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In solving the Hydrogen problem, one has not take into consideration P_theta and P_phi at all.

Quantum River

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- Thread starter Quantum River
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In summary: This is because the Lagrangian is a function of the radial and angular velocities, which are related to the canonical momenta through the definition of the Lagrangian. In summary, the canonical momenta for theta and phi in spherical coordinates are defined as derivatives of the Lagrangian with respect to the corresponding velocities. In solving the Hydrogen problem, these momenta are not considered, but they do appear in the Hamiltonian as squared terms in the kinetic energy. In quantum mechanics, the canonical momenta are promoted to derivative operators, with p_theta becoming -i*hbar*partial/ partial theta. The form of the momenta depends on the form of the Lagrangian, and it is important for the Lagrangian to be non

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In solving the Hydrogen problem, one has not take into consideration P_theta and P_phi at all.

Quantum River

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[tex]p_i \equiv \frac{\partial L}{\partial \dot{q}_i}.[/tex]

So you have to get the Lagrangian out of the Hamiltonian, which is given by

[tex]L(q_i, \dot{q}_i) = \sum_i p_i q_i - H(p_i, q_i),[/tex]

where you substitute for [itex]\dot{q}_i[/itex] wherever you find a [itex]p_i.[/itex]

Remember that the canonical momenta in Q.M. are promoted to derivative operators, so that

[tex]p_\theta \rightarrow -i\hbar \frac{\partial}{\partial \theta}.[/tex]

This has to be because the canonical commutation relations hold for co-ordinates and their canonical momenta (since in classical mechanics, the corresponding relations hold for the Poisson bracket).

So these momenta do in fact appear in the Hamiltonian for the hydrogen atom: they usually appear as squared terms in the [itex]\nabla ^2[/itex] term (i.e. the kinetic energy).

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Quantum River said:

In solving the Hydrogen problem, one has not take into consideration P_theta and P_phi at all.

Quantum River

[tex] p_{r}=:\frac{\partial L}{\partial \dot{r}} [/tex]

[tex] p_{\theta}=:\frac{\partial L}{\partial \dot{\theta}} [/tex]

[tex] p_{\phi}=:\frac{\partial L}{\partial \dot{\phi}} [/tex]

The form depends on the form of the lagrangian, of course. It's important the lagrangian be nondegenerate, else the quantization will be difficult.

Daniel.

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The Schrodinger equation in spherical coordinates is a mathematical expression that describes the behavior of quantum particles in a three-dimensional space. It takes into account the angular momentum and radial distance of the particle, and allows for the prediction of its wave function.

The Schrodinger equation is expressed in spherical coordinates because it is a more natural and convenient way to describe the motion of particles in a three-dimensional space. It takes into account the spherical symmetry of the system and simplifies the mathematical calculations.

The Schrodinger equation in spherical coordinates is typically solved using separation of variables, where the radial and angular components of the equation are solved separately. The solutions are then combined to form the complete wave function.

The Schrodinger equation in spherical coordinates allows us to understand the behavior of quantum particles in a three-dimensional space. It helps us to predict the probability of finding a particle at a certain location and time, and provides insights into the fundamental nature of matter.

The Schrodinger equation in spherical coordinates is closely related to the uncertainty principle, which states that it is impossible to simultaneously know the exact position and momentum of a particle. The Schrodinger equation allows us to calculate the probability of finding a particle at a certain location, but it does not provide information about its exact position or momentum.

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