Stern-Gerlach experiment

Description

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Explanation

Coming from probability theory#Text 2.
Consider first a classical set up, and the we will jump to the quantum behavior. Nevertheless, the classical behavior would be valid, even in QM, for observables that are "totally independent" (own terminology), for example, $x$ position and $y$ position. I think the technical name is commuting observables.

Suppose that after measuring an electron with the first device we obtain $r$ . Then we use the second one and obtain $b$ . In a classical setup we would expect that if we use device 1 again, we obtain $r$ again, but this is not what happens with Stern-Gerlach devices. We obtain $r$ and $g$ with certain frequencies (although fixed).

The logical explanation (at least from the point of view of QM) is to accept that electron is in a certain state $Ψ \in H_{1}$ before the use of any machine (see quantum superposition). Until now, we have assumed that the vector in the Hilbert space encodes probabilities. But now we are saying that the electron is the vector (or, at least, a part of the electron). Every time we use a machine, the vector $Ψ$ moves to a new location. For example: we use machine 1 and get $r$ or $0^{'} 7$ (with certain probability), so the state vector has traveled to $r$ . If we use machine 1 again we obtain the same result: probability of obtain $r$ again is the length of the projection of $r$ on $r$ , that is, 1!! If we use machine 2, depending on the value obtained, the state vector travels to $b$ or $s$ but they are still inside $H_{1}$ .

That is, we have machines (or one machine with different positions) that extract information of the system. The system is codified in a Hilbert space and the machines in operators. Every operator has privileged states with good behavior respect to it (a "classic behavior" respect to it, we can say). That is the reason why we cannot consider any operator, but the corresponding to Hermitian matrices.
If the system is in a "classical state" for the operator, the operator will only change the scale of the state vector, and the measurement arises when we check the length of the transformed vector

⟨ r | F (r) ⟩ .

If the system is in a "non-classical" state for this machine, it goes to the "most similar" classical state for the machine (not for sure, but with a probability that depends on the similitude), and then the above is applied.

Can we find the expression of the privileged states of a machine with respect of the ones of the other? That is, who are $b$ and $s$ (canonical base for machine 2) in the canonical base of $H_{1}$ ?

Suppose the machines are $S G_{z}$ and $S G_{x}$ in the Stern-Gerlach experiments. According to the measured probabilities in Stern-Gerlach experiment we could assign the coefficients:

b = \frac{1}{\sqrt{2}} r + \frac{1}{\sqrt{2}} g

s = \frac{1}{\sqrt{2}} r - \frac{1}{\sqrt{2}} g

The choice of the coefficients is arbitrary but subject to the measured probabilities.
Observe that, so far, we have no need of complex numbers.
Of course, the operator $S G_{z}$ has the matrix expression in "its associated basis" is (we are assuming now that the outputs are 1 and -1 instead of 0'7 and 1'8):

S G_{z} \equiv (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})

What is the matrix expression for the $S G_{x}$ machine in this basis? Since $S G_{x} b = b$ and $S G_{x} s = - s$ , solving a very easy linear system we find that

S G_{x} \equiv (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

But we can think like if we have an infinite number of machines $S G_{θ}$ . They are all the same that our original $S G_{z}$ but rotated an angle $θ$ from the $z$ axis towards the $x$ axis. For the moment, we will be pretending that the universe is of dimension 2. With each of these machines we get two results, say $M$ and $B$ , and empirically obtain that, once our system is in state $M$ , $S G_{z}$ yields an expected value of $c o s (θ)$ , what is reasonable even from the classical point of view. If we write $M = α r + β g$ and $B = γ r + δ g$ , since

⟨ M | S G_{z} (M) ⟩ = c o s (θ)

(expected value) and $M$ is normalized, i.e.,

⟨ M | M ⟩ = 1

we obtain

M = c o s (\frac{θ}{2}) r + s i n (\frac{θ}{2}) g

In a similar way, since the expected value for $B$ is $- c o s (θ)$ and $M$ and $B$ must be orthogonal we can conclude

