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I have a simple (I think !) question about the representations of boson operators and how they are related. First of all let's define two conjugate observables $Q$ and $P$ (i.e. $\left[Q,P\right]=i$ and $Q^\dagger=Q$, $P^\dagger=P$). If we further define: \begin{equation} a=\sqrt{\frac{\alpha}{2}}\left(Q+\frac{i}{\alpha}P\right)~~~~~~~~~~~~ a^\dagger=\sqrt{\frac{\alpha}{2}}\left(Q-\frac{i}{\alpha}P\right)~~~~~~~\alpha\in \mathbb{C}, \end{equation} (as in the harmonic oscillator problem) we have that $\left[a,a^{\dagger}\right]=1$. We can there identify $a^{(\dagger)}$ as boson annihilation (creation) operators. However we can also define: \begin{equation} b=\sqrt{Q}e^{iP}~~~~~~~~~~~~~~~~b^\dagger=e^{-iP}\sqrt{Q} \end{equation} which will verify $\left[b,b^{\dagger}\right]=1$ (this requires a bit more algebra though).

Question : Is there a relation between these two representations ? These are specific examples, but one could probably think of other representations. Since these representations implement the same commutation relations, does it mean that there are related by some transformation (a unitary transformation in particular) ?

(I give here specific examples for bosonic operators, but I guess one may extend the discussion to any type of operator satisfying some commutation relation).

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Your non standard representation does not produce a well-behaved canonical theory.

The most evident and direct way to face it, barring theoretical remarks based on the absence of rigorous hypotheses sufficient to apply some theorem (Stone von Neumann, Nelson, FS^3, Dixmier...), is the following.

To construct a representation of your bosonic theory (a) you have to build up the orthonormal set of occupation numbers states $\{|n\rangle\}_{n=0,1,2,\ldots}$ and (b) you have to prove that this set is complete (i.e., maximal)($^*$).

By definition, where $C_n \neq 0$ is a normalization coefficient: $$|n\rangle := C_n(b^\dagger)^n|0\rangle \qquad (1)$$ with: $$b|0\rangle =0\quad\mbox{and}\quad \langle 0|0\rangle =1\:.\qquad (2)$$ The former equation in (2), making explicit the form of the operator $b$ in the Hilbert space of the theory, $L^2(\mathbb R)$, and writing down the equation using the wavefunction $\psi_0$ of $|0\rangle$ in position representation, reads: $$\sqrt{x}\psi_0(x+1)=0 \quad \mbox{(almost everywhere)}\:,\qquad (3)$$ where I have exploited the fact that $\{e^{-i\lambda P}\}_{\lambda \in \mathbb R}$ is the unitary representation of the group of $x$-translations.

The only $L^2$ solution of (3) is trivially: $$\psi_0(x) = 0 \quad \mbox{almost everywhere.}$$ Consequently the latter condition in (2) is untenable and all the construction aborts here.


footnotes

$(^*)$ Technically speaking, these vectors are consequently analytic vectors for all the involved operators and this a guarantee for the validity of several crucial properties like essentially self-adjointness of the new canonical variables.

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  • $\begingroup$ Okay, so the theory is not consistent. But I still don't understand something : the representation (1) can be used to solve the harmonic oscillator problem. Yet you show here that the theory is not consistent... what is going on !?? Moreover, let's quote PM Dirac on the harmonic oscillator problem : "This means that the dynamic system consisting of an assembly of similar bosons is equivalent to the dynamic system consisting of a set of oscillators - the two systems are just the same system looked at from two different points of view [...] $\endgroup$ Feb 2 '14 at 20:58
  • $\begingroup$ (cont.) [...] There is one oscillator associated with each independent boson state. We have here one of the most fundamental results of quantum mechanics, which enables a unification of the wave and corpuscular theories of light to be effected..........Thus a set of harmonic oscillators is equivalent to an assembly of bosons in stationary states with no interactions between them. If an oscillator of the set is in its n'th quantum state, there are n' bosons in the associated boson state" $\endgroup$ Feb 2 '14 at 20:58
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    $\begingroup$ What I am saying is that the representation in terms of states $|n>$ makes sense if $X$ and $P$ are the standard ones in $L^2(R)$ or, more weakly, if there is a unitary operator translating them into the standard ones. This is not the case for the new operators $X'$ and $P'$ associated with $b$ and $b^\dagger$. $\endgroup$ Feb 2 '14 at 21:07
  • $\begingroup$ In other words, not all classical canonical transformations that in classical physics produce equivalent descriptions of the physical system, do the same job in quantum physics. $\endgroup$ Feb 2 '14 at 21:09
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    $\begingroup$ The general problem is the following one. Assume the configuration manifold is not necessarily $\mathbb R^n$, so that, for instance momentum operators are not defined (no translational symmetry) while coordinate operators, in principle are defined. What is the basic class of observables (I mean self-adjoint operators, not only Hermitian or symmetric) one can exploit to construct all observables of the system? This problem was studied by several mathematical physicists finding out interesting solutions. $\endgroup$ Feb 3 '14 at 15:06
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There is only one unitary representation for the algebra of bosonic operators. Given a set of creation and annihilation operators,

