Elementary question about energy

[jonjacson]

Hi folks,

I just want to check if I understand this.

We start the calculation using the electric potential energy, which is a continuous function that for every value of the distance r, tells you the value of the potential energy "spent" to get the system to that state. We use this function in the Hamiltonian, together with the kinetic energy term, and integrating the Schrodinger differential equation we conclude that the electron is only able to have certan discrete energy values, and for everyone of those eigenvalues, there is a corresponding "eigenstate", which is a probability distribution.

But, How is that possible?

I mean, the electric potential energy function we used as an input gives for every value of r, a value of the energy. But we conclude after integrating the equations that there is a definite energy, associated with an eigenstate, in this particular eigenstate the electron could have different positions around the nucleus. (I have seen this distributions in several websites).

Can somebody explain this?

 

[Orodruin]

The potential energy is not the only contribution to the hamiltonian.

 

[houlahound]

Quantisation occurs due to the boundaries of the system.

 

[Orodruin]

Quantisation occurs due to the boundaries of the system.

This system has no boundary. Your answer has nothing to do with the question.

 

[secur]

We're talking about a spherically symmetric differential equation whose solutions are the spherical harmonics. The "condition" imposed is that the (angular) solution at a point (x) must equal the solution at point (x+2 pi). This can be viewed as a "boundary condition", and as the reason for the discrete energy values.

 

[Orodruin]

We're talking about a spherically symmetric differential equation whose solutions are the spherical harmonics. The "condition" imposed is that the (angular) solution at a point (x) must equal the solution at point (x+2 pi). This can be viewed as a "boundary condition", and as the reason for the discrete energy values.

By this logic the entire spectrum should be discrete, which is not the case.

It is also not a question of 2pi periodic functions, but spherical harmonics.

Furthermore, the actual question is not "why are the energy levels quantised?" but "why do the energy levels have definite energy although they have a distribution over positions with different potential?" These are completely different questions.

 

[secur]

Furthermore, the actual question is not "why are the energy levels quantized?" but "why do the energy levels have definite energy although they have a distribution over positions with different potential?"

Okay, reading the OP, I see what you mean. It's not clear; I'd say the discreteness is also relevant, but let's see what OP says.

Of course, we get the same energy level with different radii because kinetic energy is also involved, not just potential.

By this logic the entire spectrum should be discrete, which is not the case.

Yes, the spectrum becomes continuous in certain ways. For one, as r goes to infinity the spectrum approaches continuity. Two, due to HUP each line actually has a finite width, within which it's continuous. Generally, any real-number physical solution, throughout all of science, cannot be precisely one number. It always covers a small neighborhood, or interval. None of these details invalidates the fact that the discreteness (such as it is) of energy eigenvalues of hydrogen are (or, can be considered) due to a boundary condition. This is just a basic fact about solutions of a very elementary differential equation.

 

[jonjacson]

Maybe the source of my misunderstanding is the kinetic energy so, I will try to explain what I thought. Kinetic energy basically is given by the momentum operator, but p is not defined like the classical momentum x2-x1/t when the increment t gets smaller, instead it is defined as a partial derivative acting on the wave function. So if the electron is in the state n=2, l=1, there is a fixed eigenstate, a fixed distribution of probabilities so I thought the momentum is zero for a fixed eigenstate. The shape of the wave is fixed, there is no group velocity, but inside this fixed distribution, several positions are allowed, Doesn't that mean the electric potential energies are different?

 

[Nugatory]

So if the electron is in the state n=2, l=1, there is a fixed eigenstate, a fixed distribution of probabilities so I thought the momentum is zero for a fixed eigenstate. The shape of the wave is fixed, there is no group velocity, but inside this fixed distribution, several positions are allowed, Doesn't that mean the electric potential energies are different?

The momentum is not zero for that eigenstate (or any of the others). The shape of the wave is fixed, but that doesn't mean that the electron's momentum is fixed, and it certainly doesn't mean that the electron isn't moving. It means that the probability of finding the electron with any particular position or momentum (when and if we measure one or the other) isn't changing over time.