B = - s i n (\frac{θ}{2}) r + c o s (\frac{θ}{2}) g

Solving a linear system we find the matrix for $S G_{θ}$ :

S G_{θ} \equiv (\begin{matrix} c o s (θ) & s i n (θ) \\ s i n (θ) & - c o s (θ) \end{matrix})

So, in conclusion, we can assume that when we rotate the $S G_{z}$ machine an angle $θ$ the well-behaved states are now

(\begin{matrix} c o s (\frac{θ}{2}) \\ s i n (\frac{θ}{2}) \end{matrix}), (\begin{matrix} - s i n (\frac{θ}{2}) \\ c o s (\frac{θ}{2}) \end{matrix})

that is, the privileged vectors are rotated an angle of $θ / 2$ . Or we can interpret that if we rotate the system (our electron gun in state $r = (\begin{matrix} 1 \\ 0 \end{matrix})$ ) an angle of $- θ$ , we are putting the electron in a state

(\begin{matrix} c o s (\frac{θ}{2}) \\ - s i n (\frac{θ}{2}) \end{matrix}) \in H

And the operator goes from

(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})

(\begin{matrix} c o s (θ) & s i n (θ) \\ s i n (θ) & - c o s (θ) \end{matrix})

This let us predict probabilities in the Stern-Gerlach experiment: what would be the probability of obtain such result if we turn the machine such angle? It is a numerical model for this family of Stern-Gerlach experiments.

But we can think in it like a geometric model, too. The different directions of space (plane, in this case) are codified in the matrices $S G_{θ}$ : $S G_{z}$ would represent the unit vector along $z$ axis, and $S G_{θ}$ is the result of a rotation of angle $θ$ . And we are led to think that spin is an "internal object" inside the electron with a special kind of symmetry without counterpart in classical-macroscopic terms.
(see section \textit{About spin})
Pasted image 20220619120550.png
So, what about our electron? After performing a rotation of angle $θ$ the privileged states of the Hilbert space have rotated $θ / 2$ so we come back to the initial configuration after a $4 π$ rotation. So we can say that electron has spin 1/2. It is a bit difficult to visualize but you can imagine an arrow with a flag attached in such a way that the flag turns half the angle of the arrow.

On the other hand, observe that the matrices

S G_{z} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}), S G_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

behaves like a orthogonal vector basis of $R^{2}$ . In fact

R^{θ} (S G_{z}) = (\begin{matrix} \cos (θ) & - \sin (θ) \\ s i n (θ) & c o s (θ) \end{matrix}) \cdot (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) =

= (\begin{matrix} \cos (θ) & \sin (θ) \\ \sin (θ) & - \cos (θ) \end{matrix}) = \cos (θ) S G_{z} + \sin (θ) S G_{x}

among other similarities. But the matrices product gives them an additional intrinsic algebra structure which the vector space $R^{2}$ lacks. What does it mean, geometrically?

There is a correspondence between the subalgebra generated by the matrices ${S G_{z}, S G_{x}}$ and the abstract Clifford algebra of dimension 2 and positive signature with orthonormal basis ${e_{1}, e_{2}}$ . Since we know that in the Clifford algebra the even subalgebra acts over $R^{2}$ like rotations (on the "sandwich way") we are led to think that internal states of the electron are elements of $G_{2}^{+}$ , that is, spinors. That is, the element

c o s (\frac{θ}{2}) 1 + s i n (\frac{θ}{2}) e_{1} e_{2} \in {Cl}_{2} (R)

is identified with the pair of elements of the Hilbert space

(\begin{matrix} c o s (\frac{θ}{2}) \\ s i n (\frac{θ}{2}) \end{matrix}), (\begin{matrix} - s i n (\frac{θ}{2}) \\ c o s (\frac{θ}{2}) \end{matrix})

One observation: the even subalgebra behaves like complex numbers. So the complex numbers have already entered the scene.

