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What is normally meant by quantizing a classical theory, versus Hardy's axioms

  1. Sep 19, 2003 #1


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    Loop and Paden have brought up Hardy's axioms of quantum theory

    what do you think is usually meant by quantizing a classical (non-quantum) theory? And how does this connect to these axioms of what a quantum theory ought to be

    Here is a mainstream summary description of what quantizing means:

    (just quoting from a June 2002 paper by Bojowald)

    "Quantization consists in turning functions on the phase space of a given classical system into operators acting on a Hilbert space associated with the quantized system.

    To construct this map one selects a set of 'elementary' observables, like (q,p) in quantum mechanics, which
    generate all functions on the phase space and form a subalgebra
    of the classical Poisson algebra. This subalgebra has to be
    mapped homomorphically into the quantum operator algebra, turning real observables into selfadjoint operators..."
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  3. Dec 6, 2011 #2
    The quote you give is indeed the usual meaning of 'first quantisation', i.e., finding a quantum mechanical analogue to a classical theory. (This is distinct from 'second quantisation', which is passing from a quantum mechanical theory to a quantum field theory by quantising again.)
    You understand what you quoted? A simple example is the classical harmonic oscillator has
    $H=p^2+q^2$ on the Euclidean plane coordinatised by $p$ and $q$, so this is a function of two variables. To quantise this, you simply take the same equation but now interpret $p$ as the operator on square integrable functions of the p-variable, i.e., one-variable functions, and the operator is multiplication by the coordinate function $p$. Then $q$ is $-i\frac\del{\del p}$ and $H$ now describes a quantum system with state space $L^2(p$-axis).

    It is not always so easy to figure out how to pick the formula for $H$...if $H$ could be written in different ways that were the same when $p$ and $q$ commute, for example, $H=p^2 + 2pq + q^2$ which of course equals $H=(p+q)^2$, then one would not be sure which quantisation to pick, because if these two formulas are interpreted as operators, they no longer agree. So the procedure of quantisation is not unambiguous. Quantisation is an art, not an algorithm. (Second quantisation is routine and unambiguous.)

    Souriau, Kostant, and Michelle Vergne devoted a good deal of attention to trying to find a geometric meaning and procedure for quantisation, but they were mathematicians and I never heard that many physicists took it up. They basically agree with what you quote but try to define it more abstractly, without a choice of coordinates, i.e., they did not mind their p's and q's.

    I have looked some at Hardy's papers and don't quite see that he is doing anything like this, although I might be mistaken. In fact, I do not even think he has quite reconstructed orthodox Quantum Mechanics except in the finite dimensional case, so I do not quite see how he can do anything like quantisation. See the answer on stack exchange to a related query, http://physics.stackexchange.com/qu...ity-distribution-the-wave-function-or-the-den
  4. Dec 7, 2011 #3


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    Historically and as an introduction to QFT this is OK, but it does not reflect modern physical methods. Looking at QCD (in canonical quantization) as an example there is no such thing as first quantization. You take a classical Lagrangian, derive the canonical fields and turn them into field operators. There is no quantum mechanical system that is quantized a second time.

    Definitly not! It is plagued by ordering ambiguities, violation of classical operator algebras (anomalies, look at quantum gravity where off-shell closure has not been proven), missing global definition of physical fields (gauge fixing issues and Gribov ambiguities, look at QCD again), infinities (strictly speaking non-existence of the quantum field theory!), ...
  5. Dec 7, 2011 #4
    Dear Tom:
    In the procedure you mention, which is more modern, you are of course right that it bypasses what the questioner and I call quantisation, or 'first quantisation', which is all that is really relevant to Hardy and the question, anyway. Precisely since it bypassed this step, and, as you point out, there is no QM system being re-quantised, the order ambiguities that would have been solved in that step are still there. All I said was that second quantisation in the sense I meant it, the older-fashioned sense, was free from such ambiguities and doing it as a process was routine. (I of course didn't mean to imply that the system you got as a result of that process was a system whose solution would be cut and dried, I was only talking about the decisions needed to appy that process.) Do QIT people ever consider the differences between the QM they seem to work with and QFT anyway? Do you think those differences might be important for what they are trying to do?
  6. Dec 7, 2011 #5


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    @andrebourbaki: we agree regarding QFT / quantization of fields in general; regarding QIT I can't make any reasonable comments
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