Probability and Energy Levels

[bob012345]

TL;DR Summary

What is the relationship between probability and energy levels in for a particle in a box?

Given a particle in a 1D box with a finite number of states $m$, is the probability a particle is in a certain state $n$ equal to the energy of that state divided by the sum of energies of all states? In other words, given $$ E_n = \dfrac{n^{2}h^{2}}{8mL^{2}}$$ is $$P_n= \frac{E_n}{\sum_{k=1}^{m}E_k}? $$

 

[PeterDonis]

the probability a particle is in a certain state $n$

...makes no sense. Probabilities in QM are not probabilities of being in a certain state; the state of the system is whatever you prepared that state to be when you set up the experiment. The probabilities QM calculates for you are the probabilities of obtaining particular measurement results, given that the system was prepared in a particular state; for example, the probability of measuring a certain value $E$ when you measure the energy of the system, given that you prepared the system to be in a particular state.

You need to rethink your question with this in mind.

 

[Vanadium 50]

In addition to Peter's excellent comment,

$$P_n= \frac{E_n}{\sum_{k=1}^{m}E_k}$$

simplifies to

$$\frac{6n^2}{m(m+1)(2m+1)}$$

Which if it means anything at all, probably does not mean what you want.

 

[bob012345]

If I understand you both, you are saying that without prior knowledge, I can know nothing about what energy level the particle occupies before measuring it and then it doesn't even tell me what state (energy level) it was in only what state it is in now as a result of the measurement. Is that even close to being correct?

If so, applying that to my original question then I would ask, If the system is prepared to be in the lowest energy state $n=1$, what is the probability that when measured it will be in one of the other states such as the $n^{th}$ state? Would that change if it was prepared in the $m^{th}$ state? Thanks.

 

[PeterDonis]

you are saying that without prior knowledge, I can know nothing about what energy level the particle occupies before measuring it

I said no such thing. Obviously if you prepare the system in an eigenstate of energy, that will mean the system is in just one of the available energy levels, and you'll know that that is the level it's in.

If you prepare the system in a state that is not an eigenstate of energy, the state will tell you the probability of getting each of the different possible results you could get when measuring its energy. So it's still not correct to say we know "nothing" about the system's energy in this case.

and then it doesn't even tell me what state (energy level) it was in only what state it is in now as a result of the measurement

Of course not, because what state it was in before measurement depends on how you prepared the system to begin with, i.e., how you set up the experiment. You don't need the measurement to tell you that; you already know it.

If the system is prepared to be in the lowest energy state $n=1$, what is the probability that when measured it will be in one of the other states

Zero. If the system is prepared to be in a single definite energy state (i.e., an eigenstate of energy), that means there is probability 1 of measuring the energy to be the energy of that state, and probability 0 of measuring any other energy. That is the definition of an eigenstate.

Would that change if it was prepared in the $m^{th}$ state?

The value of energy that there would be a probability 1 to measure would be whatever the energy of the $m^\text{th}$ state was, instead of the energy of the $n = 1$ state.

 

[PeterDonis]

applying that to my original question

Note that everything I have said assumes that the system in question is isolated and does not interact with any other systems (except the interaction required to measure its energy). Of course no real system is like that. For example, if the system in question is a hydrogen atom, strictly speaking, if the atom is prepared in any state other than the ground state (say the 2s state, where the electron is in the $n = 2$ energy level with zero orbital angular momentum, instead of the $n = 1$ ground state), there will be a nonzero probability for the atom to emit a photon and drop to the ground state, and that has to be taken into account when calculating the probabilities for different possible values we could measure for the atom's energy; those probabilities will change depending on how long we wait after preparing the atom before we measure it (since the longer we wait, the higher the probability that the atom will emit a photon).

Another way of looking at the above is that the states we normally call the "energy level" states of the hydrogen atom (1s, 2s, 2p, etc.) are not actually eigenstates of the full Hamiltonian (energy operator) of the atom plus the electromagnetic field. They are only eigenstates of the part of the Hamiltonian that describes the atom by itself, in isolation from the electromagnetic field. But when we measure energy, we can only measure the full Hamiltonian, so that's what we have to use to calculate probabilities.

