# 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