- #1

- 4

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( -1/2 -1 1

( -2 -1.5 2

(-1/2 2 3

( 1 3 1 )

2 -2 -1

Any pointers on where to go from here would be greatly appreciated. External links are helpful too.

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- Thread starter kenneththo85431
- Start date

In summary, the author plotted out the trajectory of an imaginary electron in 3D, then represented it's points with the matrix A(x1 y1 z1). Next, he asked for pointers on where to go from here.

- #1

- 4

- 0

( -1/2 -1 1

( -2 -1.5 2

(-1/2 2 3

( 1 3 1 )

2 -2 -1

Any pointers on where to go from here would be greatly appreciated. External links are helpful too.

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- #2

- 4

- 0

Both

- #3

Mentor

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- #5

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One way to do this is to start with a position vector in the x-y plane ##\vec{x}=(x,y)## and use a 2x2 matrix to change the position by multiplication so that ##\vec{x}_{n+1}=M\vec{x}_n##,kenneththo85431 said:

( -1/2 -1 1

( -2 -1.5 2

(-1/2 2 3

( 1 3 1 )

2 -2 -1

Any pointers on where to go from here would be greatly appreciated. External links are helpful too.

For instance

[tex]

M= \pmatrix{\cos\left( a\right) & \sin\left( a\right) \cr -\sin\left( a\right) & \cos\left( a\right) }

[/tex]

so that

[tex]

M\vec{x}= \pmatrix{\cos\left( a\right) & \sin\left( a\right) \cr -\sin\left( a\right) & \cos\left( a\right) }\vec{x}=\pmatrix{\sin\left( a\right) \,y+\cos\left( a\right) \,x\cr \cos\left( a\right) \,y-\sin\left( a\right) \,x}

[/tex]

If you start with position (-1,0) and choose a small ##a##, say 0.05 radians, then applying the matrix successively moves the point in a circle with radius 1 and center 0.

- #6

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Wow! Thank you so much!Mentz114 said:One way to do this is to start with a position vector in the x-y plane ##\vec{x}=(x,y)## and use a 2x2 matrix to change the position by multiplication so that ##\vec{x}_{n+1}=M\vec{x}_n##,

For instance

[tex]

M= \pmatrix{\cos\left( a\right) & \sin\left( a\right) \cr -\sin\left( a\right) & \cos\left( a\right) }

[/tex]

so that

[tex]

M\vec{x}= \pmatrix{\cos\left( a\right) & \sin\left( a\right) \cr -\sin\left( a\right) & \cos\left( a\right) }\vec{x}=\pmatrix{\sin\left( a\right) \,y+\cos\left( a\right) \,x\cr \cos\left( a\right) \,y-\sin\left( a\right) \,x}

[/tex]

If you start with position (-1,0) and choose a small ##a##, say 0.05 radians, then applying the matrix successively moves the point in a circle with radius 1 and center 0.

The A matrix is used in quantum mechanics to describe the position and momentum of an electron in three-dimensional space. It represents the wave function of the electron, which is a mathematical function that describes the probability of finding the electron at a certain position in space.

The A matrix is used to calculate the electronic orbit of an electron in a quantum system. It contains information about the energy, position, and momentum of the electron, which are all important factors in determining the shape and location of the electronic orbit.

The A matrix can be used to describe any type of electronic orbit, as long as the orbit is in three-dimensional space. This includes orbitals such as s, p, d, and f orbitals, which are commonly seen in chemistry and physics.

Scientists use the A matrix to solve the Schrödinger equation, which is the fundamental equation of quantum mechanics. By solving this equation, they can determine the wave function of an electron and use it to calculate the electronic orbit in 3D space.

While the A matrix is a powerful tool for describing electronic orbit in 3D space, it does have some limitations. It assumes that the electron is a point particle and does not take into account the effects of relativity. It also does not account for the interactions between multiple electrons in a system.

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