A Finding the Hermitian generator of a Symplectic transformation

[andresB]

Consider a set of $n$ position operators and $n$ momentum operator such that
$$\left[q_{i},p_{j}\right]=i\delta_{ij}.$$
Lets now perform a linear symplectic transformation
$$q'_{i} =A_{ij}q_{j}+B_{ij}p_{j},$$
$$p'_{i} =C_{ij}q_{j}+D_{ij}p_{j}.$$
such that the canonical commutation relations are maintained

$$\left[q'_{i},p'_{j}\right]=i\delta_{ij}.$$

Any such symplectic transformation should be unitarily implemented due to the Stone-von Neumann theorem (right?)

$$U^{-1}q_{i}U =q'_{i},$$
$$U^{-1}p_{i}U =p'_{i}.$$

The question is: Assuming the coefficients A,B,C and D are given , is there a systematic way to calculate the generator $G$ of the unitary transformation $U=e^{-iG}$?

 

[MathematicalPhysicist]

I guess you can plug $U:=e^{-iG}=\sum_{n=0}^\infty \frac{(-iG)^n}{n!}$, or use https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula after plugging $q_i',p_j'$ into the commutation relations.

I am not sure how painful the calculations are, haven't done them myself.

 

[fresh_42]

The generators are the left-invariant vector fields, so you can consider all paths from $1$ to $U$ and differentiate them.

 

[strangerep]

Consider a set of $n$ position operators and $n$ momentum operator such that
$$\left[q_{i},p_{j}\right]=i\delta_{ij}.$$ Let's now perform a linear symplectic transformation $$q'_{i} =A_{ij}q_{j}+B_{ij}p_{j},$$ $$p'_{i} =C_{ij}q_{j}+D_{ij}p_{j}.$$
such that the canonical commutation relations are maintained $$\left[q'_{i},p'_{j}\right]=i\delta_{ij}.$$

This is essentially a "Bogoliubov" transformation, although in QFT one usually mixes annihilation and creation operators (but this is equivalent to mixing positions and momenta).

Any such symplectic transformation should be unitarily implemented due to the Stone-von Neumann theorem (right?)

No. Many Bogoliubov transformations cannot be implemented unitarily. Instead, we map between unitarily inequivalent Hilbert spaces.

$$U^{-1}q_{i}U =q'_{i},$$ $$U^{-1}p_{i}U =p'_{i}.$$
The question is: Assuming the coefficients A,B,C and D are given , is there a systematic way to calculate the generator $G$ of the unitary transformation $U=e^{-iG}$?

Yes, provided you understand that the mapping goes between inequivalent Hilbert spaces. If fresh42's cryptic hint is not enough, let me know and I'll try to dig out my old computations for this stuff. (It's been a while...)

 

[PeterDonis]

Any such symplectic transformation should be unitarily implemented due to the Stone-von Neumann theorem

Note that the Stone-von Neumann theorem only applies to the case of finitely many degrees of freedom.

Many Bogoliubov transformations cannot be implemented unitarily.

This would be for transformations in QFT, for example, where there are infinitely many degrees of freedom, correct?

 

[strangerep]

This would be for transformations in QFT, for example, where there are infinitely many degrees of freedom, correct?

Yes.

 

[andresB]

Note that the Stone-von Neumann theorem only applies to the case of finitely many degrees of freedom.

When you say finite degrees of freedom you mean the Irreducible set of operators is finite? So, there should be finite position and momentum operators (ignoring spin and other internal degrees of freedom)? that's my case.

 

[andresB]

This is essentially a "Bogoliubov" transformation, although in QFT one usually mixes annihilation and creation operators (but this is equivalent to mixing positions and momenta).

No. Many Bogoliubov transformations cannot be implemented unitarily. Instead, we map between unitarily inequivalent Hilbert spaces.

Yes, provided you understand that the mapping goes between inequivalent Hilbert spaces. If fresh42's cryptic hint is not enough, let me know and I'll try to dig out my old computations for this stuff. (It's been a while...)

I don't know about Bogoliubov transformation and inequivalent Hilbert spaces, so I don't know what to say about it.

In any case, even if not related to the Stone- von Neumann theorem, I think that linear symplectic transformation can always be given unitarily https://aip.scitation.org/doi/10.1063/1.1665805

 

[PeterDonis]

When you say finite degrees of freedom you mean the Irreducible set of operators is finite?

Not quite the set of operators, because complementary operators, such as position and momentum along the same spatial dimension, represent just one degree of freedom. For example, for the case of a single spin-zero particle moving in three spatial dimensions, the system has three degrees of freedom. But in QFT, there are an infinite number of degrees of freedom, because there is, roughly speaking, at least one degree of freedom for each spacetime point.

 

[andresB]

Not quite the set of operators, because complementary operators, such as position and momentum along the same spatial dimension, represent just one degree of freedom. For example, for the case of a single spin-zero particle moving in three spatial dimensions, the system has three degrees of freedom. But in QFT, there are an infinite number of degrees of freedom, because there is, roughly speaking, at least one degree of freedom for each spacetime point.

Well, I'm mostly interested in a fixed $n$ non-relativistic particles, so, not QFT.

