Figureshows the Stern-Gerlach apparatus. It reveals that spin-$\frac{\mathbf{1}}{\mathbf{2}}$particles have just two possible spin states. Assume that when these two beams are separated inside the channel (though still near its centreline). we can choose to block one or the other for study. Now a second such apparatus is added after the first. Their channels are aligned. But the second one is rotated about the-axis by an angle \(\phi\) from the first. Suppose we block the spin-down beam in the first apparatus, allowing only the spin-up beam into the second. There is no wave function for spin. but we can still talk of a probability amplitude, which we square to give a probability. After the first apparatus' spin-up beam passes through the second apparatus, the probability amplitude is$\mathbf{cos}\mathbf{(}\mathbf{\varphi }\mathbf{/}\mathbf{2}\mathbf{)}{\mathbf{\uparrow }}_{\mathbf{2}\mathbf{\text{nd}}}\mathbf{+}\mathbf{sin}\mathbf{(}\mathbf{\varphi }\mathbf{/}\mathbf{2}\mathbf{)}{\mathbf{\downarrow }}_{\mathbf{2}\mathbf{nd}}$where the arrows indicate the two possible findings for spin in the second apparatus.

(a) What is the probability of finding the particle spin up in the second apparatus? Of finding it spin down? Argue that these probabilities make sense individually for representative values of$\mathit{\varphi }$and their sum is also sensible.

(b) By contrasting this spin probability amplitude with a spatial probability amplitude. Such as$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}{\mathbf{te}}^{\mathbf{2}}}$. Argue that although the arbitrariness of$\mathbf{\varphi }$gives the spin cases an infinite number of solves. it is still justified to refer to it as a "two-state system," while the spatial case is an infinite-state system.

Step by step solution

01

Given information: 

The probability amplitude is $\cos \left(\frac{\mathrm{\varphi }}{2}\right){\left|{}_{2\text{nd}}+\sin \left(\frac{\mathrm{\varphi }}{2}\right)\right|}_{2\mathrm{nd}}$.

02

Concept of spin up and spin down

The expression for probability of spin up is given by ${\mathbf{P}}_{\mathbf{\text{up}}}\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\varphi }}{\mathbf{2}}\mathbf{\right)}$

The expression for probability of spin down is given by ${\mathbf{P}}_{\mathbf{\text{down}}}\mathbf{=}{\mathbf{sin}}^{\mathbf{2}}\left(\frac{\varphi }{2}\right)$

03

Evaluate spin up probability

(a)

The spin up probability is calculated as,

${\mathrm{P}}_{\text{up}}={\cos }^2\left(\frac{\mathrm{f}}{2}\right)={\cos }^2\left(\frac{{90}^{°}}{2}\right)={\cos }^2\left({45}^{°}\right)=\frac{1}{2}$

The spin down probability is calculated as,

${\mathrm{P}}_{\text{down}}={\sin }^2\left(\frac{\mathrm{f}}{2}\right)={\sin }^2\left(\frac{{90}^{°}}{2}\right)={\sin }^2\left({45}^{°}\right)=\frac{1}{2}$

04

Two-state system justify

(b)

It is a two-state system because it contains the superposition of the two states. Each of these states corresponds to different physical outcome and the possible outcomes when a measurement is made of spin orientation are one of the two state spin.

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