spin components = atoms?

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In a measurement on a BEC of non-entangled atoms, on average half of the atoms are found in the ground state (spin downwards), the other half in the excited state (spin upwards). "Deviations from this mean value that occur from measurement to measurement, lead to a quantum noise that is evenly distributed among the spin components orthogonal to the mean spin," adds Pascal Böhi, another doctoral student.

In order to investigate the influence of the state-dependent potential on the quantum noise the scientists determined the noise for each spin component using yet another microwave pulse. As they could clearly demonstrate, for one spin component the noise could be "squeezed" below the limit given by the Heisenberg uncertainty relation. From the observed noise reduction the scientists concluded that inside the BEC clusters of at least four atoms are entangled.

More:

http://www.sciencedaily.com/releases/2010/04/100401101810.htm

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@oristarA,

Atoms can spin on each axis. You can therefore break down the overall spin into spin components, where each spin component is the spin for an individual axis.

http://en.wikipedia.org/wiki/Spin_%28physics%29

Quote:

Spin vector

For a given quantum state, it is possible to describe a spin vector \lang S \rang whose components are the expectation values of the spin components along each axis, i.e., \lang S \rang = [\lang s_x \rang, \lang s_y \rang, \lang s_z \rang]. This vector describes the "direction" in which the spin is pointing, corresponding to the classical concept of the axis of rotation.

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@DrewDad,

I've read through your reply, but still couldn't get it ...

Thanks anyways.

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spin components = atoms?

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