Is Probability in Quantum States Proportional to Energy Levels?

In summary: If that is the case, then no, you cannot say anything about the probabilities of being in the different states.
  • #1
bob012345
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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 $$P_n= \frac{E_n}{\sum_{k=1}^{m}E_k}? $$
 
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  • #2
bob012345 said:
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.
 
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  • #3
In addition to Peter's excellent comment,

[tex]P_n= \frac{E_n}{\sum_{k=1}^{m}E_k}[/tex]

simplifies to

[tex]\frac{6n^2}{m(m+1)(2m+1)}[/tex]

Which if it means anything at all, probably does not mean what you want.
 
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  • #4
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.
 
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  • #5
bob012345 said:
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.

bob012345 said:
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.

bob012345 said:
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.

bob012345 said:
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.
 
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  • #6
bob012345 said:
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.
 
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  • #7
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.
 
  • #8
bob012345 said:
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?

bob012345 said:
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.
 
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  • #9
PeterDonis said:
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!
 
  • #10
bob012345 said:
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.
 
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  • #11
Nugatory said:
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?
 
  • #12
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.
 
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  • #13
hutchphd said:
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?
 
  • #14
bob012345 said:
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.
 
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  • #15
bob012345 said:
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.
 
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  • #16
bob012345 said:
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?
 
  • #17
bob012345 said:
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.
 
  • #18
Vanadium 50 said:
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.
 
  • #19
bob012345 said:
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.
 
  • #20
PeterDonis said:
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.
 
  • #21
bob012345 said:
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.
 
  • #22
Vanadium 50 said:
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.
 
  • #23
bob012345 said:
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.
 
  • #24
Vanadium 50 said:
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?
 
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  • #25
bob012345 said:
We were beyond the Boltzmann discussion.

Were we? I don't see where. I also can't tell what you are arguing any more.
 
  • #26
Vanadium 50 said:
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?
 
  • #27
I still don't know what you are arguing.
 
  • #28
Vanadium 50 said:
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.
 
  • #29
There is nothing special about Boltzmann other than that it is an example of a specific preparation.
 
  • #30
Vanadium 50 said:
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.
 
  • #31
bob012345 said:
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".

bob012345 said:
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.

bob012345 said:
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).
 
  • #32
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.
 
  • #33
bob012345 said:
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.
 
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Related to Is Probability in Quantum States Proportional to Energy Levels?

1. What is the concept of probability in relation to energy levels?

The concept of probability in relation to energy levels is based on the likelihood of a particle or system being in a certain energy state. In quantum mechanics, energy levels are described as discrete and can only take on certain values. The probability of a particle being in a specific energy level is determined by its wave function, which represents the probability amplitude of the particle's position or energy state. The higher the probability, the more likely the particle is to be found in that energy level.

2. How are energy levels determined in a quantum system?

Energy levels in a quantum system are determined by the Hamiltonian operator, which is a mathematical representation of the total energy of the system. The Hamiltonian operator acts on the wave function to determine the allowed energy levels and their corresponding probabilities. The energy levels are discrete and can only take on specific values, which are determined by the properties of the system.

3. How does the concept of probability relate to the uncertainty principle?

The uncertainty principle states that it is impossible to know with certainty both the position and momentum of a particle at the same time. This is because the act of measuring one property will inevitably affect the other property. In relation to energy levels, the uncertainty principle means that the more precisely we know the energy of a particle, the less certain we are about its position and vice versa. This is due to the wave-particle duality of quantum systems, where particles behave like waves and have a range of possible energy states.

4. Can the probability of a particle being in a certain energy level be manipulated?

Yes, the probability of a particle being in a certain energy level can be manipulated through external factors such as temperature, pressure, and electromagnetic fields. For example, increasing the temperature of a system can increase the probability of particles being in higher energy levels. Additionally, applying an external electromagnetic field can cause particles to jump between energy levels, altering their probabilities.

5. How is the concept of probability used in practical applications of quantum mechanics?

The concept of probability is used in various practical applications of quantum mechanics, such as in quantum computing and quantum cryptography. In quantum computing, the probabilities of different energy states are used to store and manipulate information in quantum bits (qubits). In quantum cryptography, the probabilities of different energy states are used to generate random cryptographic keys that are unbreakable using classical computing methods.

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