U1L5e.txt

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We've talked about when we have the observables for spin
in the x-direction, y-direction, and z-direction for one qubit.
I want to talk about what happens
when you have these observables for two qubits--
S sub x, S sub y, S sub z for two qubits.

Well, the spin for two qubits should
be the spin for the first qubit plus the spin
for the second qubit.
At least, it [AUDIO OUT] if the qubits are in a tensor product
state.

So S sub x is equal to sigma x tensor I plus I tensor sigma
x, and ditto for y and z.

I'm not going to write that out.

And maybe I'll write this out.
So sigma x tensor I is 1, 1, 1, 1 because you take sigma
x, which is just 0, 1, 1, 0, 0, and you tensor it with I,
which is--
identity here and identity here--
plus 1, 1, 1, 1.

And actually, I'm leaving out a factor of 2
because the observable for spin was 1/2 sigma x
because the spin values were--
spin 1/2 particles have either one half pointing down
or one half point up.
So that is equal to 1/2, 1, 1, 1, 1.

And let's do S sub z equals 1/2--

1, minus 1, 1, minus 1--

plus 1/2-- 1, 1, minus 1, minus 1--
is equal to 1, 0, 0, 1.

And I guess this is I tensor sigma x.
And this is sigma x tensor I.
z.
z.
z.
Thank you.
Yeah?
But isn't it sigma x tensor sigma x and sigma x tensor
sigma z?
Oh.
Well, so we're trying to figure out their combined bit
of two particles is.
So to get the combined spin of two particles,
you take the first particle and take its spin,
and then take the second particle and add its spin.
So if we take the first--
and this is the spin of the first particle,
sigma x tensor I over 2.
And this is the spin of the second particle
in the x-direction and in the z-direction.

So if we multiplied them, we would
be taking the spin of the first particle
and multiplying it by the spin of the second particle, which
would give us an observable.
But it's not the observable for spin.