Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Help to understand the wave function for atom (gas)

  1. Jul 14, 2016 #1

    KFC

    User Avatar

    Hi there,
    I took the course of quantum mechanics long time ago. From there I learn how to describe an atom with wave function. For example, Hydrogen has the wave function in (spherical coordinates) space. In the book they consider a reduced mass for the nucleus and the only external electron so the Hamiltonian simplified to a single term for momentum and potential. My first question is is this momentum refer to the motion of the center of mass of the system? If I solve the Schrodinger equation and get the momentum of the center of mass, how can I figure out what the momentum of the electron and/or nucleus separately?

    Honestly, solving the Hydrogen problem is not trivial for me though it is more or less understandable. But if we consider a more complicate case like an atom with more than 1 electron. What's the form of the wave function for that kind of atom look like? I know in that case, we cannot use the reduce mass to simplify the Hamiltonian, so it contains one term for the nucleus momentum, n momentum terms for n electrons, n potential terms for potential energy between each electron to the nucleus, C(n,2) terms for potential between any 2 electrons. I am not sure if this system is solvable but it is very complicate. My question is if it is possible to separate the system into two: one for nucleus alone and one for electrons, we solve the two systems separately to get two wave functions and then multiply the results to get the complete wave function?

    My next question is about an even complicate system. Let's say I have n atoms instead of one. How generally people solve these systems if each of them have more than 1 electron? Or let's make it simple, consider n hydrogen atoms, each has 1 electron. If we could mix those n atoms but ignore interaction between atoms. Does it mean I can solve the problem for 1 atom to get wave function ##|\psi_1\rangle## so the complete wave function for the whole system is ##\prod_n|\psi_n\rangle##

    My last question is about including the spin for electron, P electrons in each atom and K atoms without interactions between them. Is it correct to write down the wave function as

    ##\prod_{k, p}|\Psi_n\rangle |\Phi_{kp}\rangle##

    with electronic wave function for the ##k##th atom as
    ##
    |\Phi_{kp}\rangle = |s_pn_pl_pm_p\rangle_k
    ##
    where ##s, n, l, m## stands for quantum number of spin, principal, angular momentum and magnetic.

    I apology for the long question. I am trying to read other material so to learn how multiple particles system is dealed. It is very hard to understand most of the content. Above statements are based on my understanding so far, they might be wrong. I am looking for a general picture how complicate system is tackled.
     
  2. jcsd
  3. Jul 14, 2016 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    The treatment of the two body system, nucleus and electron, begins in the classical theory with the resolution of motion about the center of mass and the use of the reduced mass in describing the remaining degrees of freedom. You have 2 particles orbiting their common COM which is 3+3 coordinate degrees of freedom (and 6+6 dynamical d.o.f.) One redefines variables to treat this the same as one particle with the total mass moving with the COM of the system and another particle with the reduced mass orbiting around a fixed central force. Call this a quasi-particle if you like and its really this quasi-particle (an inverse weighted average of the two particles so mostly light electron but a wee bit of heavy nucleus) which is described by the spherical harmonic orbitals in the Hydrogen atom solution.

    Past that you've touched on one of the big conundrums of QM and even Classical Mechanics in that the many body problem is intractable. Classically you get chaotic behavior except for special unstable cases and in QM you just get a huge intractable PDE to solve. One must resort to statistical mechanics (classical or quantum) to talk about aggregate behavior of many atoms.

    What we can do for single atoms is make qualitative predictions assuming the electrons behave similarly in the presence of other electrons to how a single electron would about a nucleus of some given positive charge. One then can speak of partial masking of the nuclear charge by the cloud of electrons and filling of orbitals. But the details and the energy levels must be established empirically. Indeed you'll note that in the heavy elements the orbital filling sequence (sequence of energy levels for various orbitals) has a few exceptions to the usual rules. Note that we can still qualitatively indicate orbitals because we still have the (near) spherical symmetry and conservation of total angular momentum. We can always resolve the angular dependencies of the electron wave function in terms of spherical harmonics (it's really just a type of Fourier expansion). But I say near spherical symmetry as there are magnetic moments breaking the nucleus' spherical symmetry. That's where you get some interesting perturbations and splitting of the otherwise degenerate spin states of electrons in given orbitals... oh yes and there's also orbital-spin magnetic coupling... just loads of fun!!!
     
