Question about an Eqn. in Shankar - wave function probability

In summary: The projection operator to the subspace ##\text{Eig}(\hat{\Omega},\omega)## is$$\hat{P}_{\omega} = \sum_{\alpha} |\omega,\alpha \rangle \langle \omega,\alpha|.$$Then the probability to find the value ##\omega## when measuring ##\Omega## when the system is prepared in a pure state represented by a normalized state vector ##|\psi \rangle## is, according to Born's postulate,$$P(\omega|\psi)=\langle \psi|\hat{P}_{\omega}
  • #1
Jacob Nie
9
4
Homework Statement
The section is about the probability of obtaining a certain ##\omega## upon measurement of operator ##\Omega.## And here it is dealing specifically with the case of a degenerate ##\Omega.##

It says: In general, one can replace in Postulate III
$$P(\omega)\propto \langle \psi | \mathbb{P}_{\omega}|\psi\rangle $$
where ##\mathbb{P}_{\omega}## is the projection operator for the eigenspace with eigenvalue ##\omega.##

(Postulate III was ##P(\omega)\propto |\langle \omega | \psi \rangle |^2.##)
Relevant Equations
n/a
I don't see why it is not ##P(\omega)\propto |\langle \psi | \mathbb{P}_{\omega}|\psi\rangle |^2.## After all, the wavefunction ends up collapsing from ##|\psi\rangle## to ##\mathbb{P}_{\omega}|\psi\rangle.##
 
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  • #2
Jacob Nie said:
It says: In general, one can replace in Postulate III
$$P(\omega)\propto \langle \psi | \mathbb{P}_{\omega}|\psi\rangle $$
where ##\mathbb{P}_{\omega}## is the projection operator for the eigenspace with eigenvalue ##\omega.##

I don't see why it is not ##P(\omega)\propto |\langle \psi | \mathbb{P}_{\omega}|\psi\rangle |^2.## After all, the wavefunction ends up collapsing from ##|\psi\rangle## to ##\mathbb{P}_{\omega}|\psi\rangle.##
What is the explicit form of ##\mathbb{P}_{\omega}## when ##\omega## is not degenerate? When this ##\mathbb{P}_{\omega}## is used in ##\left< \psi \right| \mathbb{P}_{\omega} \left|\psi \right> ##, what results?
 
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Likes Jacob Nie
  • #3
George Jones said:
What is the explicit form of ##\mathbb{P}_{\omega}## when ##\omega## is not degenerate? When this ##\mathbb{P}_{\omega}## is used in ##\left< \psi \right| \mathbb{P}_{\omega} \left|\psi \right> ##, what results?

The explicit form would be ##\mathbb{P}_{\omega} = |\omega\rangle \langle \omega | \psi \rangle.## So, ##\langle \psi | \mathbb{P}_{\omega}| \psi \rangle = \langle \psi | \omega \rangle \langle \omega | \psi \rangle = |\langle \omega | \psi \rangle |^2.##

Thank you very much for the helpful hint.
 
  • #4
Now, a projection operator is an operator, not a vector. Let ##\Omega## be an observable, ##\hat{\Omega}## its representing self-adjoint operator. For an eigenvalue ##\omega## there may be several linearly independent eigenvectors ("degeneracy"). They form a subvector space of the Hilbert space, ##\text{Eig}(\hat{\Omega},\omega)##, and you always can choose an orthonormal basis ##|\omega,\alpha \rangle##. For simplicity I assume the ##\alpha## are just discrete labels and we deal with true normalizable eigenvectors. The generalization if you have continuous generalized eigenvectors "normalized to a ##\delta## distribution" is straight forward.

Then the projection operator to the subspace ##\text{Eig}(\hat{\Omega},\omega)## is
$$\hat{P}_{\omega} = \sum_{\alpha} |\omega,\alpha \rangle \langle \omega,\alpha|.$$
Then the probability to find the value ##\omega## when measuring ##\Omega## when the system is prepared in a pure state represented by a normalized state vector ##|\psi \rangle## is, according to Born's postulate,
$$P(\omega|\psi)=\langle \psi|\hat{P}_{\omega}|\psi \rangle=\sum_{\alpha} \langle \psi|\omega,\alpha \rangle \langle \omega,\alpha|\psi \rangle = \sum_{\alpha} |\psi(\omega,\alpha)|^2.$$
 

1. What is the Shankar equation and how is it related to wave function probability?

The Shankar equation is a mathematical equation used in quantum mechanics to describe the evolution of a particle's wave function over time. It is related to wave function probability because the square of the wave function gives the probability of finding the particle at a particular location.

2. How do you solve the Shankar equation for a specific system?

The Shankar equation can be solved using various techniques, such as separation of variables or perturbation theory. The specific method used depends on the system being studied and the level of accuracy required.

3. Can the Shankar equation be used to describe all quantum systems?

No, the Shankar equation is a simplified version of the Schrödinger equation and can only be used to describe non-relativistic systems. It cannot be used for systems involving high speeds or energies, such as those found in particle physics.

4. How does the wave function probability change over time according to the Shankar equation?

The Shankar equation describes the time evolution of the wave function using the Hamiltonian operator. This operator acts on the wave function to determine how it changes over time, taking into account the potential energy of the system.

5. What are the limitations of using the Shankar equation to study quantum systems?

The Shankar equation is limited in its applicability to non-relativistic systems and cannot account for certain phenomena, such as particle spin. It also assumes a single particle system and cannot be used for systems with multiple particles or interactions between particles.

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