I Particle in a box and quantization of energy

Alan Ezra

Greetings,

In the scenario of a particle in an infinite potential well, there are discrete energy levels, i.e.$E=\hbar ^2 n^2 \pi ^2/ (2 m L^2)$ where L is the width of the potential well, and n takes on positive integers. But what will happen if I put a particle of energy $E_i$ that is not a multiple of $E=\hbar ^2 \pi ^2/ (2 m L^2)$ into the potential well? I am thinking that the answer may be that there is some uncertainty in the $E_i$ so the particle can always takes on an energy level in the potential well within this allowable range of energy. Is this correct? What exactly is will happen in this case? Thank you!

Best,
Alan

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Sclerostin

Your question interests me, also. I am a layman where this is concerned. But Brian Cox writes that "every similar fermion (eg, think electrons with like spin) has a slightly different energy level from every other in the Universe". You would think you would begin to move into a different energy state.
And the Universe is likely bigger than we can tell, as distant information can't reach us.

Alan Ezra

Greetings,

In the scenario of a particle in an infinite potential well, there are discrete energy levels, i.e.$E=\hbar ^2 n^2 \pi ^2/ (2 m L^2)$ where L is the width of the potential well, and n takes on positive integers. But what will happen if I put a particle of energy $E_i$ that is not a multiple of $E=\hbar ^2 \pi ^2/ (2 m L^2)$ into the potential well? I am thinking that the answer may be that there is some uncertainty in the $E_i$ so the particle can always takes on an energy level in the potential well within this allowable range of energy. Is this correct? What exactly is will happen in this case? Thank you!

Best,
Alan
Or will it just emit a photon and go to the next lower energy level?

Mentor

PeterDonis

Mentor
what will happen if I put a particle of energy $E_i$ that is not a multiple of $E=\hbar ^2 \pi ^2/ (2 m L^2)$ into the potential well?
How are you going to "put" the particle in? If it's an infinite potential well, there is no way for a particle to exist outside it in the first place.

Nugatory

Mentor
Or will it just emit a photon and go to the next lower energy level?
Short answer: there is some uncertainty until you measure the energy and get one of the allowed results. Energy will be conserved because there has to be some exchange of energy between the particle and the measuring device; no matter the measurement result the total energy of the entire system (particle, measuring device, whatever apparatus you used to get the particle into the well in the first place) will be conserved.
(It's worth noting that the idea of introducing a particle into an infinite square well is seriously problematic - where did it come from? Better to imagine that it was there all along and we excited it with an arbitrary amount of energy).

The energy eigenfunctions that you found by solving the time-independent Schrodinger equation, $\psi_n(x)=\sqrt{2/L}\sin\frac{n\pi{x}}{L}$ each with a particular discrete energy $E_n$, are not the possible states for the particle; they are building blocks for the state and they give us the possible results of an energy measurement.

In general, the state of the particle will be a solution of the time-dependent Schrodinger equation; a bit of algebra will satisfy you that any linear combination (called a "superposition") of the functions $\psi_n(x,t)=\psi_n(x)e^{-iE_nt/\hbar}$ is such a solution. When we measure the energy, the result will be one of the $E_n$, the probability of getting any particular one is given by the Born rule, and after we get the result $E_n$ the wave function will have collapsed to $\psi_n(x,t)$.

Sclerostin

re Post 4,
Brian Cox book ISBN 978-0-241-95270-2
The Quantum Universe
Chapter 8

The wave "leaks" outside the well of Alan's width L, and so has a possibility of a variation on its energy ---> a slightly different set of sin waves.
But can all the Fourier series involved allow this super-vast individuality?

Nugatory

Mentor
re Post 4,
Brian Cox book ISBN 978-0-241-95270-2
The Quantum Universe
Chapter 8
That's a popularization, as opposed to a serious textbook. It will give you a feel for how quantum mechanics works but you have to be very cautious about drawing any further conclusions from it.
The wave "leaks" outside the well of Alan's width L, and so has a possibility of a variation on its energy ---> a slightly different set of sin waves.
@Alan Ezra specifically stated "infinite square well", and there is no leakage in that case - $\psi(x)$ is zero at the boundaries and outside the well.

