(Alpha)Expectation Value

[Green Destiny]

[Edited]

So, I am trying to understand something. The need for the expectation value I seem to be learning - the example I have been learning involves spin.

I have the equation $$|S_{xu}>= \frac{1}{\sqrt{2}}[|S_{zu}>+|S_{zd}>]$$ with properties of

$$<S_{zu}|S_{zu}>=<S_{zd}|S_zd>=1$$

and

$$<S_{zu}|S_{zd}>= 0$$

If there is a measurement made on $$|S_{xu}>$$ it will yield for every measurement on the x-direction will be a spin up. But the z-direction is undefined since it has a superpositing in the z-direction which exhibits a spin up and a spin down in equal probability.

So the z-direction has an undefined spin, and is this the reason for the expectation value in this particular area? Since $$S_{xu}$$ does not have a particular spin direction for z, it is still possible to calculate a mean value over large periods of time. So the expectation is $$<S_{xu|\hat{S}_z|S_{xu}>$$.

What I do not understand, is if the spin direction (up or down) cannot be defined for z, why is an expectation value required, what does it do exactly?

 

[1100f]

So, I am trying to understand something. The need for the expectation value I seem to be learning - the example I have been learning involves spin.

I have the equation $$|S_{xu}>= \frac{1}{\sqrt{2}}[|S_{zu}>+|S_{zd}>]$$ with properties of

$$<S_{zu}|S_{zu}>=<S_{zd}|S_zd>=1$$

and

$$<S_{zu}|S_{zd}>= 0$$

If there is a measurement made on $$|S_{xu}>$$ it will yield for every measurement on the x-direction will be a spin up. But the z-direction is undefined since it has a superpositing in the z-direction which exhibits a spin up and a spin down in equal probability.

The z direction is perfectly well defined, it in... the z direction.

So the z-direction has an undefined spin, and is this the reason for the expectation value in this particular area?

The reason for an expectation value of the spin operator in the z direction is that you measure the spin operator in the z-direction

Since $$S_{xu}$$ does not have a particular spin direction for z, it is still possible to calculate a large mean value. So the expectation is $$<S_{xu|\hat{S}_z|S_{xu}>$$.

You cannot have a large mean value because it will be zero.
Indeed $$<s_x+|\hat{S}|s_x+> = 0$$

What I do not understand, is if the spin direction cannot be defined for z, why is an expectation value required, what does it do exactly?

I really don't understand your question (and I guess that also you don't understand it)

 

[Green Destiny]

The z direction is perfectly well defined, it in... the z direction.



The reason for an expectation value of the spin operator in the z direction is that you measure the spin operator in the z-direction


You cannot have a large mean value because it will be zero.
Indeed $$<s_x+|\hat{S}|s_x+> = 0$$


I really don't understand your question (and I guess that also you don't understand it)

I didn't mean the z-direction per se, I meant it's respective spins. As for when I said a large mean value, that sounds so wrong. I meant over a large period of time, totally my fault.

But as for my question, I understand it only to a point. I am asking why an expectation value is used in this example? Why would we want one to explain this?

 

[Green Destiny]

I edited it, so it was a bit more clear.

 

[Green Destiny]

I'll restate my question, since no one has answered it.

Why is it important that the z-spin is undefined so that we must resort to an expectancy value? And on that, what does it necesserily do?

 

[CptBork]

When you start with a particle whose spin is well-defined on the x-axis, its spin cannot be well-defined along the y and z-axes. There's a kind of uncertainty principle for spins just like there's an uncertainty principle for position and momentum. This means if I start with a particle that is known to be spin up on the x-axis, and then measure its spin along the z-axis, I get a 50/50 chance of measuring it as either spin up or spin down along z. You can only know the precise spin of a particle along one axis at a time.

The expectation value is not what you'd measure after waiting a long time with a single particle. It's the average value of the z-spins you'd measure, after setting up large numbers of particles all initially spinning up along the x-axis. Since a particle with spin up/down along x has a 50/50 chance of later being measured as spin up/down along z, you get an expectation value for the z-spins of zero. On the other hand, if your particles started off spin up along an axis 30 degrees off the z-axis, and then you measured their spins directly along z, the expectation value for this is nonzero.

Recommended prerequisites to understand this properly: Discrete and continuous probability (after first learning multivariable calculus), abstract linear algebra, and classical Newtonian angular momentum including phenomena such as spin precession.

 

[Green Destiny]

When you start with a particle whose spin is well-defined on the x-axis, its spin cannot be well-defined along the y and z-axes. There's a kind of uncertainty principle for spins just like there's an uncertainty principle for position and momentum. This means if I start with a particle that is known to be spin up on the x-axis, and then measure its spin along the z-axis, I get a 50/50 chance of measuring it as either spin up or spin down along z. You can only know the precise spin of a particle along one axis at a time..

I didn't know this. So, essentially, spin can only be determined along one axis at a time. That's pretty interesting.

The expectation value is not what you'd measure after waiting a long time with a single particle. It's the average value of the z-spins you'd measure, after setting up large numbers of particles all initially spinning up along the x-axis. Since a particle with spin up/down along x has a 50/50 chance of later being measured as spin up/down along z, you get an expectation value for the z-spins of zero. On the other hand, if your particles started off spin up along an axis 30 degrees off the z-axis, and then you measured their spins directly along z, the expectation value for this is nonzero..

Oh I see, I think.

Thank you.

 

[1100f]

But as for my question, I understand it only to a point. I am asking why an expectation value is used in this example? Why would we want one to explain this?

In quantum mechanics, all the physical parameters are operators. when you want to measure something, the outcome of the measurements are the expectation values