$$ [b,b^\dagger] = 1, $$

you can define a set of canonical position and momentum operators,

$$ [Q,P] = iC,\quad [C,P] = [C,Q] = 0. $$

which is known as Heisenberg algebra. $C$ is the center of this algebra. There is only one unitary representation for Heisenberg algebra (Stone-von Neumann theorem).

As for the "new representation" you mentioned, those ($P'$ and $Q'$) are just action-angle variables. $Q'$ in the "new representation" is the amplitude of oscillation, whereas $P'$ is roughly the phase angle. More specifically (set $\alpha=1$),

$$ Q = \sqrt{2Q'}\cos(P');\quad P = \sqrt{2Q'}\sin(P'). $$

$P'$ and $Q'$ are the "new" variables. They also form canonical conjugate pair. You can check this for both classical and quantum oscillators.

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  • $\begingroup$ What do you mean by action-angle variables ? I did not assume anything in this representation (just the CCR of $Q$ and $P$). $\endgroup$ Feb 2 '14 at 1:18
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    $\begingroup$ I mean the Jacobi action-angle variables in the classical mechanics. The "new" $Q$ and $P$ are related to the old ones through a canonical transformation. For a discussion on this subject, see this. $\endgroup$ Feb 2 '14 at 1:20
  • $\begingroup$ Can you clarify a bit the following : "unitary rep. of an algebra" ? -- I know what a unitary rep. is. However, was is a rep. of an algebra (in simple terms) ? $\endgroup$ Feb 2 '14 at 4:25
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    $\begingroup$ I do not understand what you mean for things like $\sqrt{Q}$, since the spectrum of $Q$ includes the negative real axis. Do you mean instead $\sqrt{|Q|}$? $\endgroup$ Feb 2 '14 at 10:54
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    $\begingroup$ It does not work generally! Taylor expansions does not commute with quantization procedure, unless you are dealing with bounded operators and uniform convergence of operators. Here, instead, operators are unbounded and the allowed topology is the strong one. The only chance to obtain anything sensible is the spectral approach or the polar decomposition theorem for unbounded closed normal operator (both related with the strong topology). $\endgroup$ Feb 2 '14 at 21:30
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Yes, your b excitations are known: they are tweaked coherent states, based on the Displacement operator of optical phase space.

For simplicity, take α=2, so that $$ Q=\frac{a+a^\dagger}{2}, \qquad iP=a-a^\dagger, $$ and hence $$ b=\sqrt{\frac{a+a^\dagger}{2}} e^{a- a^\dagger}= \sqrt{\frac{a+a^\dagger}{2}} D^\dagger(1),\\ b^\dagger= e^{a^\dagger -a} \sqrt{\frac{a+a^\dagger}{2}} =D(1)\sqrt{\frac{a+a^\dagger}{2}} ~, $$ where the displacement operator is defined as $D(1)= e^{a^\dagger -a}$.

Then, evidently, $$[b,b^\dagger]=\frac{a+a^\dagger}{2} - D(1)\frac{a+a^\dagger}{2}D^\dagger(1)=1 ~. $$ Acting on the Fock vacuum annihilated by a, different α than the above!, the displacement operators define the coherent state $D(\alpha=1)|0\rangle=|\alpha=1\rangle$, the eigenstate of the annihilation operator, but I am not sure of the drift of the rest of your question.

Slight modifications of these maps are popular in deformed oscillator algebras, section 4.g).

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