Calculating Probability of Spin Measurement in System of Particles

Click For Summary
The discussion focuses on calculating the probability of measuring the same spin value for a system of two 1/2 particles at time t=0. The initial state is given as a superposition of three configurations, each with an equal probability of 1/3. It is noted that two out of the three states have matching spin values for both particles. Consequently, the probability of measuring the same spin for the two particles is determined to be 2/3. The conversation highlights the application of conditional probability in this context.
greisen
Messages
75
Reaction score
0
Hey,

I am a little confused with this system 1/2 particles spin at t=0

|PSI(0)> = 1/sqrt(3) (|+>_1|+>_2 + |+>_1|->_2 + |->_1|->_2

the values of S_1z and S_2z are measured at t=0. Then I have to calculate the probability that the same value will be found for the two particles.

Any help apreciated
 
Physics news on Phys.org
The probability of measuring each of the three states is 1/3.
How many of the states have s_1z=s_2z?
 
hmmm. I get that the probability for the three states is 1/3
P = |<a|a>|^2 for each of the vectors. So the probability for measuring the same state for two particles can be calculated using conditional probability ?

Thanks for your reply
 
Two states have the same spin,, so the prob is 2/3.
 

Similar threads

Replies
13
Views
2K
Replies
19
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K