Quantum -- Will the probability to provide a negative value of a Sz 1/2 spin system always be 0?

[ellenb899]

Homework Statement:

Will the probability to provide a negative value of a Sz 1/2 spin system always be 0? If lambda 1 = hbar/2 and lambda 2 = -h bar/2 ?

Relevant Equations:

P1(Sz = hbar/2) = |c1|^2

Will the probability to provide a negative value of a Sz 1/2 spin system always be 0? If lambda 1 = hbar/2 and lambda 2 = -h bar/2 ?

 

[DrClaude]

The question is not clear. Can you post the full statement?

Also, PhysicsForums requires you to provide an attempt at a solution.

 

[ellenb899]

Given particle in spin state: wavefunction in bra-ket notation = 3N|1> + i4N|2> (1/2 spin state in z axis)

Q. What is the probability that a measurement of Sz will provide negative value?

My attempt at solution is using the equation I provided, a negative value cannot be obtained as it must be squared. Is this correct?

 

[DrClaude]

Probabilities are always positive or zero, but it has nothing to do with the sign of what will be measured.

In other words, the question asks for the probability of measuring the spin as spin-down.

Given particle in spin state: wavefunction in bra-ket notation = 3N|1> + i4N|2> (1/2 spin state in z axis)

I don't understand what the states $\ket{1}$ and $\ket{2}$ correspond to.

I guess you will also have to figure out what the value of $N$ is.

 

[vanhees71]

For a spin 1/2 the eigenvalues of $\sigma_z$ are $\pm \hbar/2$. A general state is
$$|\psi \rangle = a |\hbar/2 \rangle+ b|-\hbar/2 \rangle, \quad |a|^2+|b|^2=1.$$
The probability to find $+\hbar/2$ when measuring $\sigma_z$ is
$$P(+\hbar/2)=|a|^2,$$
and the probability to find $-\hbar/2$ is
$$P(-\hbar/2)=|b|^2.$$
So what's the question?

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