Question about an Eqn. in Shankar - wave function probability

[Jacob Nie]

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 $\mathbb{P}_{\omega}|\psi\rangle.$

 

[George Jones]

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?

 

[Jacob Nie]

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.

 

[vanhees71]

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,
$$P(\omega|\psi)=\langle \psi|\hat{P}_{\omega}|\psi \rangle=\sum_{\alpha} \langle \psi|\omega,\alpha \rangle \langle \omega,\alpha|\psi \rangle = \sum_{\alpha} |\psi(\omega,\alpha)|^2.$$