4 replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/ edited Apr 13 '17 at 12:39 Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this questionthis question. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. 3 removed apparently wrong paragraph, h/t Solonedon Paradoxus edited Apr 7 '17 at 10:32 ACuriousMind♦ 74.4k18137338 Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. Finally, I am not an expert on loop quantum gravity but I believe the spin networks or loop quantum gravity do not constitute a different "method of quantization", since their notions are rather standard: You have constraints, operators, Wilson loops...it's "just" a particular attempt at a quantum theory of gravity, not a new method to generate quantum theories from classical theories, which is what a quantization method is. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. Finally, I am not an expert on loop quantum gravity but I believe the spin networks or loop quantum gravity do not constitute a different "method of quantization", since their notions are rather standard: You have constraints, operators, Wilson loops...it's "just" a particular attempt at a quantum theory of gravity, not a new method to generate quantum theories from classical theories, which is what a quantization method is. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. 2 deleted 3 characters in body edited Apr 6 '17 at 18:18 ACuriousMind♦ 74.4k18137338 Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for thereus to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. Finally, I am not an expert on loop quantum gravity but I believe the spin networks or loop quantum gravity do not constitute a different "method of quantization", since their notions are rather standard: You have constraints, operators, Wilson loops...it's "just" a particular attempt at a quantum theory of gravity, not a new method to generate quantum theories from classical theories, which is what a quantization method is. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for there to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. Finally, I am not an expert on loop quantum gravity but I believe the spin networks or loop quantum gravity do not constitute a different "method of quantization", since their notions are rather standard: You have constraints, operators, Wilson loops...it's "just" a particular attempt at a quantum theory of gravity, not a new method to generate quantum theories from classical theories, which is what a quantization method is. Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $$\mathbb{R}^{6N}$$ phase space dynamics to be recoverable from the thermodynamic limit $$N\to\infty$$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one. Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $$f(x,p)\mapsto \hat{f}$$. The Groenewold-van Hove theorem says that there is no such map such that $$f\mapsto \hat{f}$$ is linear. $$[\hat{f},\hat{g}] = \mathrm{i}\hbar\widehat{\{f,g\}}$$ holds for all observables $$f,g$$. Observables that commute with everything are multiples of the identity, meaning the representation of the algebra of observables is irreducible. $$p(\hat{f}) = \hat{p(f)}$$ for all polynomials $$p$$, meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $$\hbar^2$$, geometric quantization instead restricts the allowed inputs $$f$$ to the quantization map and drops the fourth property. Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $$f\mapsto \hat{f}$$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question. Finally, I am not an expert on loop quantum gravity but I believe the spin networks or loop quantum gravity do not constitute a different "method of quantization", since their notions are rather standard: You have constraints, operators, Wilson loops...it's "just" a particular attempt at a quantum theory of gravity, not a new method to generate quantum theories from classical theories, which is what a quantization method is. 1 answered Apr 6 '17 at 18:11 ACuriousMind♦ 74.4k18137338