- #1
da_willem
- 599
- 1
Why is it canonical momentum that is replaced by the momentum operator in quantum mechanics? What is the physical significance of canonical momentum anyway?
da_willem said:Why is it canonical momentum that is replaced by the momentum operator in quantum mechanics? What is the physical significance of canonical momentum anyway?
da_willem said:Why is it canonical momentum that is replaced by the momentum operator in quantum mechanics? What is the physical significance of canonical momentum anyway?
da_willem said:Does this mean the above is only correct in cases where the hamiltonian is [itex]p^2/2m+V[/itex] with p the canonical momentum?
kakarukeys said:To find the Hamiltonian Operator, we replace canonical momentum by the momentum operator, coordinate by position operator.
How do we find the operator for a variable like [tex]xp_x[/tex]? coordinate multiplied by momentum.
We do not qeustion about the physical meaning of the variable at this moment. If we simply do the substitution: [tex]\hat{x}\hat{p_x}[/tex]
There are two problems:
(1) How do we define multiplication of operators? In the expression, x-operator is multiplied by p-operator, but only composition of operators is well-defined.
(2) The operator obtained is not Hermitian.
sifeddin said:...
Finally a younger fellow instructor asked me a related question that I could not get a conclusive answer so try to give me your best answers to this:
Are the operators of the quantum mechanics, e.g. the momentum operator, are unique and why? I appreciate if you give link to a formal proof if you can.
sifeddin said:So here are my two questions:
1- Could someone experienced tell me the difference in the physical meaning (NOT in the use in equations) between the two momenta?
2- Does it really make any difference in the following results ,e.g. the absorption rate become the gain rate, if the sign of qA were positive?
sifeddin said:I appreciate your debate in answering this question but these two questions seem to be untouched would someone give a response on them please
kakarukeys said:Hi,
When you said the 2nd postulate of QM, I assume you meant the postulate: every operator that represents a physical variable must be Hermitian? (which book you referred to? The 2nd postulate on my book says only the eigenvalue equation.)
That postulate does not give any hint "how to make it Hermitian"
The way you suggested [tex] xp_{x} \rightarrow \frac{1}{2} (xp_{x}+p_{x}x) [/tex]
is only one of the countless number of ways to make it Hermitian. How do you ensure that the operator you obtained is the correct one?
da_willem said:Why is it canonical momentum that is replaced by the momentum operator in quantum mechanics? What is the physical significance of canonical momentum anyway?
Haelfix said:So like I said, the limiting factor is the commutation relations, which are unique by Stone Von Neumann, *nearly always*.
Haelfix said:... If your classical symplectic space is topologically nontrivial (say it has nonvanishing homology classes), the correspondance can and will break down, and you have to do things by hand. Also, there can be dirac constraints imposed on the system.
dextercioby said:Well,give a counterexample!
pmb_phy said:Your confusion seems to rest in what you think Schrodinger's equation is, i.e.
[tex]i\hbar\frac{\partial\Psi}{\partial t}=\frac{\hat{p}^2}{2m}\Psi+V\Psi[/tex]
That isn't Schrodinger's equation exactly. That is an example of Schrodinger's equation when the canonical momentum is the linear mechanical momentum. Schrodinger's equation is
[tex]\hbar \frac{d}{dt}|\Psi(t)> = H(t)|Psi(t)>[/tex]
The momentum that appears in the Hamiltonian is the canonical momentum. I.e. the source of the Hamiltonian comes from an integral of motion in classical mechanics.
chris1234 said:This is so that things like momentum conservation can apply - if I take a particle and fire it into an electric field it will accelerate, and the particle's momentum is not constant. But its canonical momentum will be constant (I think).
kakarukeys said:Counter-example for you:
[tex]x^2p[/tex]
[tex]x^2p = xpx \rightarrow \hat{x}\hat{p}\hat{x} [/tex]
[tex]x^2p = \frac{1}{2}(xxp + pxx) \rightarrow \frac{1}{2}(\hat{x}\hat{x}\hat{p} + \hat{p}\hat{x}\hat{x}) [/tex]
[tex]x^2p = \frac{1}{3}(xxp + xpx + pxx) \rightarrow \frac{1}{3}(\hat{x}\hat{x}\hat{p} + \hat{x}\hat{p}\hat{x} + \hat{p}\hat{x}\hat{x}) [/tex]
There are 3 possible Hermitian operators for the classical variable [tex]x^2p[/tex]
The last one is called Weyl Ordering.
Which one is the correct operator to use?
da_willem said:Can someone confirm this?
kakarukeys said:Me too, but I can't prove it (beyond my present ability).
It happens that if we don't use Weyl Ordering in some physics problem, we will get weird results, like infinite energy. The simple harmonic oscillator Hamiltonian is a Weyl Ordering of creation and annihilation operator.
When you say an operator is unique, it is always up to a unitary transformation. I think weyl ordering is no exception, unitary transformation won't turn it into other ordering (I don't know how to prove it, but feel that it must be so, single p alone has only one ordering, so unitary transformation won't turn it into another ordering)
sifeddin said:Forgive my ignorance :shy: (at least with the terms, as in my faculty Mech. Professors are interested with the treatment rather than the nomnclature); what is symplectic means? I run through it in various "free preprint" papers but I never seen a definition, is it equivalent to commutative, in contrary to noncommutative? and what are the "homology classes"?
Canonical momentum, also known as conjugate momentum, is a fundamental concept in quantum mechanics that describes the momentum of a particle in a quantum system. It is a mathematical operator that is used in the Schrödinger equation to determine the time evolution of a quantum system.
In classical mechanics, momentum is defined as the product of an object's mass and velocity. In quantum mechanics, however, momentum is described using the canonical momentum operator, which takes into account the wave-like nature of particles. This allows for a more accurate description of a particle's momentum in a quantum system.
Canonical momentum is important in quantum mechanics because it is a fundamental property of particles in quantum systems. It is used to calculate the energy of a particle and to determine the behavior of particles in quantum systems, such as their position and momentum at a given time.
According to the uncertainty principle, it is impossible to know both the position and momentum of a particle with 100% accuracy. The canonical momentum operator plays a crucial role in determining the uncertainty in a particle's momentum, as it is used to calculate the standard deviation of the momentum in a quantum system.
Yes, canonical momentum can be measured in experiments using various techniques such as the Compton scattering experiment or the Stern-Gerlach experiment. These experiments allow for the measurement of a particle's momentum in a quantum system, providing valuable insights into the behavior of particles at the quantum level.