What To Project Upon When Doing QM Measurements?

[flyusx]

Homework Statement

Let's say we have a set of eigenvalues for a Hamiltonian of a system: h1, h2 and h3 where the eigenvectors are |h1>, |h2> and |h3>. Let's also say we have another operator of a system A (non-commuting with the Hamiltonian) where its eigenvalues are a1, a2 and a3 with eigenvectors |a1>, |a2> and |a3>.

Relevant Equations

$$P(a)=\langle frac{|\langle a|ψ\rangle|^{2}}{\langleψ|ψ\rangle}$$

I have read that if one measures the Hamiltonian and receives a value of h2, then the quantum state will be in $|h2\rangle$. Finding the probability of a1 is done by projecting $|a1\rangle$ upon $|h2\rangle$ divided by $\langle h2|h2\rangle$. In other words: $$\frac{|\langle a1|h2\rangle|^{2}}{\langle h2|h2\rangle}$$

I have also seen elsewhere that if we are given a state $|ψ\rangle$ with the Hamiltonian and operator A, then measuring h2 will change the quantum state into a new state $|ψ'\rangle$ given by $|h2\rangle\langle h2|ψ\rangle$. To find the probability of a1, the textbook explicitly projects $|a1\rangle$ upon $|ψ'\rangle$ and not $|h2\rangle$. In other words: $$\frac{|\langle a1|ψ'\rangle|^{2}}{\langleψ'|ψ'\rangle}$$

When I expand both out, I get the same answer:
$$\frac{|\langle a1|ψ'\rangle|^{2}}{\langleψ'|ψ'\rangle}=\frac{\langleψ'|a1\rangle\langle a1|ψ'\rangle}{\langleψ'|ψ'\rangle}=\frac{\langleψ|h2\rangle\langle h2|a1\rangle\langle a1|h2\rangle\langle h2|ψ\rangle}{\langleψ|h2\rangle\langle h2|h2\rangle\langle h2|ψ\rangle}=\frac{|\langle a1|h2\rangle|^{2}}{\langle h2|h2\rangle}$$

I see that both methods work, but I'm confused as to whether the quantum state is left in $|h2\rangle$ or $|ψ'\rangle$ after measuring an energy (eigenvalue) of h2. When solving a measurement problem, is there a reason to calculate $|ψ'\rangle$ after measuring H, then using that for calculations with A? Or is it always reasonable to say that a state is in $|h2\rangle$ if h2 is measured, and then use $|h2\rangle$ for further calculations with A?

 

[DrClaude]

I much prefer to work with normalized kets only, so I don't like the $\ket{\psi'} = \ket{h2}\braket{h2 | \psi}$ approach, which results in an unnormalized $\ket{\psi'}$.

 

[flyusx]

I completely agree (especially since we don't need the <ψ|ψ> at the bottom). But both ways are valid, right? Some problems I work out use one method and other problems use the other, so I just want to make sure I'm not just assuming two methods are identical.

 

[BvU]

I have also seen elsewhere that if we are given a state $|ψ\rangle$ with the Hamiltonian and operator A, then measuring h2 will change the quantum state into a new state $|ψ'\rangle$ given by $|h2\rangle\langle h2|ψ\rangle$.

'seen elsewhere' is vague. Where ?

'A new state $|ψ'\rangle$ given by $|h2\rangle\langle h2|ψ\rangle$ ' would be unnormalized, right ?

$\$

 

[flyusx]

From Zettili's QM textbook, I have seen the |ψ'> representation used to solve Problem 3.8 whereas Problem 3.7 doesn't use this method. I've attached a PDF file of those two problems.

In Problem 3.7(b), he projects each $a$ state onto $\|φ_2\rangle$ since that's the eigenvector associated with $-\varepsilon_0$. In 3.8(b,c) he uses |ψ'> (though he names it |φ>,|χ>), perhaps since the original state was not normalised. But I do agree that normalised kets are always easier to work with.

 

[DrClaude]

I completely agree (especially since we don't need the <ψ|ψ> at the bottom). But both ways are valid, right? Some problems I work out use one method and other problems use the other, so I just want to make sure I'm not just assuming two methods are identical.

Yes.

 

[flyusx]

So if a state $|ψ\rangle$ is measured and a value h2 is recorded, the state will be in $|ψ'\rangle=|h2\rangle\langle h2|ψ\rangle$? But the probability calculation is identical when calculating with $|h2\rangle$?

 

[DrClaude]

So if a state $|ψ\rangle$ is measured and a value h2 is recorded, the state will be in $|ψ'\rangle=|h2\rangle\langle h2|ψ\rangle$? But the probability calculation is identical when calculating with $|h2\rangle$?

$|ψ'\rangle=|h2\rangle\langle h2|ψ\rangle$ is nothing but $|ψ'\rangle= c |h2\rangle$ with $c$ some complex number. Physical states are defined up to an arbitrary complex scalar constant, so it is the same state.

 

[flyusx]

Ah, so they're identical with the exception of a constant (the inner product of h2 and ψ).

 

[DrClaude]

Ah, so they're identical with the exception of a constant (the inner product of h2 and ψ).

Correct. It will be useful to remember that $\braket{|}$ correspond to complex numbers and can be moved around equations as needed.