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Translation operator 
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#1
Feb2113, 02:01 PM

P: 192

[tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...=\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\frac{d^n}{dx^n}[/tex]
Why this is translational operator? ##e^{\alpha\frac{d}{dx}}f(x)=f(x+\alpha)## 


#3
Feb2113, 02:30 PM

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P: 2,685

Consider alpha to be an infinitesimal translation. Expand [itex]f(x+\alpha )[/itex] for small [itex]\alpha[/itex] to first order.
Do the same for the LHS of the equation and you should see that the equality is true for infinitesimal translations. We say that the operator [itex]\frac{d}{dx}[/itex] (technically [itex]\frac{d}{idx}[/itex]) is the 'generator' of the translation. EDIT: Beaten to the punch by TT! 


#4
Feb2113, 04:19 PM

P: 192

Translation operator
I have a problem with that. So
[tex]f(x+\alpha)=f(x)+\alpha f'(x)+...[/tex] My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##. 


#5
Feb2113, 04:25 PM

P: 8




#6
Feb2113, 04:35 PM

P: 316

In Taylor series x is fixed, while in ##\frac{df}{dx}## ##x## isn't fixed. Well you suppose that is.



#7
Feb2113, 04:36 PM

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if you prefer, write f(x_{o} + α) = f(x_{o}) + α∂f/∂x_{xo} + … 


#8
Feb2113, 05:06 PM

P: 192

[tex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}(\frac{df}{dx})_{x_0}[/tex] and how to expand now [tex]e^{\alpha\frac{d}{dx}}[/tex] 


#9
Feb2113, 05:20 PM

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no, it's [itex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}[/itex]



#10
Feb2213, 02:21 AM

P: 192

I made a mistake. But I'm asking when you get that ##(\frac{d^n}{dx^n})_{x_0}##? Please answer my question if you know. In
[tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^ 2}{dx^2}+...[/tex] you never have ##x_0##. 


#11
Feb2313, 01:55 AM

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[tex]\left(e^{\alpha\frac{d}{dx}}(f(x))\right)_{x_o}[/tex]
[tex]= \left(\left(1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+.. .\right)(f(x))\right)_{x_o}[/tex] [tex]= \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x _0}[/tex] 


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