 

[vanhees71]

Not quite the set of operators, because complementary operators, such as position and momentum along the same spatial dimension, represent just one degree of freedom. For example, for the case of a single spin-zero particle moving in three spatial dimensions, the system has three degrees of freedom. But in QFT, there are an infinite number of degrees of freedom, because there is, roughly speaking, at least one degree of freedom for each spacetime point.

In other words in the socalled "1st-quantization formalism" you have a finite set of fundamental observables that generate the complete operator algebra. E.g., for a single particle in non-relativistic QT you have $\hat{\vec{x}}$ and $\hat{\vec{p}}$ with the "Heisenberg algebra",
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}, \quad [\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0.$$
If the particle has in addition spin $s \in \{1/2,1,\ldots \}$ there are in addition the spin-angular momentum operators $\hat{s}_j$, commuting with all components $\hat{\vec{x}}$ and $\hat{\vec{p}}$, and fulfilling the usual angular-momentum commutation relations,
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l.$$
In contradistinction to that case in QFT you have a continuous set of such "generators" of the observable algebra like the field operators $\hat{\psi}(\vec{x})$, where you have to read $\vec{x}$ as a continuous label for the independent generators.

 

[A. Neumaier]

Many Bogoliubov transformations cannot be implemented unitarily. Instead, we map between unitarily inequivalent Hilbert spaces.

This is the case if $n$ is infinite, i.e., in qft. But in the OP's question, $n$ is finite, and by Stone's theorem, all Bogoliubov transformations are unitarily implemented..

Lets now perform a linear symplectic transformation
$$q'_{i} =A_{ij}q_{j}+B_{ij}p_{j},$$
$$p'_{i} =C_{ij}q_{j}+D_{ij}p_{j}.$$
such that the canonical commutation relations are maintained
$$\left[q'_{i},p'_{j}\right]=i\delta_{ij}.$$

The question is: Assuming the coefficients A,B,C and D are given , is there a systematic way to calculate the generator $G$ of the unitary transformation $U=e^{-iG}$?

Yes. You need the unitary representation theory of the symplectic group $SP(2n)$. Explicit formulas can be given in terms of the representation by squeezed coherent states. You should study the very thorough survey paper

• Zhang, W., Feng, D. H., & Gilmore, R., Coherent states: Theory and some applications. Reviews of Modern Physics 62 (1990), 867.

 

[Nullstein]

Independent of whether the Bogoliubov transform can be implemented unitarily or not, you can only expect a (reasonable) generator to exist if you have a continuous group of Bogoliubov transformations that induced a continuous group of unitaries, parametrized by a parameter $t$. In that case, you get the generator by differentiation with respect to $t$ (Stone's theorem). If you are satisfied with an operator G such that $U = e^{-iG}$, you may try to take the logarithm of $U$, but in general, you won't get a unique answer.

 

[A. Neumaier]

Independent of whether the Bogoliubov transform can be implemented unitarily or not, you can only expect a (reasonable) generator to exist if you have a continuous group of Bogoliubov transformations

The group of Bogoliubov transformations under discussion is a finite-dimensional symplectic group, which obviously satisfies your constraints. The generators are those of the corresponding Lie algeba. It is well-known how to represent the latter on a Hilbert space with operators satisfying the canonical commutation relations by means of homogenous quadratic expressions in the canonical operators. This is done by going over to creation and annihilation operators and using normally ordered quadratics in these.

 

[Nullstein]

The group of Bogoliubov transformations under discussion is a finite-dimensional symplectic group, which obviously satisfies your constraints. The generators are those of the corresponding Lie algeba. It is well-known how to represent the latter on a Hilbert space with operators satisfying the canonical commutation relations by means of homogenous quadratic expressions in the canonical operators. This is done by going over to creation and annihilation operators and using normally ordered quadratics in these.

There is no group of Bogoliubov transformations under discussion though. The OP is talking about a single Bogoliubov transformation. Depending on which one-parameter subgroup of the full group of Bogoliubov transformations this single Bogoliubov transformation is part of, the generator will be different. It makes no sense to ask for a generator without specifying a one-parameter group. The one-parameter groups $e^{i t}$ and $e^{2 i t}$ both give the element $1$ at $t=0$, but the generators are $1$ and $2$ respectively.

 

[strangerep]

This is the case if $n$ is infinite, i.e., in qft. But in the OP's question, $n$ is finite, and by Stone's theorem, all Bogoliubov transformations are unitarily implemented.

It's good to see you're still on the ball...

 

[strangerep]

There is no group of Bogoliubov transformations under discussion though. The OP is talking about a single Bogoliubov transformation.

I read it as a set of Bogoliubov transformations parameterized by the 4 matrices A,B,C,D.

(But, yes, I know one should probably refrain from debating what an OP "really" meant.)

 

[Nullstein]

I read it as a set of Bogoliubov transformations parameterized by the 4 matrices A,B,C,D.

(But, yes, I know one should probably refrain from debating what an OP "really" meant.)

Well, those four matrices specify one Bogoliubov transformation. In order to obtain a generator, one needs a parametrization in terms of a single parameter, e.g. $A(t),\ldots,D(t)$, which induce a unitary one-parameter group $U(t)$, so one can differentiate wrt. $t$. The way I read the OP, he only has one transformation and I just wanted to make him aware that he might have a misconception about what a generator is.