Can anyone provie me the proof of $Dq-qD=1$ where $D=\frac{\partial }{\partial q}$ refers to the differential operator?
Or if it's something special to quantum mechanics, why is it?
Is this following from $[\hat{q},\hat{p}] =i\hbar ~{\bf 1}$?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ i think you mean to say D is the derivative operator, not any differential operator. $\endgroup$– PrathyushCommented Oct 18, 2012 at 10:51
-
$\begingroup$ what's the difference between two? $\endgroup$– RRRRCommented Oct 18, 2012 at 10:53
-
2$\begingroup$ a differential operator can be anything like (d/dx)^2, but derivative operator is d/dx $\endgroup$– PrathyushCommented Oct 18, 2012 at 11:15
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
a hint
$$ Dxf(x)= f(x)+xDf(x) $$
$$ xDf(x) $$
take the diference and you get $ f(x) $ or $ 1.f(x)$
answered Oct 18, 2012 at 10:58
$\begingroup$
$\endgroup$
3
$$[D,q]f(q)=Dqf(q)-qDf(q) =\frac{d}{dq}(qf)-\frac{qd}{dq}$$
$$\Rightarrow [D,q]f(q)=q \frac{df}{dq}+f \frac{dq}{dq}-q \frac{df}{dq}$$
$$\Rightarrow [D,q]f(q)=\frac{dq}{dq}=1$$
$$\Rightarrow [D,q]f(q)=f$$
Now,remove $f$ from both side,
$$[D,q]=1$$
-
1$\begingroup$ @Manikanta If you're going to LaTeXify, please do it thoroughly. If you want, use this tool to help you convert plain math to TeX. $\endgroup$ Commented Nov 26, 2012 at 15:43
-
1$\begingroup$ Generally, we only provide hints to homework questions and not full answers--you may want to edit. (See homework policy) $\endgroup$ Commented Nov 26, 2012 at 15:46
-
1$\begingroup$ I would ordinarily temporarily delete this, but it's an old question anyway so I'll just leave it here. ali, just keep the homework policy in mind when you post answers in the future. And welcome to Physics Stack Exchange! $\endgroup$– David ZCommented Nov 26, 2012 at 18:16