Unitary translation operator and taylor expansion

[FatPhysicsBoy]

Homework Statement

I have quite a straightforward question on the taylor expansion however I will try to provide as much context to the problem as possible:

$T(a)$ is unitary such that $T(-a) = T(a)^{-1} = T(a)^{\dagger}$ and operates on states in the position basis as $T(a)|x\rangle = |x+a\rangle$

It has already been shown that $\langle x | T(a) | \psi \rangle = \psi(x-a)$ using the continuous form of the closure relation.

Now it must be shown that for an infinitesimal translation $\delta a$ such that $\langle x | T(\delta a) | \psi \rangle = \psi(x-\delta a)$, via a taylor expansion of $\psi(x-\delta a)$ that the generator of infinitesimal translations acting on $\psi(x)$ takes the form:

$T(\delta a) = I - i\frac{\delta a}{\hbar}\tilde{p} + O(\delta a ^{2})$, where $\tilde{p} = -i\hbar \frac{d}{dx}$

where $I$ is the identity operator.

Homework Equations

1) $f(x) = f(x_{0}) + \frac{df(x_0)}{dx}(x-x_0) + \frac{d^{2}f(x_0)}{dx^{2}}(x-x_0)^{2} + ..$

(This is what is used in the solutions, shouldn't there be a $1/2!$ factor in the 3rd term or is it just ignored for approximation purposes..?)

The Attempt at a Solution

I am confused as to the motivations used in the solution to this problem where a parameter $\Delta x = x - x_0$ is defined and the taylor series above is recast in the following form:

2) $f(x_0 + \Delta x) = f(x_0) + \frac{df(x_0)}{dx}(\Delta x) + O((\Delta x)^{2})$,

the identification is then made that in our case $\Delta x = -\delta a$ and therefore:

3) $\psi(x - \delta a) = \psi(x) - \frac{d\psi(x)}{dx}(\delta a) + O((\delta a)^{2}) = \psi(x) - i\frac{\delta a}{\hbar}\tilde{p}\psi(x)$.

I understand the motivation to choose $\Delta x = x - x_0$ but only in the context of 'steering' the algebra towards the desired solution since we can observe that $T(a)$ increases in powers of $\delta a$. I see how this then works algebraically, $x = x_0 - \delta a$ and so we recover 2) with $\Delta x = - \delta a$.

Although it works algebraically, I'm just slightly confused by what's actually happening. My understanding of the taylor expansion only stretches as far as 'If we want to analyse $f(x)$ around $x = a$, we expand in terms of derivatives of $f(a)$' applying that to 2) doesn't make sense though 'If we want to analyse $f(x_0 - \delta a)$ around $\delta a$ we expand in terms of derivatives of $x_0$?!'

So that's the problem, I don't want to let the fact that the algebra worked out lobotomise my understanding of what's going on with the taylor expansion. I feel like I'm yet to understand something further..

Thanks in advance for any insights! :)

 

[BruceW]

Although it works algebraically, I'm just slightly confused by what's actually happening. My understanding of the taylor expansion only stretches as far as 'If we want to analyse $f(x)$ around $x = a$, we expand in terms of derivatives of $f(a)$' applying that to 2) doesn't make sense though 'If we want to analyse $f(x_0 - \delta a)$ around $\delta a$ we expand in terms of derivatives of $x_0$?!'

no, what is happening is we want to analyse $f(x_0-\delta a)$ around $x_0$, so we expand in terms of derivatives at $x_0$. And for this reason, $-\delta a$ is the perturbation. And also for this reason, the 'zeroth' term is $f(x_0)$, i.e. the value of the function at the point we are expanding around. One last thing, is that in the problem, they have written $x$ instead of $x_0$ but the only difference is the notation, they could have used any other symbol like $y$ or whatever, it doesn't matter.

 

[Delta2]

yes i think too that the confusion is mainly due to the fact that you try to expand $\psi(x-\delta\alpha)$ while for the sake of symbolism and clearer understanding it would be better to expand $\psi(x_0+\Delta x)$ and then do the replacement $\Delta x=-\delta\alpha$.

At the end you will get (3) with $x_0$ instead of $x$ but it would be more clear on how you apply taylor's theorem and then just noticing that $x_0$ can be anything, can be replace by any symbol, x,y or z.P.S Yes there should be a 2! in the denominator there but when we decide to use O-notation we can ignore all constants :)

 

[FatPhysicsBoy]

no, what is happening is we want to analyse $f(x_0-\delta a)$ around $x_0$, so we expand in terms of derivatives at $x_0$. And for this reason, $-\delta a$ is the perturbation. And also for this reason, the 'zeroth' term is $f(x_0)$, i.e. the value of the function at the point we are expanding around. One last thing, is that in the problem, they have written $x$ instead of $x_0$ but the only difference is the notation, they could have used any other symbol like $y$ or whatever, it doesn't matter.

yes i think too that the confusion is mainly due to the fact that you try to expand $\psi(x-\delta\alpha)$ while for the sake of symbolism and clearer understanding it would be better to expand $\psi(x_0+\Delta x)$ and then do the replacement $\Delta x=-\delta\alpha$.

