# Tagged Questions

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### Examples of Weyl transforms of nontrivial operators

I've been able to find examples of Weyl transforms of operators like $\hat{x}$,$\hat{p}$, and $\hat{1}$, but not anything more complicated. Are there derivations of the Weyl transforms of more ...
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### Hamiltonian Operator Interpretation of Quantum Anomaly

We can see the definition of quantum anomaly in terms of Lagrangian path integral formulation. What is the definition of quantum anomaly in terms of Hamiltonian operator approach or even more directly ...
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### Why don't we use Hamilton-Jacobi method in QM?

In classical mechanics, we usually try to find a set of coordinates by Hamilton-Jacobi method to transform the Hamiltonian to zero such that the coordinates are conservations. However, we never try ...
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### Does the Renormalization of QFT Contradict Canonical Quantization?

Does the renormalization of QFT contradict canonical quantization? In canonical quantization, you take the classical fields and canonical momenta and turn them into operators, and you require that ...
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### Quantum mechanical analogue of conjugate momentum

In classical mechanics, we define the concept of canonical momentum conjugate to a given generalised position coordinate. This quantity is the partial derivative of the Lagrangian of the system, with ...
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### Weyl Ordering Rule

While studying Path Integrals in Quantum Mechanics I have found that [Srednicki: Eqn. no. 6.6] the quantum Hamiltonian $\hat{H}(\hat{P},\hat{Q})$ can be given in terms of the classical Hamiltonian ...
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### Operator Ordering Ambiguities

I have been told that $$[\hat x^2,\hat p^2]=2i\hbar (\hat x\hat p+\hat p\hat x)$$ illustrates operator ordering ambiguity. What does that mean? I tried googling but to no avail.
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### Ordering Ambiguity in Quantum Hamiltonian

While dealing with General Sigma models (See e.g. Ref. 1) $$\tag{10.67} S ~=~ \frac{1}{2}\int \! dt ~g_{ij}(X) \dot{X^i} \dot{X^j},$$ where the Riemann metric can be expanded as, \tag{10.68} ...
Is there any significance to: $\langle[H,\hat{O}]\rangle =0$ (which can easily be shown) where $H$ is the Hamiltonian, $\hat{O}$ is an arbitrary operator? Thanks.