# Stark effect, mixing state of 2s and 2p

• luxiaolei
In summary, the conversation discusses the meaning of the mixing state of 2p and 2s in a hydrogen atom when an external em field is applied. It is explained that the degenerate states split into three classes and that the upper and lower energy states can be interpreted as symmetric and antisymmetric combinations of the initial states. It is also mentioned that the energy splitting is due to the non-vanishing dipole moment of the states.
luxiaolei
hi all, can anyone help me understanding what is the meaning of the mixing state of 2p and 2s? thanks in advance.

when an electron in n=2 sate in a hydrogen atom, experience a external em field, then the degenerate states split into 3 classes, which are:

1/sqrt2( 210 + 200 ) with higher energy
2 1 +1 and 2 1 -1 with the same energy as before
1/sqrt( 210 - 200) with lower energy

my question is how to interpretate the upper and lower energy states?
for upper level, i think it as there is a half probability to find the electron in 210 state and half in 200 state. but if i think in terms of this way, then the lower level becomes half probability in 210 state(which contridicts with it can also in uper level) and -1/2 probability in 200 state( which is obviously wrong).

thanks for any helps.

luxiaolei said:
my question is how to interpretate the upper and lower energy states?
for upper level, i think it as there is a half probability to find the electron in 210 state and half in 200 state. but if i think in terms of this way, then the lower level becomes half probability in 210 state(which contridicts with it can also in uper level) and -1/2 probability in 200 state( which is obviously wrong).

No, it does not work that way. Note that you are adding the wavefunctions and then square to get the probability density. Adding the already squared wavefunctions would have the meaning you are thinking about. But that will not give the eigenstates of the problem.

Now for visualization: Your Hamiltonian couples two degenerate levels. As a rough analogy think of a common spring pendulum in classical mechanics. If you have two identical ones of those and couple them with another spring, one of the standard problems in classical mechanics is to solve the coupled problem via some matrix eigenvalue equation and to show that it is a bad idea to express the solution in terms of the eigenstates of the bare single pendulum. Instead these two can exchange energy and the commonly chosen basis for describing all kinds of motion of the coupled pendulum are the states where the two masses oscillate in phase or exactly out of phase corresponding to the two masses showing symmetric or antisymmetric motion.

So in the case of the linear Stark effect, the eigenstates are also the symmetric and antisymmetric combinations of the initial states. If you sketch them, you will find that they cannot be eigenstates of the parity operator and have a non-vanishing dipole moment. The latter also is the reason for the energy splitting.

Cthugha said:
No, it does not work that way. Note that you are adding the wavefunctions and then square to get the probability density. Adding the already squared wavefunctions would have the meaning you are thinking about. But that will not give the eigenstates of the problem.

Now for visualization: Your Hamiltonian couples two degenerate levels. As a rough analogy think of a common spring pendulum in classical mechanics. If you have two identical ones of those and couple them with another spring, one of the standard problems in classical mechanics is to solve the coupled problem via some matrix eigenvalue equation and to show that it is a bad idea to express the solution in terms of the eigenstates of the bare single pendulum. Instead these two can exchange energy and the commonly chosen basis for describing all kinds of motion of the coupled pendulum are the states where the two masses oscillate in phase or exactly out of phase corresponding to the two masses showing symmetric or antisymmetric motion.

So in the case of the linear Stark effect, the eigenstates are also the symmetric and antisymmetric combinations of the initial states. If you sketch them, you will find that they cannot be eigenstates of the parity operator and have a non-vanishing dipole moment. The latter also is the reason for the energy splitting.

Thank you very much indeed, now I am happy now. Great answer.

## 1. What is the Stark effect?

The Stark effect is the splitting of spectral lines in the presence of an electric field. This phenomenon was first observed by Johann Stark in 1913.

## 2. How does the Stark effect occur?

The Stark effect occurs due to the interaction between the electric field and the charged particles in an atom or molecule. This interaction causes the energy levels of the particles to shift, resulting in the splitting of spectral lines.

## 3. What is the mixing state of 2s and 2p in the Stark effect?

In the Stark effect, the 2s and 2p orbitals of an atom or molecule are mixed together, resulting in a new set of energy levels. This mixing is caused by the interaction between the electric field and the charged particles.

## 4. How does the mixing of 2s and 2p affect the spectral lines?

The mixing of 2s and 2p in the Stark effect results in the spectral lines becoming split into multiple components. The number and spacing of these components depend on the strength of the electric field and the properties of the atom or molecule.

## 5. What are the applications of the Stark effect?

The Stark effect has many practical applications, such as in laser technology, atomic clocks, and spectroscopy. It is also used in studying the properties of atoms and molecules and in the development of new materials.

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