Neutron Interferometry

Let

Rn^(θ)=e−iθJ^n^/ℏR_{\hat{n}}(\theta) = e^{-i \theta \hat{J}_{\hat{n}} / \hbar}

This is how you rotate a quantum state by an angle θ\theta around n^\hat{n} where J^n^\hat{J}_{\hat{n}} is the angular momentum operator along n^\hat{n}.

We experimentally observed that

Rx(2π)=e−i2πS^x/ℏ=−I,Rx(4π)=IR_x(2\pi) = e^{-i 2\pi \hat{S}_x / \hbar} = -I, \qquad R_x(4\pi) = I

Given a Sx^\hat{S_x} operation, turning it θ=2π\theta=2\pi causes there to be a negative sign infront of the II. Rotating it another 2π2\pi causes it to return to the same state. This is definitely not how it works in classical physics, where if you rotate something 2π2\pi you’d get right back to where it started. This actually means we live in a 4π4\pi-world through experimental verification.

This was tested by using a neutron interferometer. Basically, imagine if a neutron at ∣ψ⟩\left\lvert \psi \right\rangle hits a silicon crystal beam splitter. The state of the neutron is its position {u,l}\{u,l\} tensor product its spin {+,−}\{+, -\}. This is just how we define it.

Basis: {∣n,+⟩, ∣n,−⟩, ∣ℓ,+⟩, ∣ℓ,−⟩}\{\left\lvert n,+ \right\rangle,\ \left\lvert n,- \right\rangle,\ \left\lvert \ell,+ \right\rangle,\ \left\lvert \ell,- \right\rangle\}

∣ψ⟩=∣position⟩⊗∣spin⟩\left\lvert \psi \right\rangle = \left\lvert \text{position} \right\rangle \otimes \left\lvert \text{spin} \right\rangle

Neutron beam splitter (silicon crystal): ∣n,s⟩\left\lvert n,s \right\rangle splits into transmitted ∣n,s⟩\left\lvert n,s \right\rangle and reflected ∣ℓ,s⟩\left\lvert \ell,s \right\rangle.

(B⊗I)∣n,s⟩=12(∣n⟩+∣ℓ⟩)⊗∣s⟩(B \otimes I)\left\lvert n,s \right\rangle = \tfrac{1}{\sqrt{2}}\bigl(\left\lvert n \right\rangle + \left\lvert \ell \right\rangle\bigr) \otimes \left\lvert s \right\rangle (B⊗I)∣ℓ,s⟩=12(∣n⟩−∣ℓ⟩)⊗∣s⟩(B \otimes I)\left\lvert \ell,s \right\rangle = \tfrac{1}{\sqrt{2}}\bigl(\left\lvert n \right\rangle - \left\lvert \ell \right\rangle\bigr) \otimes \left\lvert s \right\rangle B=12(111−1)B = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

The lower path passes through a static magnetic field BB, which applies a spin rotation Rx(θ)R_x(\theta). The two paths recombine at a second beam splitter, with detectors D0D_0 and D1D_1 on the two outputs.

The full evolution is

∣ψout⟩=(B⊗I) U (B⊗I) ∣u,s⟩\left\lvert \psi_{\text{out}} \right\rangle = (B \otimes I)\, U\, (B \otimes I)\, \left\lvert u, s \right\rangle

where the magnetic field implements the conditional rotation

U=∣u⟩⟨u∣⊗I+∣ℓ⟩⟨ℓ∣⊗Rx(θ).U = \left\lvert u \right\rangle\left\langle u \right\rvert \otimes I + \left\lvert \ell \right\rangle\left\langle \ell \right\rvert \otimes R_x(\theta).

Case θ=2π\theta = 2\pi:

∣u,s⟩→B⊗I12(∣u⟩+∣ℓ⟩)⊗∣s⟩\left\lvert u, s \right\rangle \xrightarrow{B \otimes I} \tfrac{1}{\sqrt{2}}\bigl(\left\lvert u \right\rangle + \left\lvert \ell \right\rangle\bigr) \otimes \left\lvert s \right\rangle →Rx(2π)=−I12(∣u⟩−∣ℓ⟩)⊗∣s⟩→B⊗I∣ℓ,s⟩\xrightarrow{R_x(2\pi) = -I} \tfrac{1}{\sqrt{2}}\bigl(\left\lvert u \right\rangle - \left\lvert \ell \right\rangle\bigr) \otimes \left\lvert s \right\rangle \xrightarrow{B \otimes I} \left\lvert \ell, s \right\rangle

⇒\Rightarrow neutron at D1D_1.

Case θ=4π\theta = 4\pi:

∣u,s⟩⟶∣u,s⟩⇒ neutron at D0.\left\lvert u, s \right\rangle \longrightarrow \left\lvert u, s \right\rangle \quad\Rightarrow\text{ neutron at } D_0.

This is fully deterministic as in general, the probability P(D0)=cos⁡2(θ/4)P(D_0) = \cos^2(\theta/4) and P(D1)=sin⁡2(θ/4)P(D_1) = \sin^2(\theta/4), so θ=2π,4π\theta = 2\pi, 4\pi are the special tunings where one probability hits 11.

This is the physical consequence of Rn^(2π)=−I≠IR_{\hat{n}}(2\pi) = -I \neq I.

This experimentally proves that we live in a 4π4\pi-world.