 

[jonjacson]

The momentum is not zero for that eigenstate (or any of the others). The shape of the wave is fixed, but that doesn't mean that the electron's momentum is fixed, and it certainly doesn't mean that the electron isn't moving. It means that the probability of finding the electron with any particular position or momentum (when and if we measure one or the other) isn't changing over time.

But why is that there are eigenfunctions with two peaks? The derivative over x must be zero in both peaks, but positions are different, doesn't this give us 2 energies?

 

[Nugatory]

But why is that there are eigenfunctions with two peaks? The derivative over x must be zero in both peaks, but positions are different, doesn't this give us 2 energies?

If we write the wave function as a function of position and it has two peaks, that tells us that there are two regions in space where the electron is likely to be found. But the energy depends on more than just the position, so the possibility of finding the electron at different locations does not necessarily imply different energies.

 

[Orodruin]

Yes, the spectrum becomes continuous in certain ways. For one, as r goes to infinity the spectrum approaches continuity.

This makes no sense. We are talking about the energy levels of a 1/r potential. The spectrum is a result of the entire Hamiltonian, including all r.

You may be referring to the fact that the bound states lie closer and closer in energy for the higher states. You are then completely forgetting the entire continuum of scattering states with positive energy. This set of states do not have discretised energy levels.

This is just a basic fact about solutions of a very elementary differential equation.

One which you are failing to realize has a continuous part and a discrete part to its spectrum. As such, your logic fails.

You can also claim the same "boundary conditions" in the absence of a potential when you only have scattering states.

None of these details invalidates the fact that the discreteness (such as it is) of energy eigenvalues of hydrogen are (or, can be considered) due to a boundary condition.

The discrete part of the spectrum arises from the radial solution, not the angular one. The angular one affects the radial differential equation, but if you take away the radial potential - the spectrum is purely continuous.

The momentum is not zero for that eigenstate (or any of the others).

To be more precise, the expectation value if the squared momentum is non-zero. The expectation value of the momentum is zero.

it certainly doesn't mean that the electron isn't moving.

I find talking about things "moving" at this level can be quite misleading and often enforces people's thinking of the electron as a small billiard ball. It also leads to the misconception that the system should be emitting radiation. It is a stationary state - but it has a non zero squared momentum expectation value.

But why is that there are eigenfunctions with two peaks? The derivative over x must be zero in both peaks, but positions are different, doesn't this give us 2 energies?

The squared momentum depends on the ratio of the second derivative of the wave function to the wave function itself, not on the first derivative. This ratio will decrease as you increase r in the bound states - by the corresponding amount as the potential increases.

 

[jonjacson]

If we write the wave function as a function of position and it has two peaks, that tells us that there are two regions in space where the electron is likely to be found. But the energy depends on more than just the position, so the possibility of finding the electron at different locations does not necessarily imply different energies.

This makes no sense. We are talking about the energy levels of a 1/r potential. The spectrum is a result of the entire Hamiltonian, including all r.

You may be referring to the fact that the bound states lie closer and closer in energy for the higher states. You are then completely forgetting the entire continuum of scattering states with positive energy. This set of states do not have discretised energy levels.One which you are failing to realize has a continuous part and a discrete part to its spectrum. As such, your logic fails.

You can also claim the same "boundary conditions" in the absence of a potential when you only have scattering states.The discrete part of the spectrum arises from the radial solution, not the angular one. The angular one affects the radial differential equation, but if you take away the radial potential - the spectrum is purely continuous.To be more precise, the expectation value if the squared momentum is non-zero. The expectation value of the momentum is zero.I find talking about things "moving" at this level can be quite misleading and often enforces people's thinking of the electron as a small billiard ball. It also leads to the misconception that the system should be emitting radiation. It is a stationary state - but it has a non zero squared momentum expectation value.The squared momentum depends on the ratio of the second derivative of the wave function to the wave function itself, not on the first derivative. This ratio will decrease as you increase r in the bound states - by the corresponding amount as the potential increases.

Well, in other words, a classical mass-spring system analyzed from the point of view of the energy is very simple. When the spring goes through the origin the speed is maximum, the kinetic energy is maximum, and potential energy is 0. When the mass is on the limits, the speed is 0, kinetic energy is 0, and all the energy is stored as elastic potential energy.