  4. Jul 14, 2016 #3
    There are two momenta: the momentum of the center of mass ## p_\text{CM} = p_1 + p_2 ## and the relative momentum ## p_r = \frac{m_2 p_1 - m_1 p_2}{m_1 + m_2} ##. There is no external force, so the solution to the equation with ## p_\text{CM}## is just a free particle. The remaining equation with ##p_r## is for the relative motion. But since the nucleus is so much heavier than the electron, the nucleus is pretty much fixed at the center of mass and the equation for the relative motion is pretty much the equation for the electron's motion.

    When we solve the The Schrödinger equation, we get wave functions of stationary states. I'm not sure how you "get the momentum," either of the whole system or separately.

    This brings us to the general point that pertains to most of you questions: you can't solve for the wave functions of individual particles and then multiply them together if there is an interaction. A solution in general will be an entangled state: a superposition (sum) of products of individual wave functions. This is why solving for complicated many-body states is very difficult. The basis of most ansatzes I've seen is the variational method.
     
  5. Jul 15, 2016 #4

    KFC

    User Avatar

    Thanks. So let assume there is no interaction just like in extreme cold environment and for boson (I read some book about these particles, called BEC?) In this situation, if I solve the wave function for one particles, can I multiple the wave function N times to find the compete wave function of the system?

    By the way, you mention that the nucleus is much heavier than than electrons. So do you mean the wave function basically describe the motion of electron instead of atom? Let's go back to the example for extreme low temperature and boson, I read a book about this boson at low temperature, it is said the wave function is a plane wave, that mean you will find the atom anywhere in space, so should we correct it to say you should see the electron anywhere?

    I read book long time ago about an quantum mechanics. It talk about how to solve the Hydrogen system and for other atoms which has K electrons (one of them does not fill the shell and the rest do), in this case, they consider K-1 electrons and the nucleus form a pseudo nucleus and apply the similar treatment for hydrogen problem to solve the system. I am not quite understand why we could do that way. For those K-1 electrons, they should have different electron cloud shape and quantum number, for what reason we could ignore the motion of K-1 electrons. Thanks.
     
  6. Jul 23, 2016 #5
    I'm out of my depth here so please take all of these with a grain of salt.

    I think you are right that the state will be close to a product state. Note that for identical particles, we have to symmetrize or antisymmetrize the wave function, which makes it look like an entangled state. But since BEC is many bosons in the same state (ground state), the symmetrized wave function will still be a product state.

    I think that's possible. For example, a wave function that is initially Gaussian will spread in time, if there is no mechanism to limit it (e.g. decoherence, which is suppressed at low temperature).

    I have not heard about iteratively solving a k electron atom that way before. Anyone who knows this stuff, please feel free to educate the OP and me.
     
  7. Jul 25, 2016 #6

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    An extreme cold environment has nothing to do with the presence or absence of interactions. [Bose-Einstein condensates have to do with Fermionic matter, when cold enough pairing up to form quasi-bosons and thence being able to form a condensed state. This is what we believe happens with low temperature superconductors and superfluids.] In the BEC case you don't even multiply, you raise the single bosonic solution to the n'th power... but again that is given no interaction. Even without physical interaction there is are statistical effects e.g. when you have two bosons in distinct states there is a symmetrized state, if fermions it is antisymmetrized. These induce "statistical interactions" which manifest e.g. as stimulated emission and BEC's for bosons, and as the "repulsion" effected by the Pauli Exclusion principle for fermions. In short "it ain't quite that simple".
    What happens is that for any (interacting) two body system you can't (even classically) easily describe it in terms of the motions of the two particles as compared to a factorization into weighted sum and difference pair of sets of degrees of freedom. The weighting is the actual masses of the constituent particles so when one is very much more massive than the other then its individual description is very close to the "weighted sum" description and likewise the light particle's description is very close to that of the "weighted difference". Since the weighted sum becomes the center of mass system it is usually ignored (by choosing it as the rest frame) and the difference factor system which is now almost describing the light particle, is the system of interest.

    This gives an approximation method useful for say calculating energy levels. It ignores the "back reaction" i.e. how the adding of the K'th electron changes the solutions for the previous K-1 electrons. But it does work well in the small K regime, Li, Be, B, etc... are modeled pretty well this way. Uranium? not so well except in all cases you get a pretty good approximation for the outer electron behavior (with some empirically driven adjustments). In fact it was in studying and trying to comprehend atomic spectra, which this sort of thing predicts, that set the stage for the development of quantum mechanics.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Help to understand the wave function for atom (gas)
Loading...