How to derive Heisenberg Uncertainty Principle relation?

  • 2
    $\begingroup$ Have you tried googling the second part of the question? $\endgroup$ – Hritik Narayan May 30 '15 at 8:34
  • 1
    $\begingroup$ Also, this link will help: physics.stackexchange.com/q/24116 $\endgroup$ – Hritik Narayan May 30 '15 at 8:35
  • $\begingroup$ Hi Pushkar Soni. Welcome to Phys.SE. You should only ask one question per post, so I removed your second question (because there already appears an answer to the first question). $\endgroup$ – Qmechanic May 30 '15 at 11:04
  • 1
    $\begingroup$ Even Wikipedia gives a proof of the general Schrödinger-Robertson uncertainty principle. Do you want something else? $\endgroup$ – ACuriousMind May 30 '15 at 13:24

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...

| cite | improve this answer | |

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 $$(\Delta\hat{x})(\Delta\hat{p})\geq\frac{\hbar}{2}$$

| cite | improve this answer | |
  • $\begingroup$ I believe that you meant '\hbar' instead of 'h' in the third line up...? $\endgroup$ – DoublyNegative Jul 2 '18 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.