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I Energy of a number of particles

  1. Mar 11, 2016 #1
    It is sometimes useful to find the average energy of a certain number [itex]N[/itex] of particles contained in a box of volume [itex]V[/itex].
    In order to find this quantity, the total energy is required and then divided by [itex]N[/itex]. The result is

    [itex]E_{average} = \displaystyle \frac{1}{N} \sum_{n = 1}^{N} \left| a_n \right|^2 E_n[/itex]

    where [itex]\left| a_n \right|^2[/itex] is the probability, for the [itex]n[/itex]-th particle, of having energy [itex]E_n[/itex].
    If the wavefunction related to the [itex]n[/itex]-th particle is [itex]\psi_n (\mathbf{r}, t)[/itex], a total wavefunction can be built:

    [itex]\psi (\mathbf{r}, t) = \displaystyle \sum_{n = 1}^N a_n \psi_n (\mathbf{r}, t)[/itex]

    and the above result is obtained by computing the energy of this (global) wavefunction, divided by [itex]N[/itex]:

    [itex]E_{average} = \displaystyle \frac{1}{N} \int_{V} \psi^* (\mathbf{r}, t) \mathcal{H} \psi (\mathbf{r}, t) dV = \displaystyle \frac{1}{N} \int_{V} \psi^* (\mathbf{r}, t) \mathcal{H} \left( \displaystyle \sum_{n = 1}^N a_n \psi_n (\mathbf{r}, t) \right) dV =[/itex]

    [itex]= \displaystyle \frac{1}{N} \int_{V} \psi^* (\mathbf{r}, t) \left( \displaystyle \sum_{n = 1}^N a_n E_n \psi_n (\mathbf{r}, t) \right) dV = \displaystyle \frac{1}{N} \int_{V} \left( \sum_{m = 1}^N a_m^* \psi_m^* (\mathbf{r}, t) \right) \left( \displaystyle \sum_{n = 1}^N a_n E_n \psi_n (\mathbf{r}, t) \right) dV[/itex]

    (and then the orthonormality of the [itex]\psi_n[/itex] must be used).

    [itex]E_n[/itex] is the energy related to the [itex]n[/itex]-th wavefunction, that is the eigenvalue of the hamiltonian

    [itex]\mathcal{H} \psi_n (\mathbf{r}, t) = E_n \psi_n (\mathbf{r}, t)[/itex]

    But if the single [itex]\psi_n (\mathbf{r}, t)[/itex] must satisfy by itself the normalization condition

    [itex]\displaystyle \int_{V} \left| \psi_n (\mathbf{r}, t) \right|^2 dV = 1[/itex]

    why an [itex]a_n[/itex] has to be used? I've not found this computation on the web, but just in some notes.
    If the total number of particles is [itex]N[/itex] and the [itex]n[/itex]-th particle must have its own wavefunction
    [itex]\psi_n (\mathbf{r}, t)[/itex], how can we consider a probability that the particle will have this wavefunction?
  2. jcsd
  3. Mar 11, 2016 #2


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    The above lines do not look fine to me. ##\psi_n## is the wavefunction of a single particle, and then you construct the wavefunction of the entire particles as a linear combination of the wavefunction of individual particles. Is it your own proposal or did you read it somewhere?
    Last edited: Mar 11, 2016
  4. Mar 11, 2016 #3
    I read it in the notes I found. The entire computation is made through the use of such a ##\psi##. I can understand your doubt. If you should solve the same problem, how could you suggest to proceed instead?
  5. Mar 11, 2016 #4


