# Understanding proof for Heisenberg uncertainty

1. Jan 17, 2016

I've uploaded a proof of the Heisenberg uncertainty principle from Konishi's QM. I just don't quite understand one part: what is the significance of the discriminant being less than or equal to 0? Wouldn't this just result in $\alpha = R \pm iZ$? Why would this be desired in this proof?

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2. Jan 17, 2016

### Samy_A

He finds an inequality that must hold for all real $\alpha$.
Simplified, it says: ${\alpha}^{2}+b\alpha +c \geq 0$
The discriminant of the equation ${\alpha}^{2}+b\alpha +c = 0$ is $D=b²-4c$.
If $D>0$, the equation will have two different real roots, $r_1$ and $r_2$, so you get ${\alpha}^{2}+b\alpha +c=(\alpha-r_1)(\alpha-r_2) \geq 0$ (still for all real $\alpha$).
But this can't be true for values of $\alpha$ lying between the two roots. Therefore $D \leq 0$.

Last edited: Jan 17, 2016
3. Jan 17, 2016

### Staff: Mentor

If I made no mistake just set $α = ħ / (2<(P-P_0)^2>)$.

4. Jan 17, 2016

### stevendaryl

Staff Emeritus
I'm not sure about the discriminate, but the equation he derives, true for any $\alpha$, is:

$A - B \alpha + C \alpha^2 \geq 0$

where $A = \langle (Q - Q_0)^2 \rangle$, $B = \hbar$, and $C = \langle (P- P_0)^2 \rangle$

So if it's true for every $\alpha$, then in particular, it's true when $\alpha = \frac{B}{2C}$. Plugging this into the inequality gives:
$A - \frac{B^2}{2C} + \frac{B^2}{4C} \geq 0$

Which implies $AC - \frac{B^2}{4} \geq 0$, or $\sqrt{A}\sqrt{C} \geq \frac{B}{2}$

Going back to the definitions of $A$, $B$ and $C$ gives us the uncertainty principle:

$\Delta Q \Delta P \geq \frac{\hbar}{2}$

where $\Delta Q = \sqrt{\langle (Q - Q_0)^2 \rangle}$ and $\Delta P = \sqrt{\langle (P - P_0)^2 \rangle}$

5. Jan 17, 2016

Okay, that makes sense since if it's between the two roots then the inequality is not satisfied. But if the discriminant is less than 0, then isn't $\alpha$ now complex and thus not real? Therefore the initial assumption that $\alpha$ is real for the inequality doesn't stand?

6. Jan 17, 2016

### Samy_A

The inequality holds for all real $\alpha$. That leads to the condition that the discriminant must be 0 or less. He doesn't care about non-real roots of the equation. Sure they will exist if the discriminant is less than 0, but the inequality has been derived for real $\alpha$.

Last edited: Jan 17, 2016
7. Jan 17, 2016

### stevendaryl

Staff Emeritus
No, in the original derivation, $\alpha$ is just declared to be an arbitrary real number. It's not the solution to the equation $\alpha^2 + b \alpha + c = 0$ (which would be complex if $b^2 - 4c < 0$).