2
$\begingroup$

Imagine that I have a singlet state:

$|s\rangle = {1 \over \sqrt2}(| \uparrow_1\downarrow_2 \rangle - |\downarrow_1\uparrow_2\rangle)$

I want to measure the spin along the z axis of the first particle. So I should apply the operator $S_{z1}$ to the singlet state:

$S_{z1}|s\rangle$ = ?

The possible answer is 50% $\uparrow_1$ and 50% $\downarrow_1$.

How should I write in the bra-ket notation the result of this measurement?

Thank you!

$\endgroup$
2
$\begingroup$

There are a few ways you can formally approach this problem.

First note that the singlet state is an element of the Hilbert space of the tensor product of two spin-1/2 particles $1$ and $2$, i.e. $| s \rangle \in \mathcal{H} = \mathcal{H}_{1} \otimes\mathcal{H}_{2}$.

To describe the system in the full space, you can form the density matrix $\rho = | s \rangle \langle s | \in D(H)$.

1st way: Since you care only about the first system, you can take the partial trace over system $2$ yielding the reduced density matrix of system $1$: \begin{align} \rho_1 = \text{tr}_2(\rho) \in D(H_1). \end{align} This density matrix will give you the correct predictions if you choose to only do measurements on $1$. In this case, you can show that \begin{align} \rho_1 = \frac{1}{2}|\uparrow\rangle_1 \langle \uparrow |_1+\frac{1}{2}|\downarrow\rangle_1 \langle \downarrow |_1, \end{align} a mixed state.

Now the operator $S_{z1}$ has the spectral decomposition \begin{align} S_{z1} = +\frac{\hbar}{2} \pi_{\uparrow 1} + (- \frac{\hbar}{2}) \pi_{\downarrow 1}, \end{align} where $\pi_{\uparrow 1}$ is the projector onto the spin-z-up axis of the 1st particle: $\pi_{\uparrow 1} = | \uparrow \rangle_1 \langle \uparrow |_1$ and similarly for the other projector.

According to the Born rule, the probability of getting $\hbar/2$ is \begin{align} \mathbb{P}(s_{z1}=\hbar/2) &= \text{tr}_1 (\rho_1 \pi_{\uparrow 1}) \nonumber \\ &= \sum_k \langle k|_1 \left[\frac{1}{2}|\uparrow\rangle_1 \langle \uparrow |_1+\frac{1}{2}|\downarrow\rangle_1 \langle \downarrow |_1 \right]| \uparrow \rangle_1 \langle \uparrow |_1 | k \rangle_1 \nonumber \\ & = \frac{1}{2} \sum_k \langle k|_1|\uparrow\rangle_1\langle \uparrow |_1 | k \rangle_1 \nonumber \\ & = \frac{1}{2}\langle \uparrow |_1 \sum_k | k \rangle_1\langle k|_1|\uparrow\rangle_1 \nonumber \\ & = \frac{1}{2} \uparrow |_1|\uparrow\rangle_1 \nonumber \\ & = \frac{1}{2}, \end{align} and similarly for $\mathbb{P}(s_{z1} = -\hbar/2)$. This is called a projective valued measurement.

2nd way: You can think of a measurement on the full system, except that you do nothing on system 2. So the relevant operator is the tensor product of the $S_{z1}$ operator with the identity operator on 2: $S_{z1} \otimes \mathbb{I}_2$, which I'll call the FULL spin operator.

The spectral decomposition is \begin{align} S_{z1} \otimes \mathbb{I}_2 = +\frac{\hbar}{2} \pi_{\uparrow 1} \otimes \mathbb{I}_2 + (- \frac{\hbar}{2}) \pi_{\downarrow 1}\otimes \mathbb{I}_2, \end{align} i.e. it has only 2 eigenvalues, but now each eigenspace is 2-dimensional.

So according to the Born rule, the probability of measuring $\hbar/2$ for the FULL spin operator is given by \begin{align} \mathbb{P}(s_{z1}^\text{full}=\hbar/2) = \text{tr} (\rho (\pi_{\uparrow 1} \otimes \mathbb{I}_2)) \end{align} which you can work out to be $1/2$ again.

Note: this is the formal way of showing these calculations, following the axioms of quantum mechanics.

Expressions like $|\langle \uparrow|_1 S_{z1} |s\rangle |^2$, should be regarded as heuristic expressions that 'give' the right answer, but they are not mathematically correct because $| \uparrow \rangle_1$ is an element of $\mathcal{H}_1$, $S_{z1}$ an element of $L(\mathcal{H_1})$, but $|s \rangle$ an element of $\mathcal{H}$, so it just doesn't quite make sense.

$\endgroup$
0
$\begingroup$

$\newcommand{bra}[1]{ \langle #1 |}$ $\newcommand{ket}[1]{ | #1 \rangle}$ $\newcommand{upone}{\uparrow_1 }$ $\newcommand{downone}{\downarrow_1 }$ $\newcommand{uptwo}{\uparrow_2 }$ $\newcommand{downtwo}{\downarrow_2 }$ $\newcommand{spin}{S_{z1} }$

You are asking what $\spin \ket{s}$ is. You say you want to do it with braket notation. Well,

$\spin \ket{s} = \spin \frac{1}{\sqrt{2}} \left( \ket{\upone \downtwo} - \ket{\downone \uptwo}\right) \\ = \frac{1}{\sqrt{2}} \left(\spin \ket{\upone \downtwo} - \spin\ket{\downone \uptwo}\right) \\ =\frac{1}{\sqrt{2}} \left(\spin \left(\ket{\upone} \otimes \ket{ \downtwo} \right) - \spin \left( \ket{\downone} \otimes \ket{ \uptwo}\right)\right)\\ =\frac{1}{\sqrt{2}} \left( \left( \spin \ket{\upone} \right) \otimes \ket{ \downtwo} - \left( \spin \ket{\downone} \right) \otimes \ket{ \uptwo}\right)\\ =\frac{1}{\sqrt{2}} \left( \left( \frac{\hbar}{2} \ket{\upone} \right) \otimes \ket{ \downtwo} - \left( -\frac{\hbar}{2} \ket{\downone} \right) \otimes \ket{ \uptwo}\right)\\ =\frac{1}{\sqrt{2}} \left( \frac{\hbar}{2} \ket{\upone} \otimes \ket{ \downtwo} + \frac{\hbar}{2} \ket{\downone} \otimes \ket{ \uptwo}\right) \\ =\frac{1}{\sqrt{2}} \left( \frac{\hbar}{2} \ket{\upone \downtwo} + \frac{\hbar}{2} \ket{\downone \uptwo}\right)\\ =\frac{\hbar}{2\sqrt{2}} \left( \ket{\upone \downtwo} + \ket{\downone \uptwo}\right).$

If you can read that you are off to a good start. If you have specific questions about any notation or steps just ask.

$\endgroup$
  • $\begingroup$ The original poster asked two questions: what is “$S_{z1}|s\rangle$” and how does that partial measurement behave. Here the first question is answered; since this answer was downvoted without explanation, Ī upvote it. $\endgroup$ – Incnis Mrsi Oct 19 '14 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.