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.