A mathematical construct used to study the effect of applying two operators in succession.

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transformations with commutators and anticommutators that generate displacements

is well known that composition of point reflections generate pure displacements. This implies that the commutator of two point reflections will be a pure displacement. Are there similar elemental ...
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989 views

Operators and Commutator Definitions

I have several problems with General Definitions of an Operator and Commutator : the product of operators is generally not commutative: $$\hat A \hat B \not= \hat B\hat A .$$ what is this means ...
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315 views

Commutators with a density matrix

The equation describing the evolution of our system is as follows: $ \dot{\rho} = u_1(t)(a^\dagger a \rho - 2a\rho a^\dagger +\rho a^\dagger a) + u_2(t)(a a^\dagger \rho - 2a^\dagger\rho a +\rho a ...
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Generalizing Heisenberg Uncertainty Priniciple

Writing the relationship between canonical momenta $\pi _i$ and canonical coordinates $x_i$ $$\pi _i =\text{ }\frac{\partial \mathcal{L}}{\partial \left(\frac{\partial x_i}{\partial t}\right)}$$ ...
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339 views

Multiplication of 3-vector operators

I've started reading "Quantum Mechanics: A Modern Development" by Leslie E. Ballentine and have some trouble understanding how to handle 3-vector operators (i.e. an operator $\mathbf{A}$ with ...
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949 views

Derivation of angular momentum commutator relations

I'm trying to understand the derivation of the angular momentum commutator relations. How is $$[zp_y, zp_x] ~=~ 0?$$ How is $$[yp_z, zp_x] ~=~ y[p_z, z]p_x?$$
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Compatible Observables

My QM book says that when two observables are compatible, then the order in which we carry out measurements is irrelevant. When you carry out a measurement corresponding to an operator $A$, the ...
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What is the connection between Poisson brackets and commutators?

The Poisson bracket is defined as: $$\{f,g\}_{PB} ~:=~ \sum_{i=1}^{N} \left[ \frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial ...
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Index Manipulation and Angular Momentum Commutator Relations

I have been trying for hours and cannot figure it out. I am not asking anyone to do it for me, but to understand how to proceed. We have the relations $$[L_i,p_j] ~=~ i\hbar\; \epsilon_{ijk}p_k,$$ ...
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845 views

The implication of anti-commutation relations in quantum mechanics

All the textbooks I saw are very clear about the implications of commutating operators in quantum mechanics. However, much less is said about anti-commutation relations. Does it have a general ...
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Why are anticommutators needed in quantization of Dirac fields?

Why is the anticommutator actually needed in the canonical quantization of free Dirac field?
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How far can you get (in quantum mechanics) with just commutation relations?

Clearly it is possible to derive a set of commutation relations from some Hamiltonian, and certainly they give useful and interesting invariants when investigating the behavior of quantum systems. ...
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Expectation of a commutation relation

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.
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Operator relation involving the logarithm of an operator?

Dirac gives the relation: $\exp(iaq)f(q,p) = f(q, p - a\hbar)\exp(iaq)$ where $\hbar$ is Planck's constant. Can anybody give me the corresponding relation when the $\exp$ function is a $\ln$?
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Trace of a commutator is zero - but what about the commutator of $x$ and $p$?

Operators can be cyclically interchanged inside a trace: $${\rm Tr} (AB)~=~{\rm Tr} (BA).$$ This means the trace of a commutator of any two operators is zero: $${\rm Tr} ([A,B])~=~0.$$ But what about ...
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How to construct the radial component of the momentum operator?

I'm having trouble doing it. I know so far that if we have two Hermitian operators $A$ and $B$ that do not commute, and suppose we wish to find the quantum mechanical Hermitian operator for the ...
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What is the Physical Meaning of Commutation of Two Operators?

I understand the mathematics of commutation relations and anti-commutation relations, but what does it physically mean for an observable (self-adjoint operator) to commute with another observable ...
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Momentum-Representations in Quantum Mechanics

Why do we get information about position and momentum when we go to different representations. Why is momentum, which was related to time derivative of position in classical physics, now in QM just a ...