Is there explained anyhwere a similar analysis for the Hydrogen atom?

I mean normally spherical armonics are presented, so we know what is the probability of finding a particle at a certain point, but what is the "momentums distribution"?

To make everything more complex Heisenberg says we can't know both at the same time, so I don't know how to do this analysis.

 

[secur]

You may be referring to the fact that the bound states lie closer and closer in energy for the higher states.

Yes, that's exactly what I'm referring to. I was trying to guess why you said there was a continuous spectrum.

You are then completely forgetting the entire continuum of scattering states with positive energy. This set of states do not have discretized energy levels.

I'm not forgetting about scattering states, I didn't know scattering was involved at all! Yes, they do include states of continuous spectrum. Please remember the OP title is "Elementary question about energy". It doesn't mention scattering. He says "... integrating the Schrodinger differential equation we conclude that the electron is only able to have certain discrete energy values", etc.

You can also claim the same "boundary conditions" in the absence of a potential when you only have scattering states.

OP says "in this particular eigenstate the electron could have different positions around the nucleus." Now, if there's a nucleus, there must be a potential. There's no such thing as a nucleus without a (positive) charge. He doesn't mention scattering.

The discrete part of the spectrum arises from the radial solution, not the angular one ...

That's true. It's the radial part that leads to the allowed, discrete, energy levels. Let's review a bit, get on the same page. In spherical coordinates, solution of Schrodinger's eqn for hydrogen is $\psi = \psi(r,\theta,\psi)$. Separate the variables, $\psi = \psi(R(r), Y(\theta,\psi))$. The solutions for $Y(\theta,\psi))$ are the spherical harmonics, which are not dependent on the Coulomb potential. That's the angular part of the function. They are characterized by quantum numbers l and m. The solutions of the radial part, $R(r)$, are characterized by n and l, and lead to discrete solutions.

Now, the situation I'm comparing this to would be a version of the equation without a spherical potential. The potential would be simply 1/z, let's say, where z is a Cartesian coordinate - height, perhaps. That has continuous spectrum of energy levels. It's in that sense that the "boundary conditions" arising from 2$\pi$ periodicity are necessary for discrete solutions - I think that's fair to say.

Ok, I didn't read the OP as carefully as I should have. For that reason, I guess, we have been talking about two different problems, which has caused confusion. But no matter how carefully I read the OP, I still don't see why you think this question involves scattering. It probably has something to do with the above-mentioned confusion and is not worth the trouble of sorting out.

 

[jonjacson]

Yes, that's exactly what I'm referring to. I was trying to guess why you said there was a continuous spectrum.
I'm not forgetting about scattering states, I didn't know scattering was involved at all! Yes, they do include states of continuous spectrum. Please remember the OP title is "Elementary question about energy". It doesn't mention scattering. He says "... integrating the Schrodinger differential equation we conclude that the electron is only able to have certain discrete energy values", etc.
OP says "in this particular eigenstate the electron could have different positions around the nucleus." Now, if there's a nucleus, there must be a potential. There's no such thing as a nucleus without a (positive) charge. He doesn't mention scattering.
That's true. It's the radial part that leads to the allowed, discrete, energy levels. Let's review a bit, get on the same page. In spherical coordinates, Schrodinger's eqn for hydrogen is $\psi = \psi(r,\theta,\psi)$. Separate the variables, $\psi = \psi(R(r), Y(\theta,\psi))$. The solutions for $Y(\theta,\psi))$ are the spherical harmonics, which are not dependent on the Coulomb potential. That's the angular part of the function. They are characterized by quantum numbers l and m. The solutions of the radial part, $R(r)$, are characterized by n and l, and lead to discrete solutions.

Now, the situation I'm comparing this to would be a version of the equation without a spherical potential. The potential would be simply 1/z, let's say, where z is a Cartesian coordinate - height, perhaps. That has continuous spectrum of energy levels. It's in that sense that the "boundary conditions" arising from 2$\pi$ periodicity are necessary for discrete solutions - I think that's fair to say.

Ok, I didn't read the OP as carefully as I should have. For that reason, I guess, we have been talking about two different problems, which has caused confusion. But no matter how carefully I read the OP, I still don't see why you think this question involves scattering. It probably has something to do with the above-mentioned confusion and is not worth the trouble of sorting out.

