Question

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=

Answer 1

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"")`