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Defining the potential generated by a particle of unknown position and velocity

  1. Jul 4, 2011 #1
    Hello,

    This isn't exactly homework, but I am attempting to learn quantum mechanics on my own and was wondering if this forum could be used for guidance in that direction. What I want to do is familiarize myself, as accurately as I can, with the specific mathematical picture of the universe given to us by quantum theory.

    1. The problem statement, all variables and given/known data

    I would like to know, in theory, how the potential V generated by a particle whose wavefunction is known is defined.

    2. Relevant equations

    My textbook states that the Schrödinger equation for a two-particle system is:

    [tex]i\hbar\frac{\partial \Psi}{\partial t}=H\Psi,[/tex]

    where

    [tex]H=-\frac{\hbar^2}{2m_1}{\nabla_1}^2-\frac{\hbar^2}{2m_2}{\nabla_2}^2+V(\mathbf{r}_1, \mathbf{r}_2,t).[/tex]

    Two questions:

    1. Does this equation apply to any two particles in the theory of quantum mechanics?

    2. Am I correct in assuming that the generalized Schödinger equation will take on the form

    [tex]i\hbar\frac{\partial \Psi}{\partial t}=V(\mathbf{r}_1,\mathbf{r}_2,\cdots,\mathbf{r}_n,t)\Psi-\sum_{j=0}^n \frac{\hbar^2}{2m_j}\nabla_j^2\Psi?[/tex]

    The "purely Coulombic" equation for an atom is given in my textbook as,

    [tex]H=\sum_{j=1}^Z\left(-\frac{\hbar^2}{2m}\nabla_j^2-\left(\frac{1}{4\pi\epsilon_0}\right)\frac{Ze^2}{r_j}\right)+\frac{1}{2}\left(\frac{1}{4\pi\epsilon_0}\right)\sum_{j\ne k}^Z\frac{e^2}{|\mathbf{r}_j-\mathbf{r}_k|},[/tex]

    where the first sum represents the kinetic plus the potential energy of the electron in the electric field of the nucleus, and the second sum represents the potential energy associated with the mutual repulsion of electrons. (Introduction to Quantum Mechanics: Second Edition, Griffiths, p. 223)

    3. The attempt at a solution

    The potential used to derive the equation of a hydrogen atom assumes that the proton is stationary and invokes Coulomb's law. My question is: what do we do when a particle isn't stationary?

    The first thing that leaps out at me in the equation above is the appearance of rj and rk. I am guessing that these position vectors refer to the locations of the electrons and are not independent variables. In that case, since the rj are constantly changing -- and indeed, if I am correct, becoming uncertain in their positions and velocities -- I do not see an obvious way to relate the Schrödinger equation for the electrons back to a specified potential.

    My intuitive guess is that each possible position of a particle, such as an electron, must be associated with its own potential, in which case we would take into account every possible potential at every possible location in calculating the wavefunction for another particle in its vicinity. However, I may be wrong. Could anyone help explain this?
     
  2. jcsd
  3. Jul 4, 2011 #2

    diazona

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    As far as your 2 numbered questions, yes, you are right, at least in nonrelativistic quantum mechanics. (Quantum field theory, which can incorporate relativity, has some more complications, but it is based on basically the same idea)
    Actually, no. That's the way it worked in classical mechanics: you had these vectors [itex]\mathbf{r}_i[/itex] that represented the positions of particles, and you would come up with differential equations that told you how the [itex]\mathbf{r}_i[/itex] changed with time.

    But in QM, the [itex]\mathbf{r}_i[/itex] have a completely different meaning. They don't refer to the location of anything; instead, they're just parameters that label the different points in space (and they are independent). In quantum mechanics you treat [itex]\mathbf{r}[/itex] the way you treated time in classical mechanics: just a (part of a) label for a specific spacetime point. Instead of finding differential equations for [itex]\mathbf{r}_i(t)[/itex], now you find differential equations for [itex]\Psi(\mathbf{r}_i, t)[/itex]. So the wavefunction is the thing that acts the way you're used to thinking about position.
    For a complex system, then yeah, you would have to do that. It's not possible in general, though. That's why most quantum systems aren't exactly soluble.

    To actually solve the Schroedinger equation for a system, you have to manipulate it in some way so that the potential doesn't depend on the wavefunction. For something like a hydrogen atom (two-particle Coulomb interaction), you can do this by transforming to center-of-mass coordinates, in which the atom looks just like a single particle moving in a static potential; the thing is, this particle doesn't correspond to the electron or proton individually. Its position actually corresponds to the difference in position between the electron and proton.
     
