonal to |\Psi\rangle, which you could have used in your solution. But it isn't necessary.

 

[fzero]

Not sure I'm following. I could just say that:

\alpha \gamma^* = -\beta \delta^*

then the following relationship would satisfy.

\gamma^* = \beta , \delta^* = -\alpha

so:
\left|\Phi\right\rangle = \beta\left|\uparrow\right\rangle - \alpha\left|\downarrow\right\rangle

This should be

\left|\Phi\right\rangle = \beta^*\left|\uparrow\right\rangle - \alpha^*\left|\downarrow\right\rangle .

Note that

\left|\Phi\right\rangle = \beta^*\left( \left|\uparrow\right\rangle - \frac{\alpha^*}{\beta^*} \left|\downarrow\right\rangle \right),

so any multiple of

\left|\uparrow\right\rangle - \frac{\alpha^*}{\beta^*} \left|\downarrow\right\rangle

is orthogonal to |\Psi\rangle.

I just thought you were having algebra trouble, so starting with a state with \gamma=1 would make things easier. However, the fact that all vectors in the Hilbert space that differ only by a rescaling correspond to the same physical state is a fundamental concept.

The extra benefit of the state that you actually found though is that it's already normalized.