What is the generator for scaling transformation in one dimension? consider the hamiltonian of 1D harmonic oscillator
$H = Px^2/2m + 1/2 kx^2$
and let $H' = Px'^2/2m + 1/2 kx'^2$
such that $x' = Ax$
and $Px' = (1/A)Px$
then the wavefunction of the two hamiltonians ( $H$ and $H'$) are related by a term . What will be the generator of this scaling transformation?
What will be the minimum energy eigenvalue of the total wavefunction($\psi A$) ie., the coherent state?
|$\psi n'\rangle = \sum  \lambda n |\psi n\rangle$
How to find the coefficient  $\lambda n$?
NOTE: The virial operator $G$ can be expressed in terms of the ladder operators as
$G = (xp+px)/2 = i\hbar{(a^\dagger)^2 - (a)^2)}/2$
and the generator of transformation becomes $\exp\{ \ln A ((a^\dagger)^2 - (a)^2) \}$
How to find the action of this generator on the ground state wave function ψn
ie., $\exp\{ \ln A ((a^\dagger)^2 - (a)^2) \} |0\rangle $?
 A: The unitary operator of scaling transformations is
$$
D_A=\exp\left\{\frac{i\ln A}\hbar G\right\},
$$
where
$$
G=\frac{xp+px}2
$$
is the virial operator, which is actually the generator of these transformations.
We can find the transformed operators
$$
x_A\equiv D_AxD^+_A,\qquad p_A\equiv D_ApD^+_A
$$
by differentiation:
$$
\frac{\partial x_A}{\partial A}=\frac\partial{\partial A}D_AxD^+_A=-D_A\frac{i[x,G]}{\hbar A}D^+_A=\frac{x_A}A,
$$
$$
\frac{\partial p_A}{\partial A}=\frac\partial{\partial A}D_ApD^+_A=-D_A\frac{i[p,G]}{\hbar A}D^+_A=-\frac{p_A}A,
$$
where simple commutation relations $[x,G]=i\hbar x$, $[p,G]=-i\hbar p$ were taken into account.
      
With the initial conditions $x_{A=1}=x$, $p_{A=1}=p$ we obtain
$$
x_A=Ax,\qquad p_A=\frac{p}A.
$$
It is also interesting to look at the transformation law for a wave function:
$$
\psi_A\equiv D_A\psi,\qquad \psi_A(x)=\sqrt{A}\psi(Ax),
$$
which can also be proved by differentiation.
From the physical point of view, the operator $D_A$ squeezes the $x$ axis $A$ times, thus the wave function also becomes squeezed, while its amplitude grows as $\sqrt{A}$ to preserve normalization.
In a $d$-dimensional case the formulas are the same except
$$
G=\frac{\mathbf{rp}+\mathbf{pr}}2=-i\hbar\left(\mathbf{r}\frac\partial{\partial\mathbf{r}}+\frac{d}2\right),\qquad \psi_A(\mathbf{r})=A^{d/2}\psi(A\mathbf{r}).
$$
A: Just as a supplement to the accepted answer, I would like to show that
$$\hat{D}_{A}\hat{x}\hat{D}_{A}^{\dagger}=A \,\hat{x} \qquad\qquad\qquad (1)$$
by direct computation, instead of by solving a differential equation.
As in the accepted answer, $\hat{D}_{A}=e^{\frac{i}{\hbar}a\,\hat{G}}$, where $a=\ln A$ and $\hat{G}=\frac{1}{2}\left(\hat{x}\hat{p}+\hat{p}\hat{x}\right)$.
Note that $\hat{D}_{A}$ is a unitary operator, since $\hat{G}$ is self-adjoint. Therefore, $\hat{D}_{A}^{\dagger}=\hat{D}_{A}^{-1}$, and so $\hat{D}_{A}\hat{D}_{A}^{\dagger}=\hat{I}$, where $\hat{I}$ is the identity operator.
The claim (1) then follows immediately from the following fact, which we will prove in moment:
$$\hat{x}\,\hat{D}_{A}^{\dagger}=e^{a}\,\hat{D}_{A}^{\dagger}\,\hat{x}\,.\qquad\qquad\qquad (2)$$
Once Eq. (2) is established, we can write
$$\hat{D}_{A}\hat{x}\hat{D}_{A}^{\dagger}= \hat{D}_{A}e^{a}\,\hat{D}_{A}^{\dagger}\,\hat{x}=e^{a}\hat{D}_{A}\,\hat{D}_{A}^{\dagger}\,\hat{x}=e^{a}\,\hat{x}=A\,\hat{x}\,.$$
Proof of Equation (2)
We write $\hat{D}_{A}^{\dagger} = e^{-\frac{i}{\hbar}a\,\hat{G}}=\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{i}{\hbar}a\,\hat{G}\right)^{n}=\sum_{n=0}^{\infty}\frac{1}{n!}(-\frac{i}{\hbar}a)^{n}\,\hat{G}^{n}$. Then $\hat{x}\,\hat{D}_{A}^{\dagger}=\sum_{n=0}^{\infty}\frac{1}{n!}(-\frac{i}{\hbar}a)^{n}\,\hat{x}\,\hat{G}^{n}$. Our hope is that $\hat{x}\,\hat{G}^{n}$ is something simple. Well, let's try it out for the first few $n$.
