Measurement of Lx: Result of Measurement?

  • Thread starter Thread starter Physgeek64
  • Start date Start date
  • Tags Tags
    Measurement
Click For Summary
The discussion revolves around measuring the operator L_x for a given wave function ψ = zf(r). The participants note that L_x can be expressed using the ladder operators L_+ and L_-, and they derive that L_x |ψ> results in a state proportional to |1,1>. However, they express uncertainty about determining the actual measurement outcome, emphasizing the need for eigenvalue equations. It is suggested that to find the expected value of L_x, one must use its differential form and integrate the wave function appropriately. Ultimately, while the expected value may be zero, the outcome of an individual measurement could differ and requires proper integration of the wave function with the operator.
Physgeek64
Messages
245
Reaction score
11

Homework Statement


If we have a wave function ##\psi =zf(r)## and we take a measurement of ##L_x## what is the result of the measurement?

Homework Equations

The Attempt at a Solution


So i know we can write ##L_x=\frac{1}{2}(L_+ + L_- )## and that ##|\psi > = g(r) |1,0> ## so ##L_x |\psi >= \frac{1}{2} \sqrt{2} g(r) |1,1>##. But i don't know what the measurement would be because this has to be an eigenvalue equation to read off the result of a measurement (I think?).

Many thanks!
 
Physics news on Phys.org
Physgeek64 said:

The Attempt at a Solution


So i know we can write Lx=12(L++L−)Lx=12(L++L−)L_x=\frac{1}{2}(L_+ + L_- ) and that |ψ>=g(r)|1,0>|ψ>=g(r)|1,0>|\psi > = g(r) |1,0> so Lx|ψ>=12√2g(r)|1,1>Lx|ψ>=122g(r)|1,1>L_x |\psi >= \frac{1}{2} \sqrt{2} g(r) |1,1>. But i don't know what the measurement would be because this has to be an eigenvalue equation to read off the result of a measurement (I think?).

Many thanks!

you have a given wave function which is a function of r and theta only
if you wish to find expected value of L(x) then one must use differential form of the operator L(x)
 
drvrm said:
you have a given wave function which is a function of r and theta only
if you wish to find expected value of L(x) then one must use differential form of the operator L(x)
But how do i find the result of a measurement. I can show that the expected value is zero, but i don't know what the outcome of an individual measurement would be.
 
Physgeek64 said:
But how do i find the result of a measurement. I can show that the expected value is zero, but i don't know what the outcome of an individual measurement would be.

In QM the measurement is defined with the help of wave function
say you have a position wavefunction then you can measure x-operator

as {Psi * (x )Psi }integrated over the whole space.
similarly L(x) is a operator which is represented in (r,theta, phi) space .

put the complex conjugate of the wave function on the left and psi on the right and L(x) in between and integrate over whole space .
the result will give you measurement and it may not be zero.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 17 ·
Replies
17
Views
904
  • · Replies 2 ·
Replies
2
Views
652
  • · Replies 1 ·
Replies
1
Views
744
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
2
Views
1K