  4. Jul 7, 2011 #3
    Thanks for the response. I just noticed that there are other forums besides the Homework Help forums, so I apologize if I posted inaccurately.

    Two corrections in my original post: V should be V and j = 0 should be j = 1.

    I think I understand about the independence of the rj. Would I be correct in saying that, just as Newtonian mechanics takes place in
    [tex]\mathbb{T}\times\mathbb{R}^3,[/tex]
    where T consists of the number of possible values of time t, non-relativistic quantum mechanics (ignoring wavefunction collapse) takes place in
    [tex]\mathbb{T}\times\mathbb{C}^{3n}?[/tex]
    If so, then would it be accurate to say in the Hugh-Everett many-worlds interpretation that space has 3n dimensions?

    P.S. I haven't quite figured out how to enter LaTeX within a paragraph without getting newlines.
     
  5. Jul 7, 2011 #4

    diazona

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    Time is an infinite continuum, so we just say [itex]\mathbb{R}^4[/itex], or perhaps [itex]\mathbb{R}^{3+1}[/itex].
    Not really; it's the wavefunction, not the space, that becomes complex.

    Here's a way to think about it:
    1. The normal dynamical variable in classical mechanics is a function [itex]\mathbf{r}: \mathbb{R}^1 \to \mathbb{R}^{3n}[/itex] which maps time to a [itex]3n[/itex]-dimensional configuration space (which can in turn be broken down into one copy of the 3D physical space for each of the [itex]n[/itex] particles).
    2. A different mathematical description of the same thing: in classical mechanics, instead of describing a system using the function [itex]\mathbf{r}: \mathbb{R}^1 \to \mathbb{R}^{3n}[/itex], you can describe it using a function [itex]\psi: \mathbb{R}^{3n+1} \to \mathbb{Z}_2[/itex] defined like this:
      [tex]\psi(t,\mathbf{x}_1,\cdots,\mathbf{x}_n) = \begin{cases}1 & \mathbf{r}_i(t) = \mathbf{x}_i\forall i \\ 0 & \text{otherwise}\end{cases}[/tex]
      In other words, [itex]\psi(t,\mathbf{x}_1,\cdots,\mathbf{x}_n)[/itex] is 1 if the position arguments [itex]\mathbf{x}_{1,\cdots,n}[/itex] match the positions of the particles in the system at time [itex]t[/itex], and 0 if they do not. This [itex]\psi[/itex] is the classical equivalent of a wavefunction, and you can think of the [itex]\mathbf{r}_i[/itex] as level sets of [itex]\psi[/itex].
    3. When you switch to quantum mechanics, the only (important) change is that the binary function [itex]\psi[/itex] is replaced with the complex-valued function [itex]\Psi: \mathbb{R}^{3n+1} \to \mathbb{C}[/itex]. Instead of only indicating "yes the system is in this state at this time" or "no the system is not in this state at this time" (1 or 0, respectively), the quantum wavefunction [itex]\Psi[/itex] gives a probability of the system being in a certain state at a certain time.
    Physical space, no. But if you're talking about an abstract configuration space, perhaps. I'm not familiar enough with the different interpretations to know offhand whether different interpretations use differently-dimensioned spaces.
    Use [itex] (and the corresponding closing tag) instead of [tex].
     
  6. Jul 11, 2011 #5
    Yes, I meant [itex]\mathbb{T}\times\mathbb{R}^{3n}[/itex]. My silly mistake.

    Looking back through my textbook, I see that the formulation of Schrödinger's equation

    [tex]i\hbar\frac{\partial \Psi}{\partial t}=V(\mathbf{r}_1,\mathbf{r}_2,\cdots,\mathbf{r}_n,t)\Psi-\sum_{j=1}^n\frac{\hbar^2}{2m_j}\nabla_j^2\Psi[/tex]

    doesn't always work when a magnetic field is involved. Here, it gives a new formulation of the Schrödinger equation,

    [tex]i\hbar\frac{\partial \Psi}{\partial t}=\left(\frac{1}{2m}\left(\frac{\hbar}{i}\nabla-q\mathbf{A}\right)^2+q\phi\right)\Psi.[/tex]

    Two questions:

    1. How do you square [itex]\frac{\hbar}{i}\nabla-q\mathbf{A}[/itex]? Do you take the dot product of the vector [itex]\mathbf{A}[/itex] with itself? What about [itex]\nabla[/itex] and [itex]\mathbf{A}[/itex]?