At the end you will get (3) with $x_0$ instead of $x$ but it would be more clear on how you apply taylor's theorem and then just noticing that $x_0$ can be anything, can be replace by any symbol, x,y or z.P.S Yes there should be a 2! in the denominator there but when we decide to use O-notation we can ignore all constants :)

Thanks for the replies! :) Why are we looking at what's happening to $f(x_0-\delta a)$ around $x_0$. It's just confusing because if we had $f(x)$ and we wanted to expand about a point $x=x_0$.. I guess I'm confused because in 1) $x_0$ doesn't appear on the left hand side, but all instances of $f(x)$ on the right hand side now have $f(x_0)$. Surely we should be able to say looking at 1) what $x$ and $x_0$ are for this problem?

Okay hmmn... So $x \rightarrow x - \delta a$ then $f(x-\delta a) = f(x_0) + \frac{df(x_0)}{dx}(x-\delta a - x_0) + \frac{d^{2}f(x_0)}{dx^{2}} + ..$, where $x_0$ is the mystery point which we're analysing and what you're saying is $x_0 = x$. This actually works, I suppose my confusion is therefore simply down to 'what' I'm analysing, I thought it was $x-\delta a$ or $\delta a$ but obviously I have no real idea why.. Like I said above, I'm used to $x_0$ being some point other than $x$ (so it doesn't appear on the lhs). Like if $f(x) = sin(x)$ then perhaps we want to know what happens around $x=\pi$ so then we expand as per 1) and choose $x_0 = \pi$ and then we'll expand and (I'm sure) show that $f(x) = 0$ about $x = x_0 = \pi$.

 

[Delta2]

You know what you started confusing me too as well lol. All i can say is that you remind me of the confusion i had in secondary high school, when my teacher was saying "the derivative of $x^2$ is $2x$" and i was insinsting "No sir its $2x_0$".

Ok well, its because when i said expand "$\psi(x_0+\Delta x)$" i was actually meaning "expand $\psi(x)$ around $x_0$ and use $\Delta x=x-x_0$".

Really I just can't read this thread any more i get confused everytime i read it lol..

 

[FatPhysicsBoy]

no, what is happening is we want to analyse $f(x_0-\delta a)$ around $x_0$, so we expand in terms of derivatives at $x_0$. And for this reason, $-\delta a$ is the perturbation. And also for this reason, the 'zeroth' term is $f(x_0)$, i.e. the value of the function at the point we are expanding around..,

See, reading this again in the context of what I posted earlier just reaffirms the thing I'm confused about. If I read your first sentence I can see why we have the $x_0$'s and $\delta a$'s where they are on the rhs of 3) but then if I try to think of a 'variable transformation' which analytically gives me the same thing I can't properly resolve it.

You know what you started confusing me too as well lol. All i can say is that you remind me of the confusion i had in secondary high school, when my teacher was saying "the derivative of $x^2$ is $2x$" and i was insinsting "No sir its $2x_0$".

Ok well, its because when i said expand "$\psi(x_0+\Delta x)$" i was actually meaning "expand $\psi(x)$ around $x_0$ and use $\Delta x=x-x_0$".

Really I just can't read this thread any more i get confused everytime i read it lol..

LOL Sorry mate! Thanks for your help though :) Yeah but that's pretty much the bit I don't get, why seeing $\psi(x_0 + \Delta x)$ means you expand $\psi(x)$ around $x_0$ and use $\Delta x=x - x_0$ especially in the context of how 1) is structured.

 

[BruceW]

I guess I'm confused because in 1) $x_0$ doesn't appear on the left hand side, but all instances of $f(x)$ on the right hand side now have $f(x_0)$. Surely we should be able to say looking at 1) what $x$ and $x_0$ are for this problem?

I'm not sure where you are getting confused. We can choose $x_0$ to be whatever we want it to be. It is just the point we are expanding around. In other words:
$$f(x) = f(x_{0}) + \frac{df(x_0)}{dx}(x-x_0) + \frac{d^{2}f(x_0)}{dx^{2}} \frac{(x-x_0)^{2}}{2} + .. = f(z) + \frac{df(z)}{dx}(x-z) + \frac{d^{2}f(z)}{dx^{2}} \frac{(x-z)^{2}}{2} + ..$$
We can expand around some point $x_0$ or some other point $z$. In both cases, we get the same function $f(x)$ (as long as the function is sufficiently 'nice').

edit: to write a differential evaluated at $x_0$ I think it is better to write:
$$\left. \frac{df(x)}{dx} \right|_{x_0}$$
This is more clear, I think.