No, I don't talk about scattering.

THe main issue is the momentum, that is my question.

We know where the probability of finding the electron is higher, but, What is the "momentum distribution"?

 

[Orodruin]

OP says "in this particular eigenstate the electron could have different positions around the nucleus." Now, if there's a nucleus, there must be a potential. There's no such thing as a nucleus without a (positive) charge. He doesn't mention scattering.

The mention of scattering states is not a reply to the OP's question. It is an answer to your (faulty) argument that expanding in spherical harmonics leads to a discrete spectrum - it does not. As shown by my other examples. The discrete part of the spectrum arises due to the radial potential.

In the case of the 1/z potential the eigenfunctions are still 2pi periodic in the angles (for any fixed r) It is just that you do not separate the radial differential equation by expanding in spherical harmonics.

But no matter how carefully I read the OP, I still don't see why you think this question involves scattering. It probably has something to do with the above-mentioned confusion and is not worth the trouble of sorting out.

In general, the same issues would arise for the scattering states. It does not treat discrete/continuous as much as any energy eigenstate. Also, the mention of scattering states is not in reply to the OP. It is in reply to your assertion that requiring the wave function to be periodic in the angles implies a discrete spectrum.

 

[Nugatory]

Well, in other words, a classical mass-spring system analyzed from the point of view of the energy is very simple. When the spring goes through the origin the speed is maximum, the kinetic energy is maximum, and potential energy is 0. When the mass is on the limits, the speed is 0, kinetic energy is 0, and all the energy is stored as elastic potential energy.

Right, but that's a classical analysis of a classical system. The classical mass has a position even if we don't measure it. It has a velocity which is defined as the change in position with respect to time whether we measure it or not, and along with that velocity it has a momentum which is the product of the velocity and the mass. It has a total energy which is the sum of the kinetic energy ($p^2/2m$, a function of the momentum) and the potential energy (a function of the position). All of these properties are connected and change together under the influence of Newton's laws.

However, none of this works with a quantum system like an electron. It is not a little ball moving around the nucleus in response to classical forces, and it does not have any position unless we actually measure the position - and note that that is "does not have any position", not "has some position but we don't know what it is". It has no position if we don't we measure, the same way that I don't have a lap except when I'm sitting down and you don't have a fist when your hand is open. Thus, we cannot treat the particle as a classical object with a position, momentum, velocity, kinetic and potential energies that are all connected and changing together.

In fact, it is impossible to simultaneously determine the position, momentum, and energy of the electron. If it is in an energy eigenstate, then a measurement of the energy will yield the corresponding energy eigenvalue with 100% certainty; but a position or momentum measurement will give us some random value according to the probability distribution you mentioned above. Furthermore, after we measure the position or the momentum, the particle will no longer be in the original energy eigenstate so we can't go back and measure one of the other properties. All quantum mechanics gives us is the probabilities of getting various measurement results for the measurements that we actually make.

To make everything more complex Heisenberg says we can't know both at the same time, so I don't know how to do this analysis.

You don't know how because no one does. It can't be done.

 

[Nugatory]

mean normally spherical harmonics are presented, so we know what is the probability of finding a particle at a certain point, but what is the "momentums distribution"?

Before you try this with the spherical harmonics in three dimensions, you should work thorugh the equivalent problem for a particle free to move in one dimension. This is a standard intro problem in first-year QM courses, and introduces the essential concept with equations much less challenging than the spherical harmonics.

A preview: In one dimension you can write the wave function as a function of position, $\psi(x)$. There's a fairly straightforward transformation (it's basically a Fourier transform) that turns this into a function of momentum, often written as $\phi(p)$ (and because we're working in one dimension, $p$ can be a signed scalar not a multi-component vector). The probability of finding the particle between $x_0$ and $x_0+dx$ is given by $\psi^*(x_0)\psi(x_0)dx$ and the probability of finding it with momentum between $p_0$ and $p_0+dp$ is $\phi^*(p_0)\phi(p_0)dp$.

 

[secur]

... we have been talking about two different problems, which has caused confusion. ... It probably has something to do with the above-mentioned confusion and is not worth the trouble of sorting out.