    2. Because the former equation does not apply to a magnetic field, I'm guessing that it does not amount to a complete formalistic description of the quantum mechanical universe (barring specifics, such as the existence of certain particles and the values of constants). Am I correct that it doesn't? If so, does the latter equation? I'm looking for a complete and succinct mathematical description of the universe as seen by quantum mechanics. (Speaking of which, I just checked out The Road to Reality by Roger Penrose. Maybe it will offer insights into some of these questions.)
     
  7. Jul 12, 2011 #6

    diazona

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    First, remember that [itex]\frac{\hbar}{i}\nabla-q\mathbf{A}[/itex] (which I will call [itex]\mathbf{p}_\text{EM}[/itex] when brevity is needed) is an operator: it acts on a function and produces another function. For example, the operator [itex]D_x\equiv\frac{\partial}{\partial x}[/itex] acts on a function [itex]\Psi(x)[/itex] to produce the function [itex]\Psi'(x)[/itex].

    When you're working with operators, you replace the normal rule of multiplication by operator composition, which is just the technical name for applying two operators in sequence. So for example, if you have two operators [itex]A[/itex] and [itex]B[/itex], the notation [itex]AB\Psi[/itex] means "start with [itex]\Psi[/itex], apply [itex]B[/itex], then apply [itex]A[/itex] to the result." Other than that, things mostly work the same as normal algebra. In particular, you "square" an operator by composing it with itself, i.e. just apply it twice. For example, to square the derivative operator [itex]D_x[/itex] I mentioned above:
    [tex]D_x^2\Psi(x) = D_x D_x \Psi(x) = \frac{\partial}{\partial x} \frac{\partial}{\partial x}\Psi(x) = \frac{\partial}{\partial x}\Psi'(x) = \Psi''(x)[/tex]
    The same applies to an operator that is a sum of terms, for example if you had [itex]F_x = D_x + a[/itex]:
    [tex]\begin{align*}F_x^2\Psi(x) &= F_x F_x \Psi(x)\\
    &= \biggl(\frac{\partial}{\partial x} + a\biggr)\biggl(\frac{\partial}{\partial x} + a\biggr)\Psi(x) \\
    &= \biggl(\frac{\partial}{\partial x} + a\biggr)(\Psi'(x) + a\Psi(x)) \\
    &= \biggl(\frac{\partial}{\partial x}\biggr)(\Psi'(x) + a\Psi(x)) + a(\Psi'(x) + a\Psi(x)) \\
    &= \Psi''(x) + a\Psi'(x) + a\Psi'(x) + a^2\Psi(x)\end{align*}[/tex]
    Note that the constant [itex]a[/itex] has really become an operator representing the action "multiply by the number [itex]a[/itex]."

    If you have a vector operator, then you use the fact that the square of a vector operator is defined to be the "dot product" of the operator with itself, the same way the square of a vector is defined to be the dot product of the vector with itself. The only difference is that, when you calculate the dot product, you compose the component operators instead of multiplying them:
    [tex]\nabla^2\Psi = \biggl(\hat{x}\frac{\partial}{\partial x} + \hat{y}\frac{\partial}{\partial y} + \hat{z}\frac{\partial}{\partial z}\biggr) \cdot \biggl(\hat{x}\frac{\partial}{\partial x} + \hat{y}\frac{\partial}{\partial y} + \hat{z}\frac{\partial}{\partial z}\biggr)\Psi
    = \biggl(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\biggr)\Psi[/tex]

    The case you're asking about puts all of this together. Actually, there is one more subtlety: [itex]\mathbf{p}_\text{EM}[/itex] is a complex operator, so instead of actually squaring it, you apply its complex conjugate: [itex]p_\text{EM}^2 \equiv \mathbf{p}_\text{EM}^*\cdot \mathbf{p}_\text{EM}[/itex]. Between that and what I mentioned above, you should be able to figure out what [itex]p_\text{EM}^2[/itex] is in detail. The one slightly odd part is the term with [itex]\nabla\cdot\mathbf{A}\Psi[/itex], but you should be able to figure out that
    [tex]\nabla\cdot\mathbf{A}\Psi = \frac{\partial}{\partial x}[A_x\Psi] + \frac{\partial}{\partial y}[A_y\Psi] + \frac{\partial}{\partial z}[A_z\Psi][/tex]
    and take it from there. Some of the cross terms cancel out in the final result.
    Neither of these equations is even close to a complete quantum description of the known universe. In fact, technically, we don't even have a complete description of everything that has been experimentally observed. If you ignore gravity, almost everything we know can be explained by the standard model of particle physics, but that requires some study of quantum field theory to understand.
     
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