Uncertainty Principle

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Two operators $A$ and $B$ are said to be compatible with each other if

  • their commutator $[A,B]=0$ or
  • they have a common basis set

In the article on Heisenberg's Microscope we derived an intuition for what it means that two observables commute and found that the poisson brackets are the classical analog of the commutator. In particular it were the canonically conjugate variables that don't commute with each other.

In practice we could find out, if two operators are compatible by measuring first $A$, this will give the eigenvalue say $a$, then we measure $B$ with say eigenvalue $b$ immediately afterwards and then again $A$. If we get the same eigenvalue $a$, the operators are compatible. If they are incompatible, they are subject to

Heisenbergs Uncertainty Principle

\begin{align} (\Delta A)(\Delta B) \geq \frac{1}{2} | \langle [A,B] \rangle | \end{align}

The uncertainty $\Delta A$ in the operator $A$ is given by the standard deviation \begin{align} \Delta A = \langle (A- \langle A \rangle )^2 \rangle = \langle A^2 -2A\langle A \rangle +\langle A \rangle^2 \rangle=\langle A^2 \rangle- 2\langle A \rangle^2+\langle A \rangle^2=\langle A^2 \rangle - \langle A \rangle^2 \end{align} where \begin{align} \langle A^2 \rangle = \langle \psi | A^2 |\psi \rangle\\ \langle A \rangle^2 = (\langle \psi | A |\psi \rangle)^2 \end{align}

We can show, that Heisebergs Uncertainty Principle is in fact true for arbitray operators $A$ and $B$. To this end we define \begin{align} A' = A- \langle A \rangle \\ B' = B- \langle B \rangle \\ \end{align} and make our argument by computing \begin{align} | \langle [A,B] \rangle |=| \langle [A',B'] \rangle |=|\langle \psi | A'B' |\psi \rangle-\langle \psi | B'A' |\psi \rangle| \end{align} but $\langle \psi | A'B' |\psi \rangle$ is just the complex conjugate of $\langle \psi | B'A' |\psi \rangle$ so it's two $i$ times the imaginary part of $\langle \psi | A'B' |\psi \rangle$ \begin{align} | \langle [A,B] \rangle |=|2 i Im (\langle \psi | A'B' |\psi \rangle)|=2 Im (\langle \psi | A'B' |\psi \rangle) \end{align} This is indeed always smaller than the modulus of both the real and imaginary part \begin{align} | \langle [A,B] \rangle |=2 Im (\langle \psi | A'B' |\psi \rangle) \leq |\langle \psi | A'B' |\psi \rangle| \end{align} If we now use the Cauchy Inequality $|\langle \phi|\psi \rangle \leq \|\chi \| \| \phi \|$ we can write \begin{align} | \langle [A,B] \rangle | \leq |\langle \psi | A'B' |\psi \rangle|\leq \|A' \psi \| \| B' \psi \| = \langle \psi | A'^{\dagger}A' |\psi \rangle\langle \psi | B'^{\dagger}B' |\psi \rangle=(\Delta A)(\Delta B) \hspace{3cm} \square \end{align}


Compute the uncertainty relation for the operators $x$ and $p$

\begin{align} (\Delta x)(\Delta p) \geq \frac{1}{2} | \langle [x,p] \rangle | \end{align} the mean of the commutator is for normalised state vectors $|\psi \rangle$ \begin{align} | \langle \psi |i \hbar \mathbb{I} |\psi \rangle |= \hbar \langle \psi |\mathbb{I} |\psi \rangle = \hbar \end{align} so we find the uncertainty relation for position and momentum \begin{align} (\Delta x)(\Delta p) \geq \frac{\hbar}{2} \end{align}