Homework Help: Griffiths Quantum 4.58

1. Sep 11, 2007

NeoDevin

[SOLVED] Griffiths Quantum 4.58

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

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

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

3. 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.

2. Sep 12, 2007

NeoDevin

Anyone?

3. Sep 13, 2007

George Jones

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

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

Last edited: Sep 13, 2007
4. Sep 13, 2007

NeoDevin

Thanks, I knew I must have overlooked something simple.

5. Sep 13, 2007

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...

6. Sep 13, 2007

NeoDevin

nevermind, got it