Why this combination and not another

[M. next]

Why did the author of the book choose the combinations in (9.66) and (9.67) as follows? Why didn't he choose some other combination; say 1/sqrt(3) and the other 2/sqrt(3) instead of 1/sqrt(2) and 1/sqrt(2)?

Please see attachment. Note that this is problem on stark effect.

 

[Matterwave]

Different combinations lead to different vectors...the author found the "eigenvectors" corresponding to the specific eigenvalues, these eigenvectors for non-degenerate eigenvalues are unique up to normalization.

 

[ChrisVer]

I guess a physical explanation is that because they both build up $\psi$ state with equal contributions...

 

[M. next]

Then it is ok to choose whatever normalization I want? I can go for the 1/sqrt(3) and the other 2/sqrt(3) here?

 

[DrClaude]

Then it is ok to choose whatever normalization I want? I can go for the 1/sqrt(3) and the other 2/sqrt(3) here?

No, all eigenstates need to have the same nomalization, and in most cases you want the norm to be 1.

 

[Bill_K]

Then it is ok to choose whatever normalization I want? I can go for the 1/sqrt(3) and the other 2/sqrt(3) here?

You need to look back in your textbook or your notes and review what an eigenvector is.

 

[M. next]

I know what an eigenvector is and I know how to normalize it. Let's say we have |+> I know it is already normalized. When I have (1 2)^T I know that 1/sqrt*5 is the normalization factor. But how would I know how to normalize |2 0 0>? |2 1 0>?
It is probably the notation of the ket including n, l, m that I am not being capable to translate into something familiar like a column vector...

 

[DrClaude]

I know what an eigenvector is and I know how to normalize it. Let's say we have |+> I know it is already normalized. When I have (1 2)^T I know that 1/sqrt*5 is the normalization factor. But how would I know how to normalize |2 0 0>? |2 1 0>?
It is probably the notation of the ket including n, l, m that I am not being capable to translate into something familiar like a column vector...

For the kets $|a,b,c \rangle$ to form a basis, they must have the same norm. Therefore, to form a new state $\psi$ as $|a,b,c \rangle + |a',b',c' \rangle$ with the same norm requires
$$
\psi = \frac{1}{\sqrt{2}} \left( |a,b,c \rangle + |a',b',c' \rangle \right)
$$

 

[Fredrik]

I know what an eigenvector is and I know how to normalize it. Let's say we have |+> I know it is already normalized. When I have (1 2)^T I know that 1/sqrt*5 is the normalization factor. But how would I know how to normalize |2 0 0>? |2 1 0>?
It is probably the notation of the ket including n, l, m that I am not being capable to translate into something familiar like a column vector...

I haven't looked at this specific case, but it's conventional to use notations like |210> only for vectors with norm 1. So these guys are almost certainly already normalized. If you're given a linear combination like $|200\rangle +|210\rangle$, and you want to normalize it, you need to keep in mind what it means to normalize a vector. It means to find a vector in the same 1-dimensional subspace that has norm 1. The vector $a|200\rangle +b|210\rangle$ isn't in the same 1-dimensional subspace as $|200\rangle +|210\rangle$ unless a=b. So you need to ask yourself this: For what values of $a$ does $a\big(|200\rangle +|210\rangle\big)$ have norm 1? The first step is of course to calculate the norm of this vector. Do you know how to do that? (Assume that |200> and |210> are already normalized).

There are infinitely many complex numbers $a$ that get the job done, but only one of them has imaginary part zero and a positive real part. It's convenient to choose that one.

 

[M. next]

Oh great! Thank you for making this so clear Dr Claude!