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

From whatever you're saying up there, all it looks like we're supposed to do is to show that #hatT_L = e^(ihatp_xL//ℏ)#. Looks like whatever place you got this question from is confused about the definition of #hatT_L#.

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 #hatT_L = e^(-LhatD)#. If we want everything to be consistent, then if #hatT_L = e^(-LhatD)#, it would have to be that #[hatD, hatx] = bb(-1)#. I've fixed the question and addressed that already.

From part 1, we had shown that for this definition (that #hatT_L -= e^(LhatD)#),

#[hatx, hatT_L] = -LhatT_L#.

Since #f(x_0 - L)# is an eigenstate of #hatT_L#, the immediate form that comes to mind is an exponential operator #e^(LhatD)#. We intuit that #hatD = +ihatp_x//ℏ#, and we'll show that that is true.

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.](https://www3.nd.edu/~bjanko/p70007/qm1hw3answers.pdf) 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 #L# is a constant, we can factor that out of the commutator. #hatx# can go in, not being index-dependent. Therefore:

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

Now, we proposed that #hatD = ihatp_x//ℏ#, and that would make sense because we know that:

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

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

so that #[hatx, hatp_x] = iℏ#. It would mean that as long as #hatT_L = e^(LhatD)#, we can finally get a CONSISTENT definition across both parts of the problem and get:

#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 #[hatx, hatp_x]#, but not necessarily #[hatx, hatp_x^n]#. You can convince yourself that

#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 #hatp_x^(n-1) = (-iℏ)^(n-1)(d^(n-1))/(dx^(n-1))#. Thus,

#[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 #n = 0# term, you should see that it goes to zero, so we omitted it. Proceeding, we have:

#= 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] = 0#.

#[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 #hatD# always commutes with itself, #[hatD^n,hatD] = 0# and therefore,

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