I was ignoring such possibilities in what I said in my previous posts, because until you have a firm grasp on the isolated case, you shouldn't even be trying to add in the complications involved in the non-isolated case.

 

[bob012345]

Thanks for the above answers!

All I am trying to get at is if you only know the system has $m$ possible energy states and you do not know what state the system is in, can you say anything about the probabilities of it being in any of the $m$ states a priori or not? Thanks.

 

[PeterDonis]

if you only know the system has $m$ possible energy states and you do not know what state the system is in

Meaning, I take it, that you did not prepare the system yourself, but have simply come across it in your travels? Or had it handed to you and the only information you are given is the set of possible energy eigenstates?

can you say anything about the probabilities of it being in any of the $m$ states a priori or not?

If what I said above is correct, the answer is no. Just knowing the possible energy eigenstates, without knowing anything about how the system was prepared, tells you nothing about probabilities.

 

[bob012345]

Meaning, I take it, that you did not prepare the system yourself, but have simply come across it in your travels? Or had it handed to you and the only information you are given is the set of possible energy eigenstates?
If what I said above is correct, the answer is no. Just knowing the possible energy eigenstates, without knowing anything about how the system was prepared, tells you nothing about probabilities.

That's exactly what I meant. Thanks!

 

[Nugatory]

All I am trying to get at is if you only know the system has m possible energy states and you do not know what state the system is in, can you say anything about the probabilities of it being in any of the m states a priori or not?

Knowing the possible energy states of a system is like knowng the number of floors in an office building. It doesn't tell you anything about which floor someone might be, just gives you a list of the possibilities.

 

[bob012345]

Knowing the possible energy states of a system is like knowng the number of floors in an office building. It doesn't tell you anything about which floor someone might be, just gives you a list of the possibilities.

But why isn't there something like a Boltzmann Distribution in statistical mechanics where probability is related to energy?

 

[hutchphd]

But then it is not an isolated system...when you say I have particle in a box, that is the "entire universe". So perhaps you need to ask a different question.

 

[bob012345]

But then it is not an isolated system...when you say I have particle in a box, that is the "entire universe". So perhaps you need to ask a different question.

What question would you suggest?

 

[Nugatory]

But why isn't there something like a Boltzmann Distribution in statistical mechanics where probability is related to energy?

One box containing one particle is neither a multiparticle system nor a statistical ensemble, so that “something like” analogy won’t go very far.

 

[PeterDonis]

But why isn't there something like a Boltzmann Distribution in statistical mechanics where probability is related to energy?

To use the Boltzmann distribution, you have to have the classical equivalent of knowledge about how the system was prepared: you have to know that you have a system containing a large number of particles whose individual energies are random but the total energy of the system is fixed. Without knowledge like that you can't apply the Boltzmann distribution.

 

[Vanadium 50]

But why isn't there something like a Boltzmann Distribution in statistical mechanics where probability is related to energy?

There is, but how do you know that the system is in a state where energy occupancies are distributed that way as opposed to some other way?

 

[f95toli]

But why isn't there something like a Boltzmann Distribution in statistical mechanics where probability is related to energy?

There is if your system is connected to some sort of thermal bath. If so, the occupancy will indeed just follow a Boltzmann distribution (or more accurately a Fermi or BE distribution).

This is quite a common situation in many solid state systems/devices; but it is not a "general" rule since it is not at all obvious that all systems will be thermalised in this way.

 

[bob012345]

There is, but how do you know that the system is in a state where energy occupancies are distributed that way as opposed to some other way?

I think that goes back to what @PeterDonis said above. If the system is prepared as a mixed state with a certain energy, the probabilities of measuring each energy eigenstate will then be known a priori. Also, if Bob prepares an ensemble of identical systems in some mixed state, Alice could deduce that state by the statistical distribution of the measured eigenvalues but would not know the probabilities a priori.

 

[PeterDonis]

If the system is prepared as a mixed state with a certain energy

A mixed state will not have a definite energy. A state with a definite energy will be an eigenstate of energy and will be a pure state.

