# 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 $\left[\hat{D} , \hat{x}\right] = \hat{1}$ To show $\left[{\hat{T}}_{L} , \hat{D}\right] = 0$ $f ' \left(x - L\right) - f ' \left(x - L\right) = 0$ 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=

Feb 24, 2018

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 ${\hat{T}}_{L}$.

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 ${\hat{T}}_{L} = {e}^{- L \hat{D}}$. If we want everything to be consistent, then if ${\hat{T}}_{L} = {e}^{- L \hat{D}}$, it would have to be that $\left[\hat{D} , \hat{x}\right] = \boldsymbol{- 1}$. I've fixed the question and addressed that already.

From part 1, we had shown that for this definition (that ${\hat{T}}_{L} \equiv {e}^{L \hat{D}}$),

$\left[\hat{x} , {\hat{T}}_{L}\right] = - L {\hat{T}}_{L}$.

Since $f \left({x}_{0} - L\right)$ is an eigenstate of ${\hat{T}}_{L}$, the immediate form that comes to mind is an exponential operator ${e}^{L \hat{D}}$. 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:

$\hat{x} \left({\hat{T}}_{L} f \left({x}_{0}\right)\right) = \left(\left[\hat{x} , {\hat{T}}_{L}\right] + {\hat{T}}_{L} \hat{x}\right) f \left({x}_{0}\right)$

$= - L {\hat{T}}_{L} f \left({x}_{0}\right) + {\hat{T}}_{L} \hat{x} f \left({x}_{0}\right)$

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 $L$ is a constant, we can factor that out of the commutator. $\hat{x}$ 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 ${\hat{T}}_{L} = {e}^{L \hat{D}}$, 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 $\left[\hat{x} , {\hat{p}}_{x}\right]$, but not necessarily $\left[\hat{x} , {\hat{p}}_{x}^{n}\right]$. You can convince yourself that

${d}^{n} / \left({\mathrm{dx}}^{n}\right) \left(x f \left(x\right)\right) = x \frac{{d}^{n} f}{{\mathrm{dx}}^{n}} + n \frac{{d}^{n - 1} f}{{\mathrm{dx}}^{n - 1}}$

and that

${\hat{p}}_{x}^{n} = {\hat{p}}_{x} {\hat{p}}_{x} {\hat{p}}_{x} \cdots$

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

so that:

$\left[\hat{x} , {\hat{p}}_{x}^{n}\right] = \hat{x} {\hat{p}}_{x}^{n} - {\hat{p}}_{x}^{n} \hat{x}$

= 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 $n \ge 1$.

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 $\left(n - 1\right)$th term becomes the $n$th term.)

As a result, we finally get:

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

$\equiv - L {e}^{L \hat{D}}$

$\equiv \textcolor{b l u e}{- L {\hat{T}}_{L}}$

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 $\left[{\hat{T}}_{L} , \hat{D}\right] = 0$.

$\left[{\hat{T}}_{L} , \hat{D}\right] = \left[{e}^{L \hat{D}} , \hat{D}\right]$

= [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 $\hat{D}$ always commutes with itself, $\left[{\hat{D}}^{n} , \hat{D}\right] = 0$ and therefore,

$\left[{\hat{T}}_{L} , \hat{D}\right] = 0$ color(blue)(sqrt"")