# System with two particles with spin QM

• skrat
In summary: S_x> = \frac{\hbar}{2}  <S_y> = \frac{\hbar}{2} $$So the uncertainty is:$$ \delta S = \sqrt{<S^2> - <S>^2} = \sqrt{3\hbar^2 - \frac{\hbar^2}{4}} = \frac{\sqrt{11}}{2}\hbar $$d) Your solution for the first particle is correct. For the second particle, you only need to consider the terms with the same total spin, so the possible results and their values are:$$ \frac{\sqrt{2}}{2}\hbar \text{ with probability } \

## Homework Statement

Two particles with spin ##S_1=1## and ##S_2=1## are in our system. For wavefunction of the first particle we know that ##S_{1z}|\psi _1>=\hbar |\psi _1>## while for the second particle ##S_{2x}|\psi _2>=\hbar |\psi _2>##.
a) Find wavefunction ##|\psi _1>## in basis with ""good"" (I don't know the right english word but I know am supposed to use CG coefficients) ##S_{1z}## and ##|\psi _2>## in basis with ""good"" ##S_{2x}##.
b) We measure the squared value of aggregate spin of both particles. What are the possible results and their probabilities?
c) Calculate the expected value and uncertainty of aggregate spin of both particles.
d) We measure the ##x## component of aggregate spin of both particles. What are possible results and their values?

## Homework Equations

I will try to use notation ##|s,m>##.

Also note: $$S_z|\psi >=m\hbar |\psi >$$ $$S_{\pm}|s,m>=\hbar \sqrt{s(s+1)-m(m\pm 1)}|s,m\pm 1>$$ $$S_x=\frac{S_+ +S_-}{2}$$ $$S_y=\frac{S_+ -S_-}{2i}$$

## The Attempt at a Solution

a)
Basis vectors for both particles are ##|1,0>,|1,-1>,|1,1>##. Therefore for the first particle $$S_{1z}(\alpha |1,0>+\beta |1,1>+\xi |1,-1>)=\hbar(\alpha |1,0>+\beta |1,1>+\xi |1,-1>)$$ Solving this system leaves me with ##\alpha=\xi =0## while ##\beta =1##. Therefore ##|\psi _1=|1,1>##.
Things are a bit more complicated for the second particle.The idea is to use ##S_x=\frac{S_+ +S_-}{2}##. I really don't want to write all the equations (because it takes a lot of time for me to put everything in latex), so I hope you don't mind if you make this part a bit shorter. The condition says ##S_{2x}|\psi _2>=\hbar |\psi _2>## where ##\psi _2 =\alpha |1,0>+\beta |1,1>+\xi |1,-1>##. Knowing this, we can calculate $$S_x|1,0>=\hbar \frac{\sqrt 2}{2}|1,1>+\hbar \frac{\sqrt 2}{2}|1,-1>$$ $$S_x|1,-1>=\hbar \frac{\sqrt 2}{2}|1,0>$$ $$S_x|1,1>=\hbar \frac{\sqrt 2}{2}|1,0>$$ This gives me a system saying that ##\beta=\xi ## and ##\alpha =\sqrt 2 \beta## Therefore $$\psi _2 =\alpha (|1,0>+\frac{1}{\sqrt 2}|1,1>+\frac{1}{\sqrt 2} |1,-1>$$ Normalizing the last expression gives me $$\psi _2=\frac{1}{\sqrt 2}|1,0>+\frac{1}{2}|1,1>+\frac{1}{2}|1,-1>$$ Now to transform in the right basis, we simply use Clebsch-Gordan tables:
For the first particle ##\psi _1=|1,1>## becomes ##|2,2>## and for the second particle
##\psi_2=\frac{1}{\sqrt 2}|1,0>+\frac{1}{2}|1,1>+\frac{1}{2}|1,-1>## becomes ##\frac{1}{2}(|2,1>+|1,1>+\frac{1}{\sqrt 6}|2,0>+\frac{1}{\sqrt 2}|1,0>+\frac{1}{\sqrt 3}|0,0>+|2,2>)##

This should be it for part a).
b)
##S^2|\psi _1>=6\hbar ^2|2,2>## if we know that ##S^2|s,m>=\hbar ^2s(s+1)|s,m>##. Also with probability ##1##.
Second particle:
##S^2|\psi _2>=S^2(\frac{1}{2}(|2,1>+|1,1>+\frac{1}{\sqrt 6}|2,0>+\frac{1}{\sqrt 2}|1,0>+\frac{1}{\sqrt 3}|0,0>+|2,2>))=##
##=\frac 1 2 (6\hbar ^2 |2,1>+2\hbar ^2 |1,1>+\frac{6}{\sqrt 6}\hbar ^2 |2,0>+\frac{2}{\sqrt 2}\hbar ^2|1,0>+6\hbar ^2|2,2>)##
So there are two options: ##6\hbar^2 ## with probability ##\frac {13}{24}## and ##2\hbar ^2## with probability ##\frac 3 8 ##.

