# Dispalcement Operator

1. Nov 2, 2011

### bikashkanungo

In Dirac's text regarding displacement operator it is given that :-
lim(δx→0)⁡[D*exp⁡(iy)-1]/δx =lim(δx→0) [D-1+iy]/δx = dx + iax
Where dx = displacement operator =lim(δx→0) [D-1]/δx
ax = lim(δx→0) y/ δx
and it is assumed that y tends to zero as δx tends to zero
can anyone explain how the equality of ⁡[D*exp⁡(iy)-1] = [D-1+iy] holds good ??

2. Nov 2, 2011

### nlieb

If y tends to zero then the taylor expansion of the exponential becomes exact. I think that's what he's getting at. Although if this were the case the limit would be D iy +D-1. Did you leave out parentheses?

3. Nov 2, 2011

### bikashkanungo

@nileb : No its exactly as given in Dirac's book , I did not leave out any paranthesis

4. Nov 2, 2011

### nlieb

I'm fairly sure there should be a D in front of the iy and iax. Perhaps a misprint? Does he use the formula ever again?

5. Nov 2, 2011

6. Nov 2, 2011

### dextercioby

I think Fitzpatrick copy-pasted from Dirac...Anyway, Dirac's unrigorous treatment looks quite dubious.

7. Dec 4, 2011

### MaJac

Hi! I was reading the Dirac's text and I was very disappointed for his treatments of the infinitesimal displacement operator, so I found this post and I want to add a reply. I don't know if what I want to say is right but is my interpretation of the equality that appears in dirac's book:

lim(δx→0)⁡[D*exp⁡(iy)-I]/δx =

Obs. I is the Identity operator

lim(δx→0) exp(iy)*[D-exp(-iy)*I]/δx =

Obs. now, in such expression I can say that the exp(iy) in front of all is an arbitrary phase factor and so is completly irrelevant (the important thing is the relative phase factor between the two operators D and I that appear in the equation)

lim(δx→0) [D-exp(-iy)*I]/δx =

lim(δx→0) [D-I+iy*I]/δx =

dx + ax

The thing that I don't understand yet is the taylor expansion for the exponential, I don't understand why he assume y → 0 as δx→0

Please anyone can write me back to tell me if I'm completly wrong?? Thanks;)