Determining orthonormal states to a non-zero inner product

[electrogeek]

Hi everyone,

I was attempting the following past paper question below:

I have found a value for the coefficient c and I think I have calculated the inner product of <x|x>. I've attached my workings below. But I'm not sure what to do next to answer the last part of the question which asks about finding an orthonormal state to |b>. Am I right to say that <a|a> and <b|b> are equal to one or is my <x|x> calculation wrong?

 

[BvU]

Is it really necessary to make this assumption ? You already made use of it in the first part, when you said $C\langle b|b\rangle = C$ !

What if $|a\rangle$ and $|b\rangle$ are not normalized ?

 

[PeroK]

What if $|a\rangle$ and $|b\rangle$ are not normalized ?

One might argue that if they are quantum states, then that implies they are normalized.

 

[electrogeek]

I've had a think about it and would I be right in saying that the state |x> / <x|x> would be orthonormal to |b>?

 

[PeroK]

I've had a think and would I be right in saying that the state |x>/<x|x> would be orthonormal to |b>?

If $|x \rangle$ is orthogonal to $|b \rangle$, then the normalized state $\frac{|x \rangle}{\langle x|x \rangle^{1/2}}$ must be orthogonal to $|b \rangle$.

Your term "orthonormal to $|b \rangle$" has no meaning.

There is, however, one exceptional case that you should have picked up on. What happens if $1 - S^2 = 0$?

 

[electrogeek]

Would |x> / sqrt(<x|x>) both normalised and orthogonal to |b> then instead?

 

[PeroK]

Would |x> / sqrt(<x|x>) both normalised and orthogonal to |b> then instead?

Yes, you're right it should be the square root there. I'll change it.

PS it's a strange question since: a) everything works with $S = 0$; b) it works perfectly well with $S, c$ as complex numbers; c) It doesn't work with $|S| = 1$, which is the case where $|a \rangle$ and $|b \rangle$ are the same state (up to a phase factor).

It shouldn't have been "non-zero $S$", it should have been $|S| \ne 1$ in the question.

 

[electrogeek]

Yes, you're right it should be the square root there. I'll change it.

PS it's a strange question since: a) everything works with $S = 0$; b) it works perfectly well with $S, c$ as complex numbers; c) It doesn't work with $|S| = 1$, which is the case where $|a \rangle$ and $|b \rangle$ are the same state (up to a phase factor).

It shouldn't have been "non-zero $S$", it should have been $|S| \ne 1$.

Ah brilliant! Thanks for the help. :)