 

[FatPhysicsBoy]

I'm not sure where you are getting confused. We can choose $x_0$ to be whatever we want it to be. It is just the point we are expanding around. In other words:
$$f(x) = f(x_{0}) + \frac{df(x_0)}{dx}(x-x_0) + \frac{d^{2}f(x_0)}{dx^{2}} \frac{(x-x_0)^{2}}{2} + .. = f(z) + \frac{df(z)}{dx}(x-z) + \frac{d^{2}f(z)}{dx^{2}} \frac{(x-z)^{2}}{2} + ..$$
We can expand around some point $x_0$ or some other point $z$. In both cases, we get the same function $f(x)$ (as long as the function is sufficiently 'nice').

edit: to write a differential evaluated at $x_0$ I think it is better to write:
$$\left. \frac{df(x)}{dx} \right|_{x_0}$$
This is more clear, I think.

Hi Bruce, so if we look at the terms either side of the first equals sign I see that as the definition of 'expanding $f(x)$ around the value $x=x_0$. So using that 'definition', if we wanted to say 'what does $f(x)$ look like around $x=z$?' we'd substitute $x_0 = z$ right? Where I've taken:

4) $$f(x) = f(x_{0}) + \frac{df(x_0)}{dx}(x-x_0) + \frac{d^{2}f(x_0)}{dx^{2}} \frac{(x-x_0)^{2}}{2} + ..$$

to be my 'definition' of the taylor expansion. The bit I'm confused about is, in the case in question, what are we substituting for $x_0$? Or in other words what is $z$ equal to here? Because now we haven't got $f(x)$ with just one term in parenthesis which to me says 'change all the $x_0$'s to the point where you want to analyse'. Now we have $f(x - \delta a)$. So I'm trying to figure out how to write down the correct expansion for $f(x-\delta a)$ given the definition for the taylor expansion for just $f(x)$ (only one term in parenthesis).

Where I'm at now (thanks to your help) is, that comparing $f(x)$ and $f(x-\delta a)$ I see that $x$ goes to $x-\delta a$ so $x \rightarrow x - \delta a$. Substituting that into 4) gives:

5) $$f(x-\delta a) = f(x_{0}) + \frac{df(x_0)}{dx}((x - \delta a)-x_0) + \frac{d^{2}f(x_0)}{dx^{2}} \frac{((x-\delta a)-x_0)^{2}}{2} + ..$$,

I think this is fine and true, I don't need to change the $x$'s in any of the $dx$'s because that means something different.. Now, finally, what you're saying and I think is the source of my confusion is that whereas in the case above we said let $x_0 = z$, now we are saying let $x_0 = x$ which returns:

6) $$f(x-\delta a) = f(x) + \frac{df(x)}{dx}(- \delta a) + \frac{d^{2}f(x)}{dx^{2}} \frac{(-\delta a)^{2}}{2} + ..$$,

which is correct. It's that last step I don't understand, how should I know that we want to look around $x_0 = x$? I see $f(x-\delta a)$ and get confused thinking that we want to look around $x_0 = x-\delta a$. Hopefully this all explains my confusion a little better. The second smaller layer of confusion is that I'm not sure whether it is even correct to think of everything this way and whether my steps from 4) through to 6) work just as a special case or luck or something.

Thanks once again for all your time and patience! :)

 

[BruceW]

ah, right. I think I understand you now. Thanks for explaining so clearly! uh, yes, so we have $f(x-\delta a)$ so what point should we expand around? $x- \delta a$ or just $x$ or some other point? In truth, we can expand around any point and still get the same answer if we use the entire Taylor series. But, if we choose to expand around $x$, then the terms of our series will go with powers of $-\delta a$. Therefore, the later terms in the series will be extremely small, in the case that $\delta a$ is small. So, the most convenient point to expand around is $x$ because of the fact that $x-\delta a$ is very close to $x$, so when we use Taylor series, we don't have to write out the entire thing, we can just keep the first few terms.

 

[FatPhysicsBoy]

ah, right. I think I understand you now. Thanks for explaining so clearly! uh, yes, so we have $f(x-\delta a)$ so what point should we expand around? $x- \delta a$ or just $x$ or some other point? In truth, we can expand around any point and still get the same answer if we use the entire Taylor series. But, if we choose to expand around $x$, then the terms of our series will go with powers of $-\delta a$. Therefore, the later terms in the series will be extremely small, in the case that $\delta a$ is small. So, the most convenient point to expand around is $x$ because of the fact that $x-\delta a$ is very close to $x$, so when we use Taylor series, we don't have to write out the entire thing, we can just keep the first few terms.

Ooooooooooooooooooooooooooooooooooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh! :D Oh! :D Oh my lord okay I understand :D That is an insanely good explanation thanks so much! :D

I guess because it is a show that question too it is clear from the result that one should expand in powers of $\delta a$ which then implies the rest. I did not realize what you said but things make so much more sense now so thanks so much for your help! :)

 

[BruceW]

glad I could be of help :)