# Measuring the spin of a particle from a singlet state

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!

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.

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

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.

• 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. Oct 19 '14 at 16:32