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    I think the system can be modeled as 3D infinite well problem. So that the Hamiltonian should look like
    H = -\frac{\hbar^2}{2m}\sum_{i=1}^N \nabla_i^2 + \sum_{i<j} k\frac{e^2}{|\mathbf{r}_i - \mathbf{r}_j|}
    (I have assumed that all particles are identical). It's very difficult, if not impossible, to solve the Schroedinger equation corresponding to the above Hamiltonian. However, if for some reason, the second term which is the interaction potential between the particles can be neglected, then the solution takes a simple form
    \psi_{n_1n_2\ldots n_N}(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N) = \prod_{i=1}^N \psi_{n_i}(\mathbf{r}_i)
    where ##\psi(\mathbf{r}_i)## is a solution of ##-\frac{\hbar^2}{2m} \nabla_i^2 \psi_{n_i}(\mathbf{r}_i)= E_{n_i}\psi_{n_i}(\mathbf{r}_i)##. Now depending on whether the particle belongs to boson or fermion, the proper wavefunction must be symmetrized or antisymmetrized.
  6. Mar 14, 2016 #5
    Yes, it is certainly neglected to simplify the problem, with no interaction between particles.
    I forgot to say that the system (the box with the particles inside) is in the stationary state. Each particle has a stable behaviour with a well-defined wavefunction and a well-defined time dependance. So, with the following two hypotheses
    • stationary state
    • negligible potential
    is it possible to have a wavefunction which is the sum of the [itex]\psi_n[/itex] instead of a product? Or in other words: when does a sum of weighted wavefunctions

    $$\psi = \displaystyle \sum_{n = 1}^N a_n \psi_n$$

    make sense in such a problem?
  7. Mar 14, 2016 #6


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    No, that's not possible. Such a wavefunction formed by summing individual particle's eigenfunctions is not a stationary state as you required. A stationary state is the solution of the equation
    -\frac{\hbar^2}{2m}\sum_{i=1}^N \nabla_i^2 \psi_{n_1n_2\ldots n_N} = E_{n_1n_2\ldots n_N} \psi_{n_1n_2\ldots n_N}
    with ##E_{n_1n_2\ldots n_N} = \sum_{i=1}^N E_{n_i}##. You can see the the summed wavefunction does not satisfy that Schroedinger equation by plugging it in.
  8. Mar 14, 2016 #7
    What process lets you know that there are N particles in the box ? Are you reversing the Rydberg atom counting process as it is populated in application of decoherence methods ?
  9. Mar 14, 2016 #8
    No, it was originally a simple example to show the average energy of a certain number [itex]N[/itex] of charge carriers (electrons, for example) enclosed in a box.
    Last edited: Mar 14, 2016
  10. Mar 14, 2016 #9
    I trust in your statements, but I can't immediately see that the summed wavefunction is not a solution of that equation. This is a first issue.
    A second issue is: what if we considered [itex]N[/itex] not as the number of the particles, but something like the number of the possible states of a single particle into the box? In that case, it would make sense having a probability [itex]a_n[/itex] that the single particle will have that state (as I specified in the first post) and maybe the summed wavefunction itself may be acceptable.
  11. Mar 14, 2016 #10


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    Consider a simple example of two, non-interacting particles. The corresponding Schroedinger equation will be
    -\frac{\hbar^2}{2m}\left( \nabla_1^2 + \nabla_2^2 \right) \psi(\mathbf{r_1},\mathbf{r_2}) = (E_{n_1}+E_{n_2})\psi(\mathbf{r_1},\mathbf{r_2})
    Now substitute ##\psi(\mathbf{r_1},\mathbf{r_2}) = a_1\psi_{n_1}(\mathbf{r_1}) + a_2\psi_{n_2}(\mathbf{r_2})## into the above equation and check if it is the solution or not.
    What do you actually want to calculate? Stationary states or just arbitrary states? Even if the problem is just that of a single particle, a linear combination of the possible stationary states is in general not a stationary state (it is when the stationary states being summed are degenerate states), it's just some wavefunction that satisfies the time-dependent Schroedinger equation.
    Moreover, if you consider ##N## as a number of states of a single particle, then the Schroedinger equation will also look entirely different. It will contain only one term in the summation of the momentum operators - you are shifting to an entirely different problem.
  12. Mar 21, 2016 #11
    Ok and thank you. It was just an attempt. I will spend some time to exactly report the notes I have, by translating them and reporting all the words. At the end of the post, I wrote my questions.

