# How to prepare the silver atoms in Stern–Gerlach experiment with total angular momentum $j_z=+\frac{1}{2}\hbar,-\frac{1}{2}\hbar$?

Total angular momentum at $z$ direction $j_z$ is summation of orbital angular momentum $l_z$ and spin angular momentum $s_z$ of many electrons in silver atom. $j^2=j(j+1)$,$j_z=-j,-j+1,...,j-1,j$

The stream of silver atoms splitting in two direction means $j=\frac{1}{2},j_z=+\frac{1}{2}or-\frac{1}{2}$,my question is why the silver atoms always at this state where they always splitting in two streams in Stern-Gerlach experiment?

The silver atoms are not "in this state". The surprise of the Stern-Gerlach experiment is that, however you orient the magnetic field gradient, the beam always splits in two and only two components, and splits along the field gradient.

In other words, if you orient the gradient along $$\boldsymbol{\hat x}$$, the beam will split along $$\boldsymbol{\hat x}$$ so that one component has $$j_x=+\frac{1}{2}\hbar$$ and the other $$j_x=-\frac{1}{2}\hbar$$. If you orient the gradient along $$\boldsymbol{\hat n}$$, the beam will split in two along $$\pm \boldsymbol{\hat n}$$ so that $$\boldsymbol{\vec j}\cdot \boldsymbol{\hat n}=\pm \frac{1}{2}\hbar$$. This is just the experimental evidence.

The interpretation is that atoms do not come out of the oven with their spin in a specified direction (i.e. their spin is NOT along any particular but unknown direction $$\boldsymbol{\hat n}$$). If it were, you could tinker with the orientation of the apparatus and eventually orient the field gradient parallel to this $$\boldsymbol{\hat n}$$ so that all atoms would be deflected in one and only one direction. This simply does not happen experimentally.

Instead, the interpretation is that the direction of the spin is determined by the field gradient once the atoms interact with the magnetic field.

So, as a direct answer to your question, to prepare silver atoms in the $$j_z=\pm \frac{1}{2}\hbar$$ states, take a magnet with a well-defined field gradient, orient it along a direction which you call $$\boldsymbol{\hat z}$$, and pass your beam through it. The output of this is what you're looking for, i.e. states "prepared" in either orientations of spin.

Note that this has been verified for other systems:

• The version with hydrogen was done by Phipps and Taylor (Phipps, T. E., and J. B. Taylor. "The magnetic moment of the hydrogen atom." Physical Review 29.2 (1927): 309.) The paper is available but behind a paywall.
• The version with polarized neutrons was done by Sherwood et al. (Sherwood, J. E., T. E. Stephenson, and Seymour Bernstein. "Stern-Gerlach experiment on polarized neutrons." Physical Review 96.6 (1954): 1546.) The paper is available but behind a paywall.

The literature on this experiment is rich and the topic is still active, as per this arXiv submission.

• .Thanks.You answer is good one with clear expression but not the answer I want to know.It seems my poor English cause the misunderstanding of my thought.Your answer could be expressed into the language of wavepacket.The silver atoms in prepared in oven with a wavepacket $\phi$.The wavepacket $\phi$ could be decomposed into $\phi=\alpha_1|\phi,j_z=\frac{\hbar}{2}>+\alpha_2|\phi,j_z=-\frac{\hbar}{2}>$ as well as $\phi=\beta_1|\phi,j_x=\frac{\hbar}{2}>+\beta_2|\phi,j_x=-\frac{\hbar}{2}>$. $|\phi,j_z=\frac{\hbar}{2}>$ and $|\phi,j_z=-\frac{\hbar}{2}>$ are overlapping each other.
– John
Oct 16, 2017 at 16:34
• Applying inhomogeneous magnetic field $B_z$ would make the two wavepackets $|\phi,j_z=\frac{\hbar}{2}>$ and $|\phi,j_z=-\frac{\hbar}{2}>$ split into two direction.$j_z$ originates from the angular momentum and spin momentum of electrons in the silver atom.The eigenstate of total angular moment $j=\frac{1}{2}$ would make eigenstate of $j_z$ to be $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$. My question is why the silver atom's total angular momentum is $j=\frac{1}{2}$ when there are so many electrons in the silver atom.How to make sure $j=\frac{1}{2}$
– John
Oct 16, 2017 at 16:44
• @John For this see the link provided in the comment to your question. Oct 16, 2017 at 17:04
• @John as to the wave packet aspect of your comment: this does not enter here because the wave packet locates the atom in space, and this does not have any dependence on spin. The aspect of the wavepacket that is relevant is that, because individual atoms may not be exactly on the axis of your system, the final spot on the detector would be slightly smeared even if the apparatus was perfect (i.e. perfectly uniform gradient etc.) Oct 16, 2017 at 17:17
• Thanks.Maybe I am not correct.I think a wavepacket could represent a moving atom.If a wavepacket was superposition of $\sqrt{a\%}$ z-spin up wavepacket and $\sqrt{(100-a)\%}$ z-spin down wavepacket.Inhomogeneous magnetic field $B_z$ would make the superposition wavepacket split and $a\%$ probability detecting the z-spin up atom at $+z$ direction.This is only my thought. My question is why the silver atom total angular momentum is $+-\frac{1}{2}$,there are so many electrons in this atom,addition of the electrons momentum at z-direction is not definite.
– John
Oct 19, 2017 at 9:41