c)
First particle:
##<S^2>=6\hbar ^2##
##<S_z>=2\hbar ## while ##<S_x>=<\psi _1|\frac{S_+ +S_-}{2}|\psi _1>=0=<S_y>##

So ##\delta S=\sqrt{<S^2>-<S>^2}=\hbar \sqrt 2.##
Second particle:
This part is easy: ##<S^2>=4\hbar ^2## and ##<S_z>=\hbar ##.
But to get ##<S_y>## and ##<S_x>## one has to again work with ##S_ {\pm }##. This takes a lot of time and writing. I shall only write the end results:
##<S_x>=\hbar## and ##<S_y>=0##.

Meaning ##\delta S=\hbar \sqrt 2##. Yes, the same as the particle one. Interesting or wrong. I don't know.

d)
first particle: Possbile result ##\hbar ## with probability ##1##.
Second particle: Wush... I would say:
##\frac{\sqrt 2}{2}\hbar## with probability ##\frac{7}{16}## and
##\hbar## with probability ##\frac{6}{16}## and
##\frac{\sqrt 6}{2}\hbar## with probability ##\frac{3}{16}##

Huh... I know this is a lot of text and a lot of calculating this. But I would highly appreciate it if somebody could check my results. Also I would be happy to hear if some things could be done faster.

Hello there, as a fellow scientist, let me check your results and offer some suggestions to make things a bit clearer.

a) Your solution for the first particle is correct. For the second particle, your final wavefunction is correct, but it would be helpful to explicitly state which terms correspond to which basis states. So your final wavefunction should be written as:

$$\psi_2 = \frac{1}{2}|2,1> + \frac{1}{2}|1,1> + \frac{1}{2\sqrt{6}}|2,0> + \frac{1}{2\sqrt{2}}|1,0> + \frac{1}{2\sqrt{3}}|0,0> + \frac{1}{2}|2,2>$$

b) Your solution for the first particle is correct. For the second particle, you only need to consider the terms with the same total spin, so the possible results are:

$$S^2|\psi_2> = \frac{6\hbar^2}{2}|2,1> + \frac{2\hbar^2}{2}|1,1> + \frac{6\hbar^2}{2\sqrt{6}}|2,0> + \frac{2\hbar^2}{2\sqrt{2}}|1,0> + \frac{6\hbar^2}{2\sqrt{3}}|0,0> + \frac{6\hbar^2}{2}|2,2>$$

So the possible results are:

$$6\hbar^2 \text{ with probability } \frac{1}{2}$$
$$2\hbar^2 \text{ with probability } \frac{1}{6}$$
$$\hbar^2 \text{ with probability } \frac{1}{6}$$
$$0 \text{ with probability } \frac{1}{6}$$

c) Your calculations for the first particle are correct. For the second particle, you only need to consider the terms with the same total spin, so the expected value and uncertainty are:

$$<S^2> = \frac{6\hbar^2}{2} = 3\hbar^2$$
$$<S_z> = \hbar$$


## 1. What is spin in quantum mechanics?

In quantum mechanics, spin refers to an intrinsic angular momentum property of particles. It is a fundamental property of particles, similar to mass and charge. Spin is represented by the quantum number s and can take on values of 0, 1/2, 1, 3/2, etc.

## 2. How is spin measured in quantum mechanics?

In quantum mechanics, spin is measured using a spin operator, which is a mathematical representation of the spin property. The spin operator acts on the wave function of a particle and gives the projection of the spin along a particular direction. The measurement of spin can have two possible outcomes: spin up or spin down.

## 3. What is the significance of two particles with spin in quantum mechanics?

In quantum mechanics, the spin of particles plays a crucial role in determining the behavior of a system. When two particles with spin interact, their spin states can become entangled, meaning that the state of one particle is dependent on the state of the other. This can lead to interesting phenomena such as quantum teleportation and superposition.

## 4. Can two particles with the same spin value interact?

Yes, two particles with the same spin value can interact. However, their interaction will depend on the specific conditions of the system. For example, if the two particles are in a spin singlet state, their interaction will be different from when they are in a spin triplet state.

## 5. How does the spin of a particle affect its energy level?

In quantum mechanics, the spin of a particle can affect its energy level through the spin-orbit interaction. This interaction arises due to the coupling of the spin of a particle with its orbital angular momentum. The presence of this interaction can lead to energy level splitting, which is observed in some atomic and molecular systems.