The probability that spin will be in -x direction

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Homework Help Overview

The problem involves a spin 1/2 particle described by a quantum state and requires determining the probability of measuring the spin in the -x direction. The context is rooted in quantum mechanics and the mathematical representation of spin states.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of probability using the projection of the state vector onto the -x spin state. Questions arise regarding the treatment of complex numbers in the probability calculation and whether to find magnitudes or use complex conjugates.

Discussion Status

Some participants have offered guidance on the correct method for calculating the probability, specifically mentioning the use of complex conjugates. There is an exploration of different methods to arrive at the probability, with some participants noting consistency in results across methods.

Contextual Notes

Participants express confusion regarding the appearance of complex numbers in the probability calculation and question the validity of their results. The discussion reflects an ongoing examination of the mathematical principles underlying quantum mechanics.

czaroffishies
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Homework Statement



A spin 1/2 particle is in the state \left| \Psi \right\rangle = \sqrt{2/3}\left|\uparrow\right\rangle + i\sqrt{1/3}\left|\downarrow\right\rangle

A measurement is made of the x-component of the spin. What is the probability that the spin will be in the -x direction?

Homework Equations



Spin states are represented as linear combinations of the spin in the +z direction and -z direction, since these spins form an orthonormal basis set:
http://en.wikipedia.org/wiki/Spin-½#Mathematical_description

In this case, spin in -x direction is represented by

\sqrt{1/2} \[<br /> \left( {\begin{array}{cc}<br /> 1 \\<br /> -1 \\<br /> \end{array} } \right)<br /> \]

The Attempt at a Solution



P = \left|\left\langle\leftarrow\left|\Psi\right\rangle\left|^{2}
= the transpose of the \leftarrow matrix, times the \Psi matrix, squared.

When calculating this straightforwardly, I will end up with a complex probability because of the i term in the \Psi matrix. That doesn't make sense!

So, do I just take this complex number and find its magnitude in the complex plane, and then square that?

Or something else?

Thanks!
 
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czaroffishies said:
P = \left|\left\langle\leftarrow\left|\Psi\right\rangle\left|^{2}
= the transpose of the \leftarrow matrix, times the \Psi matrix, squared.

When calculating this straightforwardly, I will end up with a complex probability because of the i term in the \Psi matrix. That doesn't make sense!
The method is correct.
So, do I just take this complex number and find its magnitude in the complex plane, and then square that?

Or something else?

Thanks!
To find the probability you multiply the projection by its complex conjugate. Did you do that or did you just square it?
 
That is a fantastic idea!

P = \left(\sqrt{1/2}*\sqrt{2/3}-i\sqrt{1/2}*\sqrt{1/3}\right)*\left(\sqrt{1/2}*\sqrt{2/3}+i\sqrt{1/2}*\sqrt{1/3}\right)

= 1/2

Is this correct?

Actually, this is the same answer I got when I used the magnitude-in-the-complex-plane method. Hrm. Coincidence, or no... I will look into it.
 
It is correct.
 

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