Uncertainty relation demonstration

[andresordonez]

HI, I will not used the template provided since this is not a textbook problem. It's a problem I have with a demonstration in Quantum Mechanics Vol 1, Cohen-Tannoudji

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Complement C III

$$[Q,P] = i \hbar$$

Consider the ket:

$$|\phi \rangle = (Q +i\lambda P)|\psi \rangle$$

where $$\lambda$$ is an arbitrary real parameter. For all $$\lambda$$, the square norm $$\langle \phi | \phi \rangle$$ is positive. This is written:

$$\langle \phi | \phi \rangle = \langle \psi | (Q - i \lambda P)(Q + i \lambda P) | \psi \rangle$$

$$=\langle \psi | Q^2 | \psi \rangle + \langle \psi | (i \lambda Q P - i \lambda P Q) | \psi \rangle + \langle \psi | \lambda ^2 P^2 | \psi \rangle$$

$$=\langle Q^2 \rangle + i \lambda \langle [Q,P] \rangle + \lambda ^2 \langle P^2 \rangle$$

$$=\langle Q^2 \rangle - \lambda \hbar + \lambda ^2 \langle P^2 \rangle \geqslant 0$$

The discriminant of this expression, of second order in $$\lambda$$, is therefore negative or zero:

$$\hbar ^2 - 4 \langle P^2 \rangle \langle Q^2 \rangle \leqslant 0$$

and we have:

$$\langle P^2 \rangle \langle Q^2 \rangle \geqslant \frac{\hbar ^2}{4}$$
"

I don't understand why the discriminant must be negative or zero for $$\lambda$$ to be real. As far as I know, if you have:

$$a x^2 + b x + c = 0$$

with a, b, c reals, then:

$$x = \frac{-b \pm sqrt{b^2 - 4 a c}}{2 a}$$

and x is real if:

$$b^2 - 4 a c \geqslant 0$$

In this case:

$$a = \langle P^2 \rangle$$

$$b = -\hbar$$

$$c = \langle Q^2 \rangle$$

so the discriminant should be positive or zero instead of negative or zero right?

 

[vela]

If f(λ)=aλ²+bλ+c ≥ 0 for all λ, then its graph either lies completely above the λ axis or is tangent to the axis, so that it has, respectively, no (real) roots or just one.

 

[andresordonez]

Right, thanks!