# Stern-Gerlach experiment for spin-1 particles

If we have a beam of spin-1 particles and let them pass through a Stern-Gerlach apparatus (oriented along z-axis, we get three output beams. Suppose we now take only the $$+\hbar$$ beam and pass it thorugh a Stern-Gerlach apparatus oriented along x-axis, we again get three states and we expect them to have equal probabilities (similar to spin-1/2 particles). Instead, we get beams with the following probabilities, $$P_{+\hbar} = \frac{1}{4} \\ P_{0\hbar} = \frac{1}{2} \\ P_{-\hbar} = \frac{1}{4}.$$ This is contrary to the case of spin-1/2 particles, where when we conduct a similar experiment, we get $$+\hbar$$ and $$-\hbar$$ beams with equal probabilities. Where does this non uniformity in the resulting beams arise from for the case of spin-1 particles?

Refer. Spin-1 System, Chapter 2, Quantum Mechanics by David McIntyre http://depts.washington.edu/jrphys/ph248A11/qmch1.pdf

Edit. (As suggested in the comments) I am completely comfortable with the mathematical formalism and deriving the probabilities. What I am interested in is a physical reasoning (if it exists) for why the probability is not equal for the three beams.

• Are you sure it's a spin 1 system (not combination of two spin $\frac{1}{2}$ system). I would be glad to know if it is not $\frac{1}{2}\otimes\frac{1}{2}\equiv 0\oplus 1$. Jun 3 '20 at 8:45
• @KartikChhajed, as far as the text goes, this has been given for a spin-1 system and not a combination of two spin-1/2 systems. This has been given as an extension to whatever math that was done with spin-1/2 systems. Jun 3 '20 at 9:31
• Did you try writing a spin +1 state in z direction in terms of spin eigenstates of spin 1 operator in x direction? Jun 3 '20 at 10:36
• @DvijD.C., That in fact gives me a clear mathematical explanation. But is there any "physical" reasoning as to why $0\hbar$ state is twice as more probable than the other outcomes or is this one of the many quantum mechanical reulsts that are far from physical reasonings? Jun 3 '20 at 11:01
• You should emphasize the fact that you understand the formalism and understand how to derive these probabilities; and that you're looking for a direct physical intuition based on which one could perhaps expect these results before explicitly calculating them. Jun 3 '20 at 11:38

The spin matrices for a spin-$$1$$ system are

$$\begin{gather} & S_x=\frac{1}{\sqrt2} \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}, \end{gather}$$

$$\begin{gather} &&& S_y=\frac{1}{\sqrt2i} \begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 1\\ 0 & -1 & 0 \end{pmatrix}, \end{gather}$$

$$\begin{gather} S_z= \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{pmatrix}. \end{gather}$$

These can be deduced similarly to how one deduces the Pauli matrices for a spin-$$1/2$$ system (remember that these matrices need to be multiplied by $$\hbar$$ when applied in the context of measurements).

Your particle just passed through a Stern-Gerlach apparatus oriented along the $$z$$-axis and was measured to have spin $$+\hbar$$. This means that the particle is in the state $$\left | +\hbar, z \right>$$, which corresponds to the eigenvector with eigenvalue $$1$$ of $$S_z$$. In other words, we can write

$$\begin{gather} \left | +\hbar, z \right>= \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}. \end{gather}$$

Now, the particle can be in any one of the states $$\left | +\hbar, x \right>$$, $$\left | 0\hbar, x \right>$$, or $$\left | -\hbar, x \right>$$. These states correspond to the eigenvectors with eigenvalues $$1,0,$$ and $$-1$$ respectively of $$S_x$$. Or, in other words we have

$$\begin{gather} \left | +\hbar, x \right>=\frac 12 \begin{pmatrix} 1\\ \sqrt2\\ 1 \end{pmatrix}, \end{gather}$$

$$\begin{gather} & \left | 0\hbar, x \right>=\frac 1{\sqrt2} \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}, \end{gather}$$

$$\begin{gather} & \left | -\hbar, x \right>=\frac 12 \begin{pmatrix} 1\\ -\sqrt2\\ 1 \end{pmatrix}. \end{gather}$$

Finally, given that the particle starts in the state $$\left | +\hbar, z \right>$$, the probabilities for it ending up in one of the states mentioned above are given by

\begin{align} P_{+\hbar} &=|\left < +\hbar, x | +\hbar, z \right>|^2 =\frac14,\\ P_{0\hbar} &=|\left < 0\hbar, x | +\hbar, z \right>|^2 =\frac12,\\ P_{-\hbar} &=|\left < -\hbar, x | +\hbar, z \right>|^2 =\frac14. \end{align}

• Thank you for your anser. I'm aware of this mathematical explanation. Rather, I'm interested in some physical reasoning. Is there any "physical" explanation as to why the $0\hbar$ is twice as more probable than the other two outcomes or is it one of the many quantum mechanical results that are far from physical reasonings? Jun 3 '20 at 10:59
• I'm not aware of any physical explanation. Your question basically reduces to "Is there a physical explanation as to why $\left | +\hbar, z \right>=\frac12 \left | +\hbar, x \right>-\frac 1{\sqrt2} \left | 0\hbar, x \right>+\frac12 \left | -\hbar, x \right>$?", which I doubt can be seen physically. Jun 3 '20 at 12:33

Excellent question! To understand the reason for this, neither expectation of $$S_x$$ or $$S_y$$ going to zero is sufficient. One could setup equations from relevant commutation relations and get the probabilities but that’s equivalent to doing the matrix algebra. Let us see if symmetry helps simplifying things.

The fact that $$\langle{S_x}\rangle=0$$ enforces the weightage between the plus and minus states along x to be equal. This means they are of the general (real) form as follows: $$|{+,z}\rangle=\alpha~ |{+,x}\rangle ~+~\sqrt{1-2\alpha^2} ~|{0,x}\rangle ~+~\alpha ~ |{-,x}\rangle \\ |{-,z}\rangle=\beta~ |{+,x}\rangle ~+~\sqrt{1-2\beta^2} ~|{0,x}\rangle ~+~\beta~ |{-,x}\rangle$$

But the symmetry of the problem dictates that if you flip the system by $$180^\text o$$, you should get the same probabilities. In other words, keeping things real, $$\alpha=\pm \beta$$ Finally, using the fact that the plus and the minus states along z are orthogonal, we get, $$-2\alpha^2 + 1-2\alpha^2=0\\ \Rightarrow \alpha^2=\frac{1}{4}$$

Where we have chosen $$\alpha=-\beta$$ as plus would imply both the states are equal.

As you can see this is, like many things in physics, ultimately an outcome of symmetry.

• Great answer! Exactly what I was looking for. Jun 6 '20 at 4:15

When physicists were discovering the quantum mechanics formalism, they were to some extent guided by their classical intuitions. In the classical case, we would expect that, for a system with total spin angular momentum 1 and with its spin in the z-direction, that a measurement of spin in the x direction would give zero with certainty.

But in the quantum case, we see that the measurement of spin in the x direction gives zero only half of the time. Why? One way to justify this is to notice that the total spin angular momentum operator $$S^2$$ has the eigenvalue $$s(s+1) = 2$$. So even in the state with definite z-spin there is a considerable amount of angular momentum perpendicular to z. This gives rise to the probability of $$1/4$$ of measuring the spin in either positive or negative x direction.

I hope this helps