Is A Self-Adjoint in Wave Mechanics?

[noblegas]

Homework Statement

Two operators , A and B , satisfy the equations

A=B^{\dagger}B+3 and A= BB^{\dagger}+1

a)Show that A is self adjoint
b)Find the commutator of [B^{\dagger},B]
c) Find the commutator of [B,B^{\dagger}]
d) Suppose \varphi is an eigenfunction of A with eigenvalue a:

A\varphi=a\varphi

show that if B\varphi =/ 0 then B\varphi is an eigenfunction of A , and find the eigenvalue.

Homework Equations

The Attempt at a Solution

I've only worked on the first part of the problem. I will address the remaining 3 parts later.

(A)^{\dagger}=(B^{\dagger}B+3)^{\dagger}=B^{\dagger}(B^{\dagger})^{\dagger}+(3)^{\dagger}=B^{\dagger}B+(3)^{\dagger}. 3 is not an operator so I don't think you can take the adjoint of it.

 

[tiny-tim]

A=B^{\dagger}B+3\text{ and }A= BB^{\dagger}+1

…
(A)^{\dagger}=(B^{\dagger}B+3)^{\dagger}=B^{\dagger}(B^{\dagger})^{\dagger}+(3)^{\dagger}=B^{\dagger}B+(3)^{\dagger}. 3 is not an operator so I don't think you can take the adjoint of it.

Hi noblegas!

(LaTeX doesn't take any notice of spaces unless you put them inside \text{} )

I assume they mean 3 times the identity, I.

 

[noblegas]

Hi noblegas!

(LaTeX doesn't take any notice of spaces unless you put them inside \text{} )

I assume they mean 3 times the identity, I.

okak then B^{\dagger}B+(3I)^{\dagger}=B^{\dagger}B+3*I^{\dagger}=B^{\dagger}B+3*I=

Therefore A^{\dagger}=A? Now proceeding to the next two parts of the problem;

b) [B,B^{\dagger}]=BB^{\dagger}-B^{\dagger}B

BB^{\dagger}=A-1, B^{\dagger}B=A-3, therefore

[B,B^{\dagger}]=BB^{\dagger}-B^{\dagger}B=A-1-(A-3)=A-1*I-(A-3*I)=2I

c)[A,B]=AB-BA=(BB^{\dagger}+1)B-B(BB^{\dagger}+1)=BB^{\dagger}B+B-BBB^{\dagger}-B=BB^{\dagger}B-BBB^{\dagger}=B(BB^{\dagger}-BB^{\dagger})=B(2I)=2BI

 

[tiny-tim]

You seem to have changed the questions …

but it looks ok ​

 

[noblegas]

You seem to have changed the questions …

but it looks ok ​

what do you mean? I was writing out my solutions for part b and c of my question; how should I start part d of the problem? Should I start by assuming that A<br /> * \varphi=a*\varphi=&gt;A*\varphi-a*\varphi=0<br />? See my OP

 

[tiny-tim]

Hi noblegas!

(have a phi: φ and a dagger: † )

what do you mean?

I meant …

b)Find the commutator of [B^{\dagger},B]
c) Find the commutator of [B,B^{\dagger}]

b) [B,B^{\dagger}]=\cdots

c)[A,B]=\cdots

For d), remember that when the examiner sets you a series of questions, the last one usually comes very easily from the preceding ones …

in this case, just use c) ​

 

[noblegas]

Hi noblegas!

(have a phi: φ and a dagger: † )

I meant …



For d), remember that when the examiner sets you a series of questions, the last one usually comes very easily from the preceding ones …

in this case, just use c) ​

oh I see then . [B^{\dagger},B]=-2*I, and [A,B]=-2BI correct?

 

[noblegas]

did you not understand my latest solution

 

[tiny-tim]

Yes, that looks ok (except that you needn't write -2BI, you can just write it as -2B).

 

[noblegas]

stuck on part d again: AB-BA=-2BI, ARe they saying B\varphi=> 2BI \neq 0

 

[tiny-tim]

stuck on part d again: AB-BA=-2BI, ARe they saying B\varphi=> 2BI \neq 0

(what happened to that φ i gave you? )

Sorry, I don't understand what you're asking.

An eigenvector cannot be zero.

d) says that if Aφ = aφ, and if Bφ ≠ 0, then A(Bφ) is a scalar multiple of Bφ …

prove it using c).​