How Do You Calculate Electron Spin Probabilities from a Wavefunction?

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To calculate the probabilities of an electron's spin state from the given wavefunction, the coefficients of the superposition must be squared. For the spin-up state, the probability is found by squaring 1/3, resulting in 1/9. For the spin-down state, the probability is obtained by squaring 2√2/3, yielding 8/9. Normalization of the wavefunction is essential to ensure the total probability equals one. This method confirms the approach to determining spin probabilities from a wavefunction.
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Simple question regarding spin :)

Homework Statement



An electron is in spin state (superposition of spin eigenstates) given by the wavefunction: \chi=\frac{1}{3}\chi_+ + \frac{2\sqrt{2}}{3}\chi_-. I need to find the probabilities that a measurement would find the electron with spin up or spin down.

Homework Equations



Uh.. you just square the coefficients, right? :)

The Attempt at a Solution



Probably a silly question but this seems too easy. I don't have this as a formula anywhere in my notes or book.. but to find the probability of spin up, you just square (1/3), and to find the probability of spin down, it's just the square of 2sqrt(2)/3... right?
 
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land said:

Homework Statement



An electron is in spin state (superposition of spin eigenstates) given by the wavefunction: \chi=\frac{1}{3}\chi_+ + \frac{2\sqrt{2}}{3}\chi_-. I need to find the probabilities that a measurement would find the electron with spin up or spin down.

Homework Equations



Uh.. you just square the coefficients, right? :)

The Attempt at a Solution



Probably a silly question but this seems too easy. I don't have this as a formula anywhere in my notes or book.. but to find the probability of spin up, you just square (1/3), and to find the probability of spin down, it's just the square of 2sqrt(2)/3... right?

If it's properly normalized, that will do it.
 
OlderDan said:
If it's properly normalized, that will do it.

Thanks! :)
 
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