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How to derive Heisenberg Uncertainty Principle relation?

Qmechanic
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Pushkar Soni
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2 Answers2

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Let's see...that's a good revision exercise:

1) To begin with, one should define the quantities involved, namely the variance or its square root the standard deviation

(As I see it, the framework is that the expectation value is the evaluation of a state $|\psi\rangle\in\mathcal{H}$ on an observable $A\in\mathcal{B}(\mathcal{H})$ self-adjoint, i.e. $ \langle A\rangle := \langle \psi|A|\psi\rangle$ (or more generally $\omega(A)$). It can really be interpreted as an expectation value in the sense of probability theory, i.e. as the integral of a random variable with some probability density over the spectrum of A which is interpreted as the possible outcomes.)

The variance of an observable $A\in\mathcal{B}(\mathcal{H})$ is defined by $$ \Delta A^2 := \left\langle (A - \langle A\rangle)^2\right\rangle = \langle A^2\rangle - \langle A\rangle^2$$ where the dependence in the state $|\psi\rangle$ is not explicitly expressed.

Let $A,B\in\mathcal{B}(\mathcal{H})$ be two observables, the Heisenberg uncertainty relation can be stated $$ \Delta A \cdot \Delta B \geq \frac{1}{2} \left|\langle [A,B]\rangle\right| $$

(Copy from some lecture notes... no references sorry) One possibility to proove it is to consider the following quantity $\left|\langle A'B'\rangle\right|^2$ (and find an inequality that holds for any two observables). One has $$A'B'=\frac{1}{2} \{A',B'\} + \frac{1}{2} [A',B']$$

  • The anticommutator is self-adjoint, so by a general property of states the expectation value $\langle \{A',B'\}\rangle \in\mathbb{R}$ (is real) and the commutator is "anti"-self-adjoint, i.e. $[A',B']^{\dagger}= -[A',B']$ so that the expectation value $\langle [A',B']\rangle \in i\mathbb{R}$ (is purely imaginary). Thus the square module of the expectation value reads $$\left|\langle A'B'\rangle\right|^2 =\frac{1}{4} \langle \{A',B'\}\rangle^2 + \frac{1}{4} \left|\langle[A',B']\rangle\right|^2 $$
  • The second essential property that one needs is the Cauchy-Schwarz inequality for states (on a C*-algebra) $$ \left|\langle A'B'\rangle\right|^2 \leq \langle A'^2\rangle \langle B'^2\rangle$$

Combining both (leaving out the anticommutator $\frac{1}{4} \langle \{A',B'\}\rangle^2$) yields $$\frac{1}{4} \left|\langle[A',B']\rangle\right|^2 \leq \langle A'^2\rangle \langle B'^2\rangle$$

  • the last ingredient is to take $A':= A-\langle A\rangle \Bbb{1},\ B':= B-\langle B\rangle 1$. One checks that $$ \Delta A^2 = \langle A'^2\rangle ,\quad \Delta B^2 = \langle B'^2\rangle \quad \text{and}\quad \langle[A,B]\rangle = \langle[A',B']\rangle$$

then take the square root.

2) I leave it to more knowledgeable people...

Noix07
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Robertson Uncertainty Relation: $$(\Delta\hat{A})^2(\Delta\hat{B})^2\geq\left(\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right)^2$$ We have $[\hat{x},\hat{p}]=\hat{x}\hat{p}-\hat{p}\hat{x}$ and $\hat{p}=-i\hbar\frac{\partial}{\partial x}$

so $\hat{x}\hat{p}g(x)=x(-i\hbar)\frac{\partial g}{\partial x}$

and $\hat{p}\hat{x}=-i\hbar\frac{\partial}{\partial x}(\hat{x}g(x))=-i\hbar g(x)-i\hbar x\frac{\partial g}{\partial x}$

This gives us $[\hat{x},\hat{p}]g(x)=\hat{x}\hat{p}g(x)-\hat{p}\hat{x}g(x)=i\hbar g(x)$

so $[\hat{x},\hat{p}]=i\hbar$

$$(\Delta\hat{x})^2(\Delta\hat{p})^2\geq\left(\frac{1}{2i}\langle\Psi|\left[\hat{x},\hat{p}\right]\Psi\rangle\right)^2$$ $$=\left(\frac{1}{2i}ih\langle\Psi|\Psi\rangle\right)^2$$ $$=\frac{\hbar^2}{4}$$ Therefore $$(\Delta\hat{x})(\Delta\hat{p})\geq\frac{\hbar}{2}$$

user142511
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