# Use the a) and b) to prove #hatT_L = e^(LhatD)# #(a)[hatT_L,hatD]=0# #(b)[hatx,hatT_L]=-LhatT_L# ?

##
Use these to prove

#[hatD,hatx]=hat1#

To show #[hatT_L,hatD]=0#

#f'(x-L)-f'(x-L) =0#

working for the first part is on https://socratic.org/questions/if-hatt-l-is-the-translation-operator-hatt-lf-x-f-x-l-and-hatx-is-hatxf-x-xf-x-t?commentevent_id=

Use these to prove

To show

working for the first part is on https://socratic.org/questions/if-hatt-l-is-the-translation-operator-hatt-lf-x-f-x-l-and-hatx-is-hatxf-x-xf-x-t?commentevent_id=

##### 1 Answer

From whatever you're saying up there, all it looks like we're supposed to do is to show that

We will end up proving that using

#hatT_L -= e^(LhatD) = e^(ihatp_xL//ℏ)#

gives

#[hatD, hatx] -= [ihatp_x//ℏ, hatx] = 1#

and **not**

From part 1, we had shown that for this definition (that

#[hatx, hatT_L] = -LhatT_L# .

Since

Recall that in the proof shown in part 1, we had written:

#hatx(hatT_L f(x_0)) = ([hatx,hatT_L] + hatT_Lhatx)f(x_0)#

#= -LhatT_Lf(x_0) + hatT_Lhatxf(x_0)#

and that is where we would have to use it. All we have to do is **Taylor expand** the exponential operator and show that the above proof still holds.

This is also shown in light detail here. I expanded it to be more thorough...

#e^(LhatD) = sum_(n=0)^(oo) (LhatD)^(n)/(n!) = sum_(n=0)^(oo) 1/(n!) L^n (hatD)^n#

Give that

#[hatx, e^(LhatD)] = sum_(n=0)^(oo) {1/(n!)(L^n) [hatx, hatD^n]}#

Now, we proposed that

#[hatx, hatp_x]f(x) = -iℏx(df)/(dx) + iℏd/(dx)(xf(x))#

#= cancel(-iℏx(df)/(dx) + iℏx(df)/(dx)) + iℏf(x)#

so that

#color(blue)([hatD", " hatx]) = [(ihatp_x)/(ℏ),hatx]#

#= -[(hatp_x)/(iℏ),hatx] = -1/(iℏ)[hatp_x,hatx]#

#= -1/(iℏ) cdot -[hatx, hatp_x]#

#= -1/(iℏ) cdot-iℏ = color(blue)(1)#

From this, we further expand the commutator:

#[hatx, e^(ihatp_xL//ℏ)] = sum_(n=0)^(oo) {1/(n!)(L^n) [hatx, ((ihatp_x)/(ℏ))^n]}#

#= sum_(n=0)^(oo) {1/(n!)((iL)/(ℏ))^n [hatx, hatp_x^n]}#

Now, we know

#d^n/(dx^n)(xf(x)) = x(d^nf)/(dx^n) + n(d^(n-1)f)/(dx^(n-1))#

and that

#hatp_x^n = hatp_xhatp_xhatp_xcdots#

#= [(-iℏ)d/(dx)]^n = (-iℏ)^n (d^n)/(dx^n)#

so that:

#[hatx, hatp_x^n] = hatxhatp_x^n - hatp_x^nhatx#

#= x cdot (-iℏ)^n (d^n f)/(dx^n) - [(-iℏ)^n d^n/(dx^n)(xf(x))]#

#= (-iℏ)^nx(d^nf)/(dx^n) - [(-iℏ)^n(x(d^nf)/(dx^n) + n (d^(n-1)f)/(dx^(n-1)))]#

#= (-iℏ)^n{cancel(x(d^nf)/(dx^n)) - cancel(x(d^nf)/(dx^n)) - n(d^(n-1)f)/(dx^(n-1))}#

#= (-iℏ)^(n-1)(-iℏ)(-n (d^(n-1)f)/(dx^(n-1)))#

#= iℏn (-iℏ)^(n-1)(d^(n-1))/(dx^(n-1))[f(x)]#

We recognize that

#[hatx, hatp_x^n] = iℏnhatp_x^(n-1)# , provided#n >= 1# .

From this, we find:

#[hatx", " e^(ihatp_xL//ℏ)] = sum_(n=0)^(oo) {1/(n!)(L^n) [hatx, ((ihatp_x)/(ℏ))^n]}#

#= sum_(n=1)^(oo) {1/(n!)((iL)/(ℏ))^n iℏnhatp_x^(n-1)}#

where if you evaluate the

#= iℏ sum_(n=1)^(oo) [n/(n!) ((iL)/(ℏ))^n hatp_x^(n-1)]#

#= iℏ sum_(n=1)^(oo) [1/((n-1)!) ((iL)/ℏ)^(n-1)((iL)/ℏ)hatp_x^(n-1)]#

Here we are simply trying to make this look like the exponential function again.

#= iℏ ((iL)/ℏ) sum_(n=1)^(oo) [((ihatp_xL)/ℏ)^(n-1)/((n-1)!)]#

(group terms)

#= -L sum_(n=1)^(oo) [((ihatp_xL)/ℏ)^(n-1)/((n-1)!)]#

(evaluate the outside)

#= -L overbrace(sum_(n=0)^(oo) [((ihatp_xL)/ℏ)^(n)/(n!)])^(e^(ihatp_xL//ℏ))#

(if#n# starts at zero, the#(n-1)# th term becomes the#n# th term.)

As a result, we finally get:

#=> color(blue)([hatx", " e^(ihatp_xL//ℏ)]) = -Le^(ihatp_xL//ℏ)#

#-= -Le^(LhatD)#

#-= color(blue)(-LhatT_L)#

**And we again get back to the original commutator**, i.e. that

#[hatx, hatT_L] = -LhatT_L color(blue)(sqrt"")#

Lastly, let's show that

#[hatT_L, hatD] = [e^(LhatD), hatD]#

#= [sum_(n=0)^(oo) ((LhatD)^n)/(n!), hatD]#

#= (sum_(n=0)^(oo) ((LhatD)^n)/(n!))hatD - hatD(sum_(n=0)^(oo) ((LhatD)^n)/(n!))#

Writing this out explicitly, we can then see it work:

#= color(blue)([hatT_L", " hatD]) = [((LhatD)^0)/(0!)hatD + ((LhatD)^1)/(1!)hatD + . . . ] - [hatD((LhatD)^0)/(0!) + hatD((LhatD)^1)/(1!) + . . . ]#

#= ((LhatD)^0)/(0!)hatD - hatD((LhatD)^0)/(0!) + ((LhatD)^1)/(1!)hatD - hatD((LhatD)^1)/(1!) + . . . #

#= [((LhatD)^0)/(0!), hatD] + [(LhatD)^(1)/(1!), hatD] + . . . #

#= L^0/(0!)[(hatD)^0, hatD] +L^1/(1!) [(hatD)^(1), hatD] + . . . #

#= color(blue)(sum_(n=0)^(oo) L^n/(n!)[(hatD)^n", " hatD])#

and since

#[hatT_L, hatD] = 0# #color(blue)(sqrt"")#