Hamiltonian Commutation | Quantum Science Philippines
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Hamiltonian Commutation

 

 

by, Jomari L. Tanghal, MS Physics I, MSU-IIT

Given the Hamiltonian operator,

\textbf{\emph{H}} = \frac{p^{2}}{2m} + V(x)

where the operators x and p satisfy the commutation relation

[\textbf{\emph{x}},\textbf{\emph{p}}] = i\hbar .

Find the following commutators:

a. [\textbf{\emph{x}}, \textbf{\emph{H}}]                           b. [\textbf{\emph{p}}, \textbf{\emph{H}}]

Solution:

Given the Hamiltonian operator,

\textbf{\emph{H}} = \frac{p^{2}}{2m} + V(x)

and

[\textbf{\emph{x}}, \textbf{\emph{p}}] = i\hbar

 a.)

\begin{array}{lll} [\textbf{\emph{x}}, \textbf{\emph{H}}]&=&\textbf{\emph{x}}\textbf{\emph{H}} - \textbf{\emph{H}}\textbf{\emph{x}}\\\\&=&x(\frac{p^{2}}{2m} + V(x)) - (\frac{p^{2}}{2m} + V(x))x\\\\&=&x\frac{p^{2}}{2m} + xV(x) - \frac{p^{2}}{2m}x - xV(x)\\\\&=&x\frac{p^{2}}{2m} - \frac{p^{2}}{2m}x\\\\&=&\frac{1}{2m}[x, p^{2}]\\\\ \end{array}

solving for  [x, p^{2}],

\begin{array}{lll} [x, p^{2}]&=&xp^{2} - p^{2}x\\\\&=&xp^{2} - pxp + pxp -p^{2}x\\\\&=&p[x,p] + [x,p]p\\\\&=&pi\hbar + pi\hbar\\\\&=&2i\hbar p\\\\ \end{array}

Thus,

\begin{array}{lll} [\textbf{\emph{x}}, \textbf{\emph{H}}]&=&\frac{1}{2m}[x, p^{2}]\\\\&=&\frac{1}{2m}(2i\hbar p)\\\\&=&\frac{i\hbar}{m}p\\\\ \end{array}

b.)

\begin{array}{lll} [\textbf{\emph{p}}, \textbf{\emph{H}}]&=&\textbf{\emph{p}}\textbf{\emph{H}} - \textbf{\emph{H}}\textbf{\emph{p}}\\\\&=&p(\frac{p^{2}}{2m} + V(x)) - (\frac{p^{2}}{2m} + V(x))p\\\\&=&\frac{p^{3}}{2m} + pV(x) - \frac{p^{3}}{2m} - V(x)p\\\\&=&pV(x) - V(x)p\\\\&=&[p, V(x)]\\\\ \end{array}

 

solving for  [p, V(x)],

 

\begin{array}{lll} [p, V(x)]f(x)&=&pV(x)f(x) - V(x)pf(x)\\\\&=&-i\hbar \frac{\partial}{\partial x}(V(x)f(x)) + V(x)i\hbar \frac{\partial}{\partial x}f(x)\\\\&=&-i\hbar \frac{\partial}{\partial x}V(x)(f(x)) - V(x)i\hbar \frac{\partial}{\partial x}f(x) + V(x)i\hbar \frac{\partial}{\partial x}f(x)\\\\&=&-i\hbar \frac{\partial}{\partial x}V(x)(f(x))\\\\ \end{array}

 

Thus,

\begin{array}{lll} [p, V(x)]&=&-i\hbar \frac{\partial}{\partial x}V(x)\\\\\end{array}

Therefore,

\begin{array}{lll} [\textbf{\emph{p}}, \textbf{\emph{H}}]&=&[p, V(x)]\\\\&=&-i\hbar\frac{\partial}{\partial x}V(x)\\\\\end{array}

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