Commutation Relations

The ordering of Hermitian Operators is important, and therefore they are not commutative. The commutator of two operators is defined as $[\hat{A},\hat{b}]=\hat{A}\hat{B}-\hat{B}\hat{A}$.

For example, (class example 1) A free particle with $H=\frac{p^2}{2m}$, determine $[\hat{p},\hat{H}]$:

$$\hat{p}=\frac{\hslash }{i}\frac{d}{dx}$$ $$\hat{H}=-\frac{\hslash }{2m}\frac{{\partial }^2}{\partial x^2}$$ $$[\hat{p},\hat{H}]=\frac{1}{2m}[\hat{p},{\hat{p}}^2]=0$$

A simple harmonic oscillator with $H=\dots$

$$[\hat{p},\hat{H}]=\left[\hat{p},\frac{{\hat{p}}^2}{2m}+\frac{1}{2}{k}_0{\hat{x}}_2\right]=\frac{1}{2}{k}_0[\hat{p},{\hat{x}}^2]$$ $$[\hat{p},\hat{H}]=-i\hslash k_{0}x$$

Why are commutation relations important?

• We need hermitian operators to have the same eigenfunctions.
• Commuting operators have the same eigenfunctions
• Non commuting operations have uncertainty.

In class assignment 2

Given $\frac{d<A>}{dt}=\frac{i}{\hslash }\int dx[(\bar{H}\Psi {)}^{\ast }\hat{A}\psi -{\Psi }^{\ast }\hat{A}\hat{H}\Psi ]$ show that $\frac{d<A>}{dt}=\frac{i}{\hslash }\int {\Psi }^{\ast }[\hat{H},\hat{A}]\Psi dx\equiv \frac{i}{\hslash }<[\hat{H},\hat{A}]>dx$ (use hermitian property of $\hat{H}$)

$$H^{†}=H$$ $$\frac{d<A>}{dt}=\frac{i}{\hslash }\int dx[(\bar{H}\Psi {)}^{\ast }\hat{A}\psi -{\Psi }^{\ast }\hat{A}\hat{H}\Psi ]$$ $$\frac{i}{\hslash }\int (\hat{H}\Psi {)}^{\ast }(\hat{A}\Psi )-{\Psi }^{\ast }\hat{A}\hat{H}\Psi dx=\frac{i}{\hslash }\int [{\Psi }^{\ast }{H}^{†}\hat{A}\Psi -{\Psi }^{\ast }\hat{A}\hat{H}\Psi ]dx$$ $$=\frac{i}{\hslash }\int [{\Psi }^{\ast }\hat{H}(\hat{A}\Psi )-{\Psi }^{\ast }\hat{A}\hat{H}\Psi ]dx$$ $$=\frac{i}{\hslash }\int {\Psi }^{\ast }[\hat{H},\hat{A}]\Psi dx=\frac{i}{\hslash }<[\hat{H},\hat{A}]>$$

In class assignment 3 For Hamiltonian $H=\frac{p^2}{2m}+V_x$ show that $\frac{d<p>}{dt}=<-\frac{\partial V}{\partial x}>$

$$\frac{d<A>}{dt}=\frac{i}{\hslash }<[\hat{H},\hat{A}]>$$ $$\frac{d<\hat{p}>}{dt}=\frac{i}{\hslash }<[\hat{H},\hat{p}]>$$ $$=\frac{i}{\hslash }<\frac{{\hat{p}}^2}{2m}+V(x),\hat{p}>=\frac{i}{\hslash }<[V(x),\hat{p}]>$$ $$[V(x),\hat{p}]\phi (x)=\frac{\hslash }{i}\left[V(x)\frac{\partial \phi }{\partial x}-\frac{\partial }{\partial x}V(x)\phi (x)\right]$$ $$=\frac{\hslash }{i}\left[\frac{V(x)\partial \phi }{\partial x}-\phi (x)\frac{\partial V(x)}{\partial x}-V(x)\frac{\partial \phi }{\partial x}\right]=-\frac{\hslash }{i}\frac{\partial V(x)}{\partial x}\phi (x)$$ $$\frac{d<p>}{dt}=\frac{i}{\hslash }<-\frac{i}{\hslash }<-\frac{\hslash }{i}\frac{\partial V(x)}{\partial x}>=<-\frac{\partial V(x)}{\partial x}>$$ $$\frac{dp}{dt}=-\frac{\partial V(x)}{\partial x}=F$$

In class assignment 4 For Hamiltonian $H=\frac{p^2}{2m}+V_x$ show that $\frac{d<x>}{dt}=\frac{<p>}{m}$

$$\frac{d<x>}{dt}=\frac{i}{\hslash }<[\hat{H},\hat{x}]\ge \frac{d<x>}{dt}$$ $$=\frac{i}{\hslash }<\left[\frac{p^2}{2m}+V(x),\hat{x}\right]>=\frac{i}{2m\hslash }<[{\hat{p}}^2,\hat{x}]>=-\frac{i}{2mh}<[\hat{x},{\hat{p}}^2]>$$ $$[\hat{x},{\hat{p}}^2]\phi =\left[x{\left(\frac{\hslash }{i}\frac{\partial }{\partial x}\right)}^2-\left(\frac{\hslash }{i}\frac{\partial }{\partial x}\right)x^2\right]\phi =-{\hslash }^2\left[x\frac{{\partial }^2}{\partial x^2}\phi -\frac{{\partial }^2}{\partial x^2}(x\phi )\right]$$ $$=-{\hslash }^2\left[x\frac{{\partial }^2}{\partial x^2}\phi -x\frac{{\partial }^2}{\partial x^2}\phi -\frac{\partial }{\partial x}\phi -\frac{\partial }{\partial x}\phi \right]=2{\hslash }^2\frac{\partial }{\partial x}\phi =2i\hslash \hat{p}\phi$$ $$\frac{d<x>}{dt}=-\frac{i}{2mh}<2i\hslash \hat{p}>=\frac{<\hat{p}>}{m}$$