    Energy of a system of particles
    Suppose that a certain volume [itex]V[/itex] contains [itex]N[/itex] electrons. Let's measure the energy of the whole system.
    Since every particle is characterized by its own probability density [itex]\left| \psi_n \right|^2[/itex], it is legitimate to suppose that the total energy can be expressed as a sum of the energies of each electron, weighted by the probability that a particle has to be in that particular energy value (that is: if [itex]|a_n|^2[/itex] is the probability that the [itex]n[/itex]-th particle has energy [itex]E_n[/itex], then the weighted sum [itex]\sum_n a_n E_n[/itex] can be used to express the total energy of the system). [1]
    Supposing that we are in the stationary state, the whole system will be in a defined state and every particle will be characterized by its own value of energy. [2]
    Let's write the wavefunction of the system [itex]\psi[/itex] as linear combination of the functions of each particle in the stationary state:

    [itex]\psi(\mathbf{r}, t) = \sum_{n = 1}^N a_n \psi_n (\mathbf{r},t)[/itex]

    [itex]|a_n|^2[/itex] is the probability that a particle is in the [itex]n[/itex]-th energy state. [3]
    Let's introduce the normalizations

    [itex]\displaystyle \int_V \left| \psi_n(\mathbf{r}, t) \right|^2 dV = 1[/itex]

    [itex]\displaystyle \int_V \left| \psi(\mathbf{r}, t) \right|^2 dV = N[/itex]

    where [itex]V[/itex] is the volume enclosing the [itex]N[/itex] particles, as said before. Because the functions [itex]\psi_n[/itex] are the eigenfunctions of the time-independent Schrödinger equation

    [itex]H \psi_n = E_n \psi_n[/itex]

    they form an orthonormal basis.
    The expected value for the energy of the whole system is defined as

    [itex]E_{average} = \left \langle E \right \rangle = \displaystyle \frac{\left \langle \psi, H \psi \right \rangle}{\left \langle \psi, \psi \right \rangle}[/itex]

    and then the integral in the first post is written.
    The obtained result is the sum of the eigenvalues (energies of the particles), weighted on the squared magnitude of each wavefunction. Differently from Classical Mechanics, here a discrete set of energy levels appears; when the number of particles is high, the discrete levels increase and become so close to each other, becoming very similar to a continuum.

    [1] and [2] seem contradicting statements. If every particle is in a stationary state, isn't its energy well-defined? I can't understand why a probability is needed here. The probability that the [itex]n[/itex]-th particle has its energy [itex]E_n[/itex] should be 1!
    [3] Again. [itex]|a_n|^2[/itex] may mean that there is no certainty that a particle is occupying the [itex]n[/itex]-th energy state. But if the total number of particles is [itex]N[/itex] and the [itex]n[/itex]-th particle is not occupying the [itex]n[/itex]-th energy state, which energy state will it occupy?!

    So, this computation is finally used to show that a great number of discrete, possible energy states can lead to a continuum, so that Quantum Mechanics gives results close to Classical Mechanics when the number of particles is high.

    Have you ever seen something similar? If you know some links or references, I'll check them out.
    Last edited: Mar 21, 2016
  13. Mar 21, 2016 #12


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    Can you mention the source from which you got that equation?
  14. Mar 21, 2016 #13
    I have it by way of notes. I just suppose that these notes are based on an Italian textbook, which I have not found in Amazon. It deals with Quantum Mechanics in order to deal with laser and optic signals transmission.
  15. Mar 21, 2016 #14


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    Do you mean you actually copied that equation from this book you are talking about into your note? Are you sure that you are copying the right thing for your present need?
  16. Mar 21, 2016 #15
    Perhaps we we need to distinguish between the general case (where the atom may be excited) and the ground state (when it is not)?
  17. Mar 22, 2016 #16
    I copied it from some photocopies. Moreover, I have compared the equation with another guy's notes and it is the same. So I am sure that the equation is the same as in the original, and yes, it is about the matter I needed.
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