 

[bob012345]

A mixed state will not have a definite energy. A state with a definite energy will be an eigenstate of energy and will be a pure state.

I understand we can only measure eigenvalues. But we can certainly know the expectation value for the energy of the system. We should get a real number which is a weighted average of eigenvalues. It would be interpreted as the average of measurements of identically prepared systems. If a system is prepared with a Ket vector where the $c_i$ are known;

$$| \psi \rangle = \sum_i c_i |{k_i}\rangle$$

An energy expectation value for the system can be written;

$$\langle E \rangle =\langle \psi |H|\psi \rangle = \frac{\sum_i c_i^2 {E_i}}{\sum_i c_i^2}$$

This is what I meant.

 

[Vanadium 50]

But we can certainly know the expectation value for the energy of the system

That does not uniquely define the state. If you have a square well with n = 2, it has <E> = 4. If you have a system that is 5/8 n = 1 and 3/8 n = 3, it will have <E> = 4.

 

[bob012345]

That does not uniquely define the state. If you have a square well with n = 2, it has <E> = 4. If you have a system that is 5/8 n = 1 and 3/8 n = 3, it will have <E> = 4.

I said above that I am specifying the coefficients $c_i$ which will define the system uniquely. Then I showed that has a certain expectation value. The fact that there are an infinite number of other possible states that might be prepared to have the same expectation value has no bearing on my argument.

 

[Vanadium 50]

I said above that I am specifying the coefficients which will define the system uniquely.

Sure but then making it Boltzman is just a way of specifying the state.

 

[bob012345]

Sure but then making it Boltzman is just a way of specifying the state.

I lost where you are going with this? We were beyond the Boltzmann discussion. How will you make it Bolzmann?

 

[Vanadium 50]

We were beyond the Boltzmann discussion.

Were we? I don't see where. I also can't tell what you are arguing any more.

 

[bob012345]

Were we? I don't see where. I also can't tell what you are arguing any more.

I agreed with @PeterDonis in #18 to your #16, he responded with #19, and I clarified with #20. Then you wrote #21 saying I wasn't uniquely defining my state which I responded I was in #22. Then you brought up Boltzmann again which is fine. Are you referring to how to prepare a system to be Boltzmann like in the classical limit?

 

[Vanadium 50]

I still don't know what you are arguing.

 

[bob012345]

I still don't know what you are arguing.

The original question about if you can figure the probability of what energy level the system will be in a priori was answered. Then we discussed cases where the system preparation was known. I think we have covered things but I would like for you to finish your point about making it Boltzmann. What did you mean? Thanks.

 

[Vanadium 50]

There is nothing special about Boltzmann other than that it is an example of a specific preparation.

 

[bob012345]

There is nothing special about Boltzmann other than that it is an example of a specific preparation.

I'm just trying to understand what that entails and how it would apply to a quantum system of a particle in a box? Does it mean a system with very high energy far above the ground state? Thanks.

 

[PeterDonis]

I understand we can only measure eigenvalues. But we can certainly know the expectation value for the energy of the system.

If what you mean is "expectation value", then you need to say "expectation value". You can't just say "energy" and expect everyone to know that you mean "the expectation value of energy in some particular state that is not an eigenstate of energy".

I'm just trying to understand what that entails

You have been told two different ways what kind of preparation is necessary for a Boltzmann distribution to apply. I did in post #15, and @f95toli did in post #17. The two descriptions we gave are equivalent.

and how it would apply to a quantum system of a particle in a box?

It wouldn't. You have already been told why (@Nugatory did that in post #14).

 

[bob012345]

The original question of this thread has been answered and I think I have synthesized in my mind all the answers given about that and the discussion of the Boltzmann distribution so I thank everyone for their answers.

 

[PeterDonis]

The original question of this thread has been answered and I think I have synthesized in my mind all the answers given about that and the discussion of the Boltzmann distribution so I thank everyone for their answers.

Thanks for the feedback! I'm glad we were able to answer your questions.