Consider Two Spin-1/2 Particles

  • Thread starter andre220
  • Start date
  • #1
75
1

Homework Statement



Consider two spin 1/2 particles. Initially these two particles are in a spin singlet state. If a measurement shows that particle 1 is in the eigenstate of ##S_x = -\hbar/2##, what is the probability that particle 2 in this same measurement is in the eigenstate of ##S_z = \hbar/2##?

Homework Equations


##\chi^{(x)}_{\pm} = \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{\pm1}{\sqrt{2}}\end{pmatrix}## and
##\chi^{(z)}_{+} = \begin{pmatrix}1\\0\end{pmatrix}##
##\chi^{(z)}_{-} = \begin{pmatrix}0\\1\end{pmatrix}##

The Attempt at a Solution


Okay so I am not entirely sure how to proceed, my instinct tells me to setup a linear combination of two measurements such that the linear combo is equal to one, i.e.:
$$
\begin{pmatrix}1\\1\end{pmatrix} = a
\begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{pmatrix} + b \begin{pmatrix}1\\0\end{pmatrix}
$$ But that gives a or b =1. So that is not correct. What am i doing wrong?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
What is the state vector for the singlet state? Which state is the second spin going to be in if you measure the first to be in state Sx = - hbar/2?
 
  • #3
75
1
The singlet state is:
$$|00\rangle = \frac{1}{\sqrt{2}}(\uparrow\downarrow-\downarrow\uparrow)$$
As I understand it, the total spin that the two particles can carry is 0. So that would mean that if the first measurement is ##-\hbar/2## then the other must yield ##+\hbar/2## for the total spin to be zero.
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
So what is the probability of a particle in the Sx = hbar/2 state to be in the Sz = hbar/2 state?
 
  • #5
75
1
So after some thought here is how I went about solving it: If particle one has ##S_x = -\hbar/2## then particle two must have ##S_x = \hbar/2## therefore the probability of measuring ##\hbar/2## for ##S_z## for particle two is: $$\left|\langle\chi^{(x)}_{+}|\chi^{(z)}_{+}\rangle\right|^2 = \begin{pmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \frac{1}{2}$$ I wanted to make sure that I could justify the answer mathematically.
 
  • #6
DEvens
Education Advisor
Gold Member
1,203
457
Just for fun, consider what the result of the probability being 1/2 means. If you had no knowledge whatever of the state of the second particle, if it was completely random up or down, you would get half. But you have been told that it is part of a singlet state, and the other particle in that state is down. And the probability is still 1/2 of it being up. Does that seem correct?

Consider exactly what it means to state that the system is in the singlet state, and to state that one of the particles is definitely in the down state. Can both of these statements be true at the same time? What does it mean to perform a measurement on the system? After the measurement, what will the system look like? (Hint: It won't be in the singlet state any more.) Think projection operators.
 
  • #7
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
26,429
9,937
therefore the probability of measuring ℏ/2\hbar/2 for SzS_z for particle two is
But what is it given that you know it already for particle one?
 
  • #8
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Consider exactly what it means to state that the system is in the singlet state, and to state that one of the particles is definitely in the down state. Can both of these statements be true at the same time?
Yes, this can (and is) definitely true as the problem states that the spin of the particles are considered in different directions. As the original state is a singlet and the first particle is found in the spin down state in the x-direction, the second particle must now be in the spin up state in the x-direction. This spin state has probability 1/2 to be in the spin up state of the z-direction, as found by the OP.
 

Related Threads on Consider Two Spin-1/2 Particles

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
3K
Replies
4
Views
3K
Replies
28
Views
4K
Replies
8
Views
3K
  • Last Post
Replies
1
Views
868
Replies
1
Views
3K
  • Last Post
Replies
12
Views
6K
Replies
4
Views
5K
Top