Question about operator and eigenvectors

[einai]

Hi, I encountered the following HW problem which really confuses me. Could anyone please explain it to me? Thank you so much!

The result of applying a Hermitian operator B to a normalized vector |1> is generally of the form:

B|1> = b|1> + c|2>

where b and c are numerical coefficients and |2> is a normalized vector orthogonal to |1>.

My question is: Why B|1> must have the above form? Does it mean if |1> is an eigenstate of B, then b=!0 and c=0? But what if |1> is not an eigenstate of B?

I also need to find the expectation value of B (<1|B|1>), but I think I got this part:

<1|B|1> = <1|b|1> + <1|c|2> = b<1|1> + c<1|2> = b

since |1> and |2> are orthogonal and they're both normalized. Does that look right?

 

[Ambitwistor]

You can decompose any vector, such as B|1>, into two components that are proportional to and orthogonal to some vector. If |1> is an eigenvector of B, then B|1> = b|1>; b can be zero or nonzero. If |1> isn't an eigenstate of B, then neither b nor c have to be zero. You correctly computed <1|B|1>.

 

[Ambitwistor]

I should add that while b can be zero if |1> is an eigenvalue of B, that would indicate that B is a non-invertible operator, which is generally taken not to be the case in quantum mechanics.

 

[einai]

Originally posted by Ambitwistor
You can decompose any vector, such as B|1>, into two components that are proportional to and orthogonal to some vector. If |1> is an eigenvector of B, then B|1> = b|1>; b can be zero or nonzero. If |1> isn't an eigenstate of B, then neither b nor c have to be zero. You correctly computed <1|B|1>.

Thank you very much. I got it :).