The mention of scattering states is not a reply to the OP's question. It is an answer to your (faulty) argument that expanding in spherical harmonics leads to a discrete spectrum - it does not. As shown by my other examples.

Ok, as I suspected, the mention of scattering is due to the initial confusion. You're giving an example of another spherically symmetric problem (scattering) without discrete spectra, to show that can't be the reason for discrete eigenvalues in this case. That makes sense, as far as it goes. Well, I explained the way I was thinking of it, still think it's legitimate. It's not about spherical harmonics. Nevertheless the discreteness can be seen as due to the fact that the problem is spherical.

In the case of the 1/z potential the eigenfunctions are still 2pi periodic in the angles (for any fixed r) It is just that you do not separate the radial differential equation by expanding in spherical harmonics.

As far as I can tell you didn't quite get my point. I'm comparing the spherical potential to a non-spherical 1/z, with no r (=radius) involved. Only z, representing height. The potential being constant in the other dimensions (x and y). That gives a continuous spectrum. The solutions are parabolic, and any energy level is possible. However when you wrap the 1/z potential around a nucleus - becoming 1/r - then you get a discrete spectrum. True, that comes from the radial solution. But the reason there are radial solutions is that the overall problem is spherical. We wouldn't even use the word "radial" otherwise. So in this sense, the discreteness is due to that sphericality, although it's not due to the angular part of the solution. Admittedly it's a subtle point, and off-topic.

 

[Orodruin]

You're giving an example of another spherically symmetric problem (scattering) without discrete spectra,

It is not a different problem. It is the same Hamiltonian and so the bound and scattering states are eigenfunctions of the same operator. It is just an operator whose spectrum has a discrete and a continuous part.

Nevertheless the discreteness can be seen as due to the fact that the problem is spherical.

No it cannot, as I have explained several times above. There are systems with bound states that do not have spherical symmetry and there are systems with spherical symmetry that do not have a discrete part of the spectrum.

As far as I can tell you didn't quite get my point. I'm comparing the spherical potential to a non-spherical 1/z, with no r (=radius) involved. Only z, representing height. The potential being constant in the other dimensions (x and y).

I got you fine. Are you claiming that you do not get the same spatial wave function back after a 2pi rotation? That would make the wave function have a branch cut which is not how we usually operate. As I said, your wave function will still be 2pi periodic - it will just not factorise in angular and radial part.

However when you wrap the 1/z potential around a nucleus - becoming 1/r - then you get a discrete spectrum.

Yes, but the symmetry is not the main reason you get a discrete spectrum. It is that there are points in the potential that are lower than the potential at infinity that allows a bound state. Not the fact that the potential has spherical symmetry.

 

[jonjacson]

The squared momentum depends on the ratio of the second derivative of the wave function to the wave function itself, not on the first derivative. This ratio will decrease as you increase r in the bound states - by the corresponding amount as the potential increases.

THanks, very interesting.

but a position or momentum measurement will give us some random value according to the probability distribution you mentioned above. Furthermore, after we measure the position or the momentum, the particle will no longer be in the original energy eigenstate so we can't go back and measure one of the other properties. All quantum mechanics gives us is the probabilities of getting various measurement results for the measurements that we actually make.

You don't know how because no one does. It can't be done.

THat is what I would like to find.

Before you try this with the spherical harmonics in three dimensions, you should work thorugh the equivalent problem for a particle free to move in one dimension. This is a standard intro problem in first-year QM courses, and introduces the essential concept with equations much less challenging than the spherical harmonics.

A preview: In one dimension you can write the wave function as a function of position, $\psi(x)$. There's a fairly straightforward transformation (it's basically a Fourier transform) that turns this into a function of momentum, often written as $\phi(p)$ (and because we're working in one dimension, $p$ can be a signed scalar not a multi-component vector). The probability of finding the particle between $x_0$ and $x_0+dx$ is given by $\psi^*(x_0)\psi(x_0)dx$ and the probability of finding it with momentum between $p_0$ and $p_0+dp$ is $\phi^*(p_0)\phi(p_0)dp$.

That is what I was looking for, thanks. I will try to read on this.