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  Is the space of coherent states generated by bounded operators empirically adequate?

+ 2 like - 0 dislike

DETLEV BUCHHOLZ, FABIO CIOLLI, GIUSEPPE RUZZI, and EZIO VASSELLI, in "The Universal C*-Algebra of the Electromagnetic Field", Lett Math Phys (2016) 106:269–285, https://arxiv.org/abs/1506.06603, construct a C*-Algebra of bounded operators effectively as \(\mathrm{e}^{\mathrm{i}\hat F_{\!f}}\), where \(f\) is a real-valued 2-form test function (the norm is \(\|\mathrm{e}^{\mathrm{i}\hat F_f}\| =\left\{\begin{array}{l}1,\|f\|=0\cr 0, \|f\|\not= 0 \end{array}\right.\)). That is, the vacuum sector is generated by complex superpositions of unitary coherent vector states. In quantum optics, however, the vacuum sector of the free EM field is constructed using both unitary and non-unitary operators, effectively generated by \(\mathrm{e}^{\mathrm{i}\hat F_{\!f}}\), where \(f\) is now a complex-valued 2-form test function.

So my question, "Is the space of coherent states generated by bounded operators empirically adequate?", asks whether the set of coherent states that are widely used in quantum optics is necessary for empirical adequacy or whether the C*-algebraic approach of using only bounded algebras is all that is needed. Is it just more convenient but unnecessary for physicists to use the larger algebra of unbounded operators, but more convenient for mathematicians to use only the algebra of bounded operators?

asked Feb 10, 2017 in Theoretical Physics by Peter Morgan (1,230 points) [ no revision ]

One can say, I guess, that all actual measurement results are finite, therefore bounded operators are surely empirically adequate, but for a free EM field observable \(\hat F_{\!f}\) in the vacuum sector the probability distribution is Gaussian, or a polynomial deformation of Gaussian, so that arbitrarily large values (in the unbounded tail) are very suppressed indeed. Some ideas of simplicity would seem to push us towards using unbounded operators, but not necessity, whereas for a mathematician the added control and simplicity of using only bounded operators seems to be more compelling. But is that all that one might say?

1 Answer

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Unitary and nonunitary group elements generate the same collection of coherent states. 

The reason are the Weyl relations for the exponentials. They allow one to write the exponential of any linear combination of creation and annihilation operators as a product of a constant factor, the exponential of a linear combination of creation operators  and the exponential of a linear combination of annihilation operators. The last factor has no effect on the vacuum. This causes a redundancy with which it is not difficult to justify my initial claim.

Thus it is enough to consider the algebra generated by the unitary elements. This algebra has the advantage of being a $C^*$-algebra. Unbounded operators can still be obtained as limits, though only in a particular unitary representation. Algebras of unbounded operators are far mor difficult to handle; many of the useful tools from $C^*$-algebras are no longer applicable.

answered Feb 10, 2017 by Arnold Neumaier (15,777 points) [ revision history ]

Many thanks, Arnold.

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