Solved: Griffiths Quantum 4.58

[NeoDevin]

[SOLVED] Griffiths Quantum 4.58

From Griffiths, Intoduction to Quantum Mechanics, 2nd Ed, p198

Homework Statement

Deduce the condition for minimum uncertainty in S_x and S_y (that is, equality in the expresion \sigma_{S_x} \sigma_{S_y} \geq (\hbar/2)|\langle S_z \rangle |), for a particle of spin 1/2 in the generic state.

Answer: With no loss of generality we can pick a to be real; then the condition for minimum uncertainty is that b is either pure real or else pure imaginary.

Homework Equations

\sigma_{S_x} = \sqrt{\langle S_x^2 \rangle - \langle S_x \rangle^2}

\sigma_{S_y} = \sqrt{\langle S_y^2 \rangle - \langle S_y \rangle^2}

The Attempt at a Solution

Without loss of generality, let a be real.

We get that:

\langle S_x^2 \rangle = \frac{\hbar^2}{4}(a^2 + |b|^2)

\langle S_x \rangle = \frac{\hbar}{2}a(b + b^*) = \hbar a Re(b)

\langle S_x \rangle^2 = \hbar^2 a^2 Re(b)^2

\langle S_y^2 \rangle = \frac{\hbar^2}{4}(a^2 + |b|^2)

\langle S_y \rangle = \frac{\hbar}{2}a(b^* - b) = \hbar a Im(b)

\langle S_y \rangle^2 = \hbar^2 a^2 Im(b)^2

So:

\sigma_{S_x} = \hbar \sqrt{\frac{1}{4}(a^2 + |b|^2) - a^2 Re(b)^2}

\sigma_{S_y} = \hbar \sqrt{\frac{1}{4}(a^2 + |b|^2) - a^2 Im(b)^2}

And:

\sigma_{S_x}\sigma_{S_y} = \hbar^2 \sqrt{\frac{1}{4}(a^2 + |b|^2)(\frac{1}{4}(a^2 + |b|^2) - a^2 Re(b)^2 - a^2 Im(b)^2) + a^4 Im(b)^2 Re(b)^2 }

For the other side of the equation we have:

\frac{\hbar}{2}|\langle S_z \rangle | = \frac{\hbar^2}{4}|a^2 - |b|^2 |

I'm not sure how to show that you need either Re(b) or Im(b) to be zero in order for these two things to be equal.

 

[NeoDevin]

Anyone?

 

[George Jones]

Hint: these expressions for the expectation values are true only if

1 = |a|^2 + |b|^2 .

 

[NeoDevin]

Thanks, I knew I must have overlooked something simple.

 

[NeoDevin]

With that substitution we get:

\sigma_{S_x}\sigma_{S_y} = \hbar^2 \sqrt{\frac{1}{4}(\frac{1}{4} - a^2 Re(b)^2 - a^2 Im(b)^2) + a^4 Im(b)^2 Re(b)^2 }

and

\frac{\hbar}{2}|\langle S_z \rangle | = \frac{\hbar^2}{4}\left| 1-2|b|^2\right|

Substituting in that either Re(b) or Im(b) = 0 and setting the two equal gives

\left| 1-2|b|^2\right| = \sqrt{1 - 4a^2 |b|^2}

which doesn't look equal to me...

 

[NeoDevin]

nevermind, got it