Suppose that a state |Ψ> is an eigenstate of operator B, with eigenvalue bi.
i. What is the expectation value of B?
ii. What is the uncertainty of B?
iii. Is |Ψi an eigenstate of B2 or not?
iv. What is the uncertainty of B2?
: Suppose, instead, |Ψ> is an eigenstate of operator B2, with eigenvalue λ.
Without loss of generality, any state can be expanded in eigenstates of B: |Ψ> =
Σci|bi>. (You can still assume that |bi> is eigenstates of operator B, so B|bi> = bi |bi>)
v. What is the uncertainty for B2 ?
vi. Starting from the expansion given above, write out B2 |Ψ> in the B basis (byoperating with B twice in a row since you know that B2|Ψ> = B(B|Ψ>).
vii. Write out the eigenvalue equation for B2in this basis.
Taking the inner product of both sides of the equation with <bj|, you can
extract information about the cj coefficients and the bj eigenvalues, given your
knowledge of λ. Write this equation and explain in words what conclusions
you can draw about the results you might get if you measured B.
The Attempt at a Solution
By the expectation formula:
<B> = <Ψ| B |Ψ> = bi <Ψ|Ψ> = bi
since for eigenstates, they are normalized, the inner product is 1.
For eigenstates, the uncertainty is simply 0.
Also we can do this mathematically as
<B2> = <Ψ| B B |Ψ> = bi <Ψ| B |Ψ> = bi2
ΔB = sqrt (<B2> - <B>2) = 0
The real problem is the rest :
I can do nothing else but operate B twice on the state
B2 |Ψ> = bi 2 |Ψ>
so I guess here bi2 is the corresponding eigenvalue for B2 here?
I feel like the answer is wrong, since if it is, the answer for part iv is zero too.
That looks bad.
So, I tried to do this in another way like
B B |Ψ> = bi B |Ψ>
so let B|Ψ> = is also an eigenstate for operator B
but |χ> = B|Ψ> = bi|Ψ> is with same basis of |Ψ>.
So B2 share a simultaneous eigenstate |Ψ> with B...................
And then I get lost of my target.
Am I thinking too much ?
iv. I cannot solve this since I cannot solve part iii.
v. Uncertainty for B2 is zero since |Ψ> is an eigenstate for B2.
vi.I tried to operate B on |Ψ>, so I get
B2|Ψ> = λ |Ψ>.
Then use the expansion:
B2 |Ψ> = Σci B2 |bi>
= Σci bi2 |bi>
<bj| B2 |Ψ> =<bj| Σci bi2 |bi>
= <bj| λ |Ψ> = λ<bj |Ψ>
= λ cj
Thus, λ = bj2.
I got this but cannot conclude anything from this.
What does this relate to the results from B operator ?