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# 物理代写|量子力学代写Quantum mechanics代考|KYA321 The Quantum State of a Thermal Beam

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## 物理代写|量子力学代写Quantum mechanics代考|The Quantum State of a Thermal Beam

Consider the apparatus illustrated in Fig. 1. A collimated beam of silver atoms is extracted from an oven, which is then passed to observer 1, who with a Stern-Gerlach apparatus measures $S_x$. In these notes we will speak of measuring spin instead of magnetic moment; these are proportional,
$$\boldsymbol{\mu}=\frac{e}{m c} \mathbf{S},$$
so measuring one implies the measurement of the other. (But physically the Stern-Gerlach apparatus really measures magnetic moment.) Of the two beams that emerge from the apparatus, the one with measured value $S_x=-\hbar / 2$ is thrown away.

In the previous set of notes we decided that the Hilbert space for the spin of a silver atom is twodimensional, and that the eigenspaces of $S_x$ (which are the eigenspaces of $\mu_x$ ) are one-dimensional. Therefore according to the postulates of quantum mechanics, whatever the state of the beam when it enters the first apparatus, it will be projected onto the one-dimensional eigenspace of $S_x$ with eigenvalue $+\hbar / 2$ when $S_x=+\hbar / 2$ is measured. Therefore the state of the $+$ beam emerging from the first measurement apparatus is described by any nonzero vector in this eigenspace. We let $\left|S_x+\right\rangle$ be such a vector, assumed to be normalized. The $+$ beam is passed to observer 2 , who measures some other observable $A$. According to the measurement postulates of quantum mechanics, observer 2 will measure an average value of $A$ given by
$$\langle A\rangle=\left\langle S_x+|A| S_x+\right\rangle .$$

## 物理代写|量子力学代写Quantum mechanics代考|Thermal Beam Not Described by a State Vector

The answer is definitely no, for a beam from a thermal source such as we have described is isotropic, and has no preferred direction of spin. For if we measure $S_x$ on the thermal beam we find $50 \%$ of the atoms with $S_x=+\hbar / 2$ and $50 \%$ with $S_x=-\hbar / 2$, for an average of $\left\langle S_x\right\rangle=0$. Similarly we find $\left\langle S_y\right\rangle=\left\langle S_z\right\rangle=0$, or $\langle\mathbf{S}\rangle=0$. Naturally, the thermal beam has no preferred direction of spin.
On the other hand, any definite quantum state (a so-called “pure” state) of the spin system is associated with a normalized ket in the 2-dimensional Hilbert space, and such a ket is not isotropic but rather necessarily “points in” some direction. To show this, let us represent an arbitrary normalized ket $|\chi\rangle$ in terms of its components with respect to the basis $|+\rangle,|-\rangle$, consisting of eigenkets of the observable $S_z$. Then $|\chi\rangle$ can be written in the form,
$$|\chi\rangle=\alpha|+\rangle+\beta|-\rangle,$$
where
$$|\alpha|^2+|\beta|^2=1 .$$

## 物理代写|量子力学代写Quantum mechanics代考|The Quantum State of a Thermal Beam

$$\boldsymbol{\mu}=\frac{e}{m c} \mathbf{S}$$

$$\langle A\rangle=\left\langle S_x+|A| S_x+\right\rangle .$$

## 物理代写|量子力学代写Quantum mechanics代考|Thermal Beam Not Described by a State Vector

$$|\chi\rangle=\alpha|+\rangle+\beta|-\rangle,$$

$$|\alpha|^2+|\beta|^2=1$$

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