Dilation generator in QM; Problem 3.1 in Ballentine

[GreyZephyr]

I am working my way through Ballentine's Quantum Mechanics and I am stuck on the following problem.

Homework Statement

Problem 3.1 from Ballentine: Quantum Mechanics

Space is invariant under the scale transformation $$x\to x'=e^cx$$
where $$c$$ is a parameter. The corresponding unitary operator may be written as $$e^{-icD}$$ where $$D$$ is the dilation generator. Determine the commutator $$\left[D,\mathbf{P}\right]$$ between the generators of dilation and space displacements.

The Attempt at a Solution

By looking at $$e^{i\epsilon D}e^{-i\epsilon P}e^{-i\epsilon D}e^{i\epsilon P}=I+\epsilon^2[P,D]$$ and comparing this with the transformations the LHS represents, i.e
$$\begin{align*} (x_1,x_2,x_3,t)&\to(x_1+\epsilon,x_2,x_3,t)\\ &\to(e^{-\epsilon}(x_1+\epsilon),e^{-\epsilon}x_2,e^{-\epsilon}x_3,t)\\ &\dots\\ &\to(x_1+\epsilon-\epsilon e^{-\epsilon},x_2,x_3,t)\\ &\to(x_1+\epsilon^2,x_2,x_3,t) \end{align*}$$
I get $$[D,P_\alpha]=iP_\alpha +?I$$ and I can then do the same thing with the angular momentum and use Jacobi identity to determine that the scale factor in front of $$I$$ is 0. However this seems very clumsy. To me it looks like the problem is just asking about the generators of the lie group. However I get confused as to what the group should be. Is it is still just the Galilei group? If this is the case, is there not some way to immediately get the answer from the commutation relations of the generators of the Galilei group without resorting to infinitesimal generators?

Given my confusion, I was wondering if anyone could explain, or recommend a book, that details how to find the generators in a more rigours fashion than Ballentine, that is, explains the relationship between the group structure and the physics. I like the clarity of the ideas, in his book, but am finding myself confused over some of the details of implementing them.

 

[arkajad]

Whenever you have space transformation L, it also acts on functions (in QM - state vectors) by U(L) according to the formula.

$$(U(L)f)(x)=f(L^{-1}x)$$

This is not a very good formula, because U(L) so defined will not be unitary. Yet here comes the fact that state vectors are defined up to a proportionality factor, so we can put a numerical factor in front.Thus we will have$$(U(L)f)(x)= a(L) f(L^{-1}x)$$

What numerical factor should we put for dilations?

For a dilation $x\mapsto \lambda x$ we want $U(\lambda)$ to preserve the norm. So we should have

$$|a(\lambda)|^2\int |f(\lambda^{-1}x)|^2 d^3x=\int |f(x)|^2 d^3x$$

We change the variables: $y=\lambda^{-1}x,\, x=\lambda y$ thus

$$d^3x=\lambda^3 d^3y$$

and we find that mu st have

$$|a(\lambda)|^2\lambda^3=1$$

with an evident solution

$$a(\lambda)=\lambda^{-3/2}$$

So we have our unitary operator

$$(U(\lambda)f)(x)=\lambda^{-3/2}f(\lambda^{-1}x)$$

Put here $\lambda=e^c$

and you get

$$(U(c)f)(x)=exp(-3c/2)f(e^{-c}x)$$

Now you can do Taylor expansion around $c=0$ to find the generator D. Once you have the generator, you can calculate its commutator with P. Of course you can also calculate the commutator without finding the generator by checking what will be the action of $U(c)T(a)U(c)^{-1}$ on wave functions, where $T(a)$ is the translation. That is easy. Then you expand in c and in a. The numerical factor discussed above will then cancel out - it is irrelevant for just getting the commutator.

P.S. What I wrote above needs to be verified. I was writing directly out of my head, without checking. And all $x$ above should be bold $\mathbf{x}$

 

[GreyZephyr]

Thank you for the reply. It helped a lot though, I have a further question after expanding around $c$ i obtain a generator of the form $$-\frac{1}{2} -\mathbf{x}\cdot D$$ where $$D$$ is the differential operator. This then gives the commutator $$[U,P]=\mathbf{x}.DP+P\mathbf{x}D$$.
and if we then choose to express $$P=iD$$ we obtain $$[U,p]=-iP$$, which almost agrees with the answer I get from expanding out $$UPU^{-1}$$. But what I don't understand, is where in the second method i have assumed the coordinate representation of $$P$$. Also, other than the fact that $$UTU^{-1}$$ corresponds to the change of basis for the transformation, and hence leads to a diagonalizable matrix representation, is there any particular reason to choose it for calculateing the commutator, for it seems I could also get the answer from expressions such as $$UTU^{-1}T^{-1}$$.


Finally sorry for the delay in my reply, but I did not have Internet access for a day.

 

[arkajad]

For symmetries coordinate representation is secondary, group properties are primary. Take for instance trnslations. They act by $x\mapsto x+a$. Therefore they should act on functions via

[tex](U(a)f)(x)=f(x-a)[/itex]

No numerical factor is needed in this case - this is a unitary transformation because Lebesgue measure is invariant under translations. If [itex]\hat{x}[/tex] is the position operator

$$(\hat{x}f)(x)=xf(x)$$

you can calculate from these definitions $U(a)\hat{x}U(a)^{-1}$, expand around a=0, and get the commutation between positions and momenta (generators of U(a)) without calculating the representation of the momentum operator as a differential operator.

P.S. In the previous note I assumed that we are in 3 dimensions. The value of the numerical factor for dilations depends on the dimensionality of the space. I should have made it clear.

 

[GreyZephyr]

Thank you for your help. That $$\mathbf{x}\in\mathbb{R}^3$$ was clear from the integral.