The case $n=0$ is trivial, since $\hat{G}^{0}=\hat{I}$.
For $n=1$, we just use the commutator: since $[\hat{x},\,\hat{G}]= i\hbar\,\hat{x}$, it follows that $\hat{x}\,\hat{G}=\hat{G}\,\hat{x}+ i\hbar\,\hat{x}=(\hat{G}+i\hbar)\,\hat{x}$.
(As usual, when a pure number appears in an operator-valued expression, it is just a shorthand notation for the product of that number and the identity operator. For example, $\hat{G}+i\hbar$ really means $\hat{G}+i\hbar\hat{I}$.)
For $n=2$, we have $\hat{x}\,\hat{G}^{2}=(\hat{x}\,\hat{G})\,\hat{G}$, and use the result for $n=1$ to write $(\hat{x}\,\hat{G})\,\hat{G}=(\hat{G}+i\hbar)\,\hat{x}\,\hat{G}$. And now we use the result from $n=1$ again: $(\hat{G}+i\hbar)\,\hat{x}\,\hat{G}=(\hat{G}+i\hbar)\,(\hat{G}+i\hbar)\,\hat{x}=(\hat{G}+i\hbar)^{2}\,\hat{x}$.
The obvious conjecture is that, for all positive integers $n$,
$$\hat{x}\,\hat{G}^{n}=(\hat{G}+i\hbar)^{n}\,\hat{x}\,.\qquad\qquad\qquad(3)$$ The proof is by induction. We have already verified the $n=1$ case. Now let's assume the conjecture is true for $n$, and show that this implies it is true for $n+1$: $$\hat{x}\,\hat{G}^{n+1}=\left(\hat{x}\,\hat{G}^{n}\right)\,\hat{G}=(\hat{G}+i\hbar)^{n}\,\hat{x}\,\hat{G}=(\hat{G}+i\hbar)^{n}\,(\hat{G}+i\hbar)\,\hat{x}=(\hat{G}+i\hbar)^{n+1}\,\hat{x}\,,$$
which proves the conjecture (3).
That was actually the hard part. The rest is easy.
$$\hat{x}\,\hat{D}_{A}^{\dagger}=\sum_{n=0}^{\infty}\frac{1}{n!}(-\frac{i}{\hbar}a)^{n}\,\hat{x}\,\hat{G}^{n}=\sum_{n=0}^{\infty}\frac{1}{n!}(-\frac{i}{\hbar}a)^{n}\,(\hat{G}+i\hbar)^{n}\,\hat{x}=\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{i}{\hbar}a\,(\hat{G}+i\hbar)\right)^{n}\,\hat{x}\,.$$
Of course, $\hat{x}$ can go out of the sum, giving
$$\left[\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{i}{\hbar}a\,(\hat{G}+i\hbar)\right)^{n}\right]\,\hat{x}=e^{-\frac{i}{\hbar}a\,(\hat{G}+i\hbar)}\,\hat{x}\,.$$
Now note that $-\frac{i}{\hbar}a\,(\hat{G}+i\hbar)=-\frac{i}{\hbar}a\,\hat{G}+a$.
Again, the last term is really the operator $a\hat{I}$. Of course, the identity operator commutes with all operators, and so $a\hat{I}$ commutes with $-\frac{i}{\hbar}a\,\hat{G}$. Now we use the fact that if two operators commute, then the exponentiation of their sum follows the same rule as if they were just ordinary numbers:
$\qquad\qquad$ For all $\hat{A}$ and $\hat{B}$, if  $[\hat{A},\,\hat{B}]=0$,  then  $e^{\hat{A}+\hat{B}}=e^{\hat{A}}e^{\hat{B}}$.
(This can be proven by expanding the exponentials in Taylor series. Alternatively, it is a corollary of the Baker–Campbell–Hausdorff formula.)
Therefore, $e^{-\frac{i}{\hbar}a\,\hat{G}+a}=e^{a}\,e^{-\frac{i}{\hbar}a\,\hat{G}}=e^{a}\,\hat{D}_{A}^{\dagger}$. Putting it all together,
$$\hat{x}\,\hat{D}_{A}^{\dagger}=e^{-\frac{i}{\hbar}a\,\hat{G}+a}\,\hat{x} =e^{a}\,\hat{D}_{A}^{\dagger} \,\hat{x}\,,$$
which proves claim (2). □
The statement that $\hat{D}_{A}\hat{p}\hat{D}_{A}^{\dagger}=(1/A) \,\hat{p}$ can be proven similarly.
