- #1

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## Main Question or Discussion Point

Hello Everyone,

I am reviewing the SG experiment. I think the experiment was set out to demonstrate that the orbital angular momentum ##L## is quantized producing a quantized magnetic momentum. But silver and hydrogen atoms have no net orbital angular momentum. Still two sub-beams, due to two different intrinsic spin magnetic moments, on which two different forces applied, were produced on the screen. Stern and Gerlarch didn't immediately know that spin was involved....

Here my question: if both orbital angular momentum and spin angular momentum are present, how many sub-beams would be present on the collecting screen? In that case, the total magnetic moment is ##\mu_{total} = \mu_{orbital} + \mu_{spin}##.

For example, if ##\ell=1##, ## m_{\ell} =-1,0,1 ## so there are three different orbital magnetic moments and three different forces (one of which is zero). Spin provides two different values: ##\frac {1}{2}## and ##- \frac {1}{2}## hence two different spin magnetic moment.

If we considered all possible six combinations ##(m_{\ell} , m_{s})##: ##(0,\frac {1}{2})##, ##(1,\frac {1}{2})##, ##(-1, \frac {1}{2})##,##(0, -\frac {1}{2})##, ##(1, - \frac {1}{2})##, ##(-1, - \frac {1}{2})##

there would seem to be 6 different possible total magnetic moments , six different forces and 6 different sub-beams generated on the screen. Or does each different orbital magnetic moment have only a single spin value? For example, are the combinations ##(1,\frac {1}{2})## and ##(1, - \frac {1}{2})## no possible together? Why?

Thanks.

I am reviewing the SG experiment. I think the experiment was set out to demonstrate that the orbital angular momentum ##L## is quantized producing a quantized magnetic momentum. But silver and hydrogen atoms have no net orbital angular momentum. Still two sub-beams, due to two different intrinsic spin magnetic moments, on which two different forces applied, were produced on the screen. Stern and Gerlarch didn't immediately know that spin was involved....

Here my question: if both orbital angular momentum and spin angular momentum are present, how many sub-beams would be present on the collecting screen? In that case, the total magnetic moment is ##\mu_{total} = \mu_{orbital} + \mu_{spin}##.

For example, if ##\ell=1##, ## m_{\ell} =-1,0,1 ## so there are three different orbital magnetic moments and three different forces (one of which is zero). Spin provides two different values: ##\frac {1}{2}## and ##- \frac {1}{2}## hence two different spin magnetic moment.

If we considered all possible six combinations ##(m_{\ell} , m_{s})##: ##(0,\frac {1}{2})##, ##(1,\frac {1}{2})##, ##(-1, \frac {1}{2})##,##(0, -\frac {1}{2})##, ##(1, - \frac {1}{2})##, ##(-1, - \frac {1}{2})##

there would seem to be 6 different possible total magnetic moments , six different forces and 6 different sub-beams generated on the screen. Or does each different orbital magnetic moment have only a single spin value? For example, are the combinations ##(1,\frac {1}{2})## and ##(1, - \frac {1}{2})## no possible together? Why?

Thanks.