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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 $[\hat{q},\hat{p}] =i\hbar ~{\bf 1}$?

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i think you mean to say D is the derivative operator, not any differential operator. –  Prathyush Oct 18 '12 at 10:51
    
what's the difference between two? –  RRRR Oct 18 '12 at 10:53
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a differential operator can be anything like (d/dx)^2, but derivative operator is d/dx –  Prathyush Oct 18 '12 at 11:15
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2 Answers 2

up vote 5 down vote accepted

a hint

$$ Dxf(x)= f(x)+xDf(x) $$

$$ xDf(x) $$

take the diference and you get $ f(x) $ or $ 1.f(x)$

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$$[D,q]f(q)=Dqf(q)-qDf(q) =\frac{d}{dq}(qf)-\frac{qd}{dq}$$

$$\Rightarrow [D,q]f(q)=q \frac{df}{dq}+f \frac{dq}{dq}-q \frac{df}{dq}$$

$$\Rightarrow [D,q]f(q)=\frac{dq}{dq}=1$$

$$\Rightarrow [D,q]f(q)=f$$

Now,remove $f$ from both side,

$$[D,q]=1$$

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@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. –  Manishearth Nov 26 '12 at 15:43
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Generally, we only provide hints to homework questions and not full answers--you may want to edit. (See homework policy) –  Manishearth Nov 26 '12 at 15:46
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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! –  David Z Nov 26 '12 at 18:16
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