Sclerostin

Thank you. I just thought it might be correct his having being one of the CERN crowd.
I (personally) wasn't drawing any conclusions, just finding his argument OK but his conclusion hard to swallow. But I'll "leave it to the experts". Thanks again.

vanhees71

Gold Member
Greetings,

In the scenario of a particle in an infinite potential well, there are discrete energy levels, i.e.$E=\hbar ^2 n^2 \pi ^2/ (2 m L^2)$ where L is the width of the potential well, and n takes on positive integers. But what will happen if I put a particle of energy $E_i$ that is not a multiple of $E=\hbar ^2 \pi ^2/ (2 m L^2)$ into the potential well? I am thinking that the answer may be that there is some uncertainty in the $E_i$ so the particle can always takes on an energy level in the potential well within this allowable range of energy. Is this correct? What exactly is will happen in this case? Thank you!

Best,
Alan
You cannot put a particle with a precisely determined energy $E_i$ into the box. This simply doesn't exist. What you can do is to prepare any other than an energy eigenstate as an initial state with a given energy expectation value $E_i$. As you intuitively got right, this implies an uncertainty in the energy value of the particle.

To argue the other way around: If your particle has a definite energy in the initial state, then it must be prepared in the corresponding energy eigenstate. Of course the energy eigenstates are also the stationary states of the system, i.e., then the system always stays in this energy eigenstate, and your energy value stays determined at this very eigenvalue.

Nugatory

Mentor
Thank you. I just thought it might be correct his having being one of the CERN crowd.
I (personally) wasn't drawing any conclusions, just finding his argument OK but his conclusion hard to swallow. But I'll "leave it to the experts". Thanks again.
It's not exactly incorrect, but it is limited by not using the math. "Every similar fermion (eg, think electrons with like spin) has a slightly different energy level from every other in the Universe" is about as good as an English-language description can get; a more rigorous explanation would involve the Hamiltonian of multiparticle systems and the behavior of fermions under particle exchange. For what it's worth, the more rigorous explanation is also much easier to swallow once you understand it.

PeterDonis

Mentor
"Every similar fermion (eg, think electrons with like spin) has a slightly different energy level from every other in the Universe" is about as good as an English-language description can get
I'm actually not clear about what, exactly, he's referring to. You mention the behavior of fermions under particle exchange, but that has nothing to do with "energy levels". If I have two electrons in a helium atom in its ground state, their wave function is antisymmetric under particle exchange, but in what sense do they have "slightly different energy levels"?

Nugatory

Mentor
I'm actually not clear about what, exactly, he's referring to. You mention the behavior of fermions under particle exchange, but that has nothing to do with "energy levels". If I have two electrons in a helium atom in its ground state, their wave function is antisymmetric under particle exchange, but in what sense do they have "slightly different energy levels"?
Cox is trying to explain how no two fermions are ever in EXACTLY the same state, without recourse to any rigorous definition of what exactly the state is or what the quantum system we're considering is. (and "about as good as it can get" is more an expression of sympathy for a good effort than a ringing endorsement).

In any case, this question is a digression from the original question, which seems to be about how we get from the eigenfunctions of the TISE to an arbitrarily prepared state.

PeterDonis

Mentor
Cox is trying to explain how no two fermions are ever in EXACTLY the same state, without recourse to any rigorous definition of what exactly the state is or what the quantum system we're considering is.
Even so, I don't see how "slightly different energy levels" is less misleading than "slightly different states". The former seems more misleading to me, not less.

PeroK

Homework Helper
Gold Member
2018 Award
Cox is trying to explain how no two fermions are ever in EXACTLY the same state, without recourse to any rigorous definition of what exactly the state is or what the quantum system we're considering is. (and "about as good as it can get" is more an expression of sympathy for a good effort than a ringing endorsement).
In a post on here a while ago, there was a link to a reasonable looking video from someone who obviously knew a lot of QM. But, they had got the idea that the Pauli exclusion principle meant that every electron in every hydrogen atom in the universe had a different energy.

Here it is:

Maybe Brian Cox got the wrong end of the stick as well? Why say "energy" when you mean "state", which includes position? And, two electrons in a Helium atom can effectively have the same position and energy. It's only when you consider the direction of the spin that they must be different.

bhobba

Mentor
Maybe Brian Cox got the wrong end of the stick as well?
I personally like Brian Cox but in his popularization's, like a number of physicists that try their hand at this difficult task, he is sometimes 'controversial':

My solution was simple - between any two real numbers is another real number so all that would happen is the energy levels of the diamond will occupy the infinite number of slots available (we will not go into that the wave-functions need to overlap for it to be an issue - it one of the niceties Brian correctly left out for a lay audience). But it did create an interesting thread on this forum at the time.

A much better book to learn QM is Susskind's - its the real deal:
https://www.amazon.com/dp/0465062903/?tag=pfamazon01-20

Thanks
Bill

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