The 3-dimensional space

But the Universe is not planar, is 3D. What happens when we rotate the machine out of the $x z$ -plane? Experiments show that the pure $y$ orientation of the machine ( $S G_{y}$ ) yields a pair of states, call it head and tail, $h$ and $t$ , with probabilities similar to those of the $S G_{x}$ ( $b$ and $s$ ) when throwed through $S G_{z}$ . But keep an eye: further experiments show that those are not the privileged states of $S G_{x}$ . This cannot be, definitively, modelled if we stay with real numbers (see page Why complex numbers in QM).
We can solve this problem by using complex coefficients, and in this way we arrive to the picture of the internal states as elements of $S U (2)$ . Indeed, the measured probabilities together with the restrictions of orthonormality of $h$ and $t$ lead us to

h = \frac{1}{\sqrt{2}} r + \frac{e^{i α}}{\sqrt{2}} g

t = \frac{1}{\sqrt{2}} r - \frac{e^{i α}}{\sqrt{2}} g

If we now measure probabilities respect to $S G_{x}$ (i.e., probabilities of $b$ and $s$ output when we feed the machine with $h$ or $t$ states) we can find that $α = π / 2$ . That is,

h = \frac{1}{\sqrt{2}} r + \frac{i}{\sqrt{2}} g

t = \frac{1}{\sqrt{2}} r - \frac{i}{\sqrt{2}} g

And with some computations we get

S G_{y} \equiv (\begin{matrix} 0 & - i \\ i & 0 \end{matrix})

The matrices $S G_{x}, S G_{y}, S G_{z}$ are called Pauli matrices.

In conclusion, we have a machine, $S G$ , that can be oriented in any spatial direction, $\hat{n}$ (unitary vector). When we fix this direction we get a matrix, $S G_{\hat{n}}$ , that encodes all the data of the machine (final states, obtained values and probabilities) with this direction. This matrix has two eigenvectors of $C^{2}$ , $| n + ⟩$ and $| n - ⟩$ , that represent the state of the electron after passing through the machine. Or we can think that the device $S G$ is fixed, and we turn the electron gun to a new spatial orientation $\hat{n}$ . Internally for the electron this has an effect: if it was at state

(\begin{matrix} 1 \\ 0 \end{matrix})

it is arriving to a new state

(\begin{matrix} α \\ β \end{matrix}) \in H

that can be computed like this:

The rotation of the gun is performed with a $3 \times 3$ orthogonal matrix $R \in S O (3)$ , that can be viewed as an element of $S U (2)$ (in fact, there are two of them, since it is a double covering).
In this assignation we can observe a kind of "miracle". The matrix $R$ can be obtained like the product of other three

R = e^{α G_{x}} \cdot e^{θ G_{y}} \cdot e^{ϕ G_{z}}

where ${G_{x}, G_{y}, G_{z}}$ are $3 \times 3$ matrices forming a basis of $s o (3)$ . To find the corresponding element of $S U (2)$ we use the isomorphism (as Lie algebras) of $s o (3)$ with $s u (2)$ , and then the exponential map over $S U (2)$ . And the miracle is that the images of $G_{x}, G_{y}, G_{z}$ are

\frac{i}{2} S G_{x}, \frac{i}{2} S G_{y}, \frac{i}{2} S G_{z}

although this matrices (without the $i$ ) appeared like operators to encode data of an observable, not like transformations!!!

The obtained element of $S U (2)$ rotates the initial "internal vector" $(\begin{matrix} 1 \\ 0 \end{matrix})$ in the usual way: complex matrix product.
If we had chosen the other element of $S U (2)$ we would have arrived to other vector in $H$ . But the states are not elements of $H$ , but of $P (H)$ .

Some considerations to further thoughts:

The element of $S U (2)$ applied in the sandwhich way to the matrices $S G_{n}$ rotates them, i.e., gives the same result that applying an usual 2x2 rotation matrix on the left. This aims to think the matrices $S G_{n}$ like vectors...
In the point 2 above we watch that $i S G_{n}$ behaves like bivectors in a Clifford algebra, so here is other reason to think that $S G_{n}$ are vectors.
All would be, therefore, simpler if we forget the Pauli matrices and go to Clifford algebras¿?
Explicit computations can be found at mathematica file \textbf