Linear Operator L with Zero Matrix Elements

sayebms
Messages
31
Reaction score
0

Homework Statement


Suppose a linear operator L satisfies <A|L|A> = 0 for every state A. Show that then all matrix elements <B|L|A> = 0, and hence L = 0.

Homework Equations


##<A|L|A>=L_{AA} and <B|L|A>=L_{BA}##

The Attempt at a Solution


It seems very straight forward and I don't know how to prove it but here is what I have tried:
##<B|L|A> \to##Using resolution of Identity ##\to \sum_{A} <B|A><A|L|A> \to <B|L|A>=0##
Is it right or do I need to write more.
 
Last edited:
Physics news on Phys.org
It is not correct, you are summing over A and at the same time assuming that ##|A\rangle## is a fixed state (i.e., the one you started with in ##\langle B|L|A\rangle##).
 
since the problems says for every state A so should I write as following ##<A_i|L|A_i>=0 \to ## then as before
## <B_j|L|A_i>=\sum_{i}<B_j|A_i><A_i|L|A_i>=0##
is it right now?
 
No, it is still wrong. You cannot let the state you are summing over in the completeness relation be denoted by the same as the state you have on the left-hand side. It is simply not correct.
 
should I solve it without the resolution of Identity?
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top