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Heuristic derivation of Schrodinger's equation for a layman

  1. Jan 10, 2010 #1
    Hi all,

    I was trying to explain to my girlfriend how you can `derive' Schrodinger's equation using the Planck relation [itex]E = \hbar \omega[/itex], the de Broglie relation [itex]p= \hbar k[/itex] and conservation of energy.

    If you assume that the fundamental wavefunction is of the form [itex]\psi = e^{i(kx - \omega t)}[/itex], then it follows easily that the momentum and energy operators have the representation [itex]\hat{p} = -i \hbar d/dx[/itex] and [itex]\hat{E} = i \hbar d/dt[/itex]. Substituting into the operator relation [itex]\hat{E} \psi = \hat{K} \psi + \hat{V} \psi[/itex] then gives the time-dependent Schrodinger equation

    [itex] i \frac{\partial \psi }{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi [/itex].

    She was getting stuck on the assumption that [itex]\psi = e^{i(kx - \omega t)}[/itex]. She is aware that a classical wave can be represented by [itex]\psi = \sin (kx - \omega t + \phi) = A \sin (kx - \omega t) + B\cos(kx -\omega t)[/itex]. I was not able to convince her of the leap to the oscillating exponential.

    Any ideas?

    Thanks
     
    Last edited: Jan 10, 2010
  2. jcsd
  3. Jan 10, 2010 #2

    diazona

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    Is she familiar with Euler's formula (I think it was Euler's)?
    [tex]e^{i\theta} = \cos\theta + i\sin\theta[/tex]
    You could try using that to show how you can decompose a sine function into a difference of imaginary exponentials.
     
  4. Jan 10, 2010 #3

    Claude Bile

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    Demonstrate to her that you can "extract" the trigonometry from the complex exponential notation by taking real and imaginary parts. This could perhaps be done by utilising complex exponentials to prove some simple trig identities.

    Claude.
     
  5. Jan 10, 2010 #4
    I was thinking along similar lines. I thought that there should be some simple physical explanation why the plane-wave solutions should be complex and not simply real as they are for electromagnetic waves.

    The best intuitive reason I could think of is that exponentials retain the same functional form (up to multiplication) after differentiation as opposed to sine or cosine.
     
  6. Jan 10, 2010 #5
    In order to obtain the final PDE equation , Schroedinguer himself had to theorize

    - for 'h' (Planck constant) being small the PDE should be the Hamilton Jacobi

    [tex] \partial _{t} S + H(x,t)=0 [/tex]

    the PDE equation is a wave equation , however if you include derivatives of second order in time, Galilean invariance does not follow (in the relativistic Klein-Gordon a similar reasoning produces the usual Wave equation for Photons )
     
  7. Jan 22, 2010 #6
    For a heuristic derivation of the Schrödinger equation, it is also worthwhile to use the state vector. When the orientation of a state vector evolves, the vector difference d(psi) of the state vector psi (= the vector joining the tips at two successive instants) is always perpendicular to the state vector itself. The perpendicularity is given by the factor i = exp(i.pi/2)

    Hence for time evolution: d(psi) = i . omega . dt . psi, which is a generalized form of the Schrödinger equation. Energy emerges multiplying both sides with the conversion factor hbar.

    More details with figures are given at http://en.wikiversity.org/wiki/Maki...vector_representing_a_quantum_system_evolves".
     
    Last edited by a moderator: Apr 24, 2017
  8. Jan 22, 2010 #7

    SpectraCat

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    The plane-wave solutions for EM are the real part of a more general complex solution as you already know. However even with E&M, the complex part comes into play when you want to explain the phenomenon of circularly polarized light, which does not follow if everything stays real. Sakurai uses this in "Modern Quantum Mechanics" to explain by analogy why the quantum states in the Stern-Gerlach experiment must have complex character as well.
     
  9. Feb 11, 2010 #8
    The simplest explanation (perhaps even proof for free space) may be the following:

    1. For nonrelativistic electron the energy is proportional to momentum square E=(p**2)/(2*m)+V

    2. Planck and Einstein found for temporal frequency E=hbar*w, De Broglie for spatial frequency p=hbar*k

    3. If you want to take out energy (local in point?) from wave-like presentation of matter
    psi=A*exp(i*(k*x-w*t)), you must get out w=E/hbar in front of exponent by taking temporal derivative that gives i*hbar*dpsi/dt=E*psi

    4. If you want to take out momentum square from this wave-like presentation you must take spatial derivative twice that gives
    -((hbar**2)/(2m))*d2psi/dx2=(p**2/(2m))*psi

    5. Writing now summary kinetic + potential energy (for every local point?), we get both
    time-dependent and time-independent forms of Schrödinger's eq.
    (i*hbar)*dpsi/dt = E*psi
    -((hbar**2)/(2m))*d2psi/dx2 + V*psi=E*psi

    Opinion for girlfriends: good old Euler exp(i*a)=cos(a)+i*sin(a) helps to handle conveniently the phase and/or
    to avoid separate sin and cos terms (like electrical engineers do for AC signals).

    THE KEY POINT IS THAT TO GET OUT ENERGY AND MOMENTUM FROM WAVE-LIKE DESCRIPTION OF MATTER YOU MUST TAKE SPATIAL AND TEMPORAL DERIVATIVES.
    THEN WRITE JUST SCHOOLBOY FORMULA E=V+(p**2)/2m AND YOU HAVE GOT SCHRÖDINGER EQUATION.
     
  10. Feb 11, 2010 #9

    Ben Niehoff

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    The reason the wavefunction must be complex is that it must be some kind of wave, while having the same magnitude everywhere. Note that by the HUP between x and p, if the particle is in a state of definite p (i.e., a plane wave), then it must be equally probable for it to be at any x. Since probability is |psi|^2, we must have

    |psi|^2 = const

    which, if psi is to exhibit any kind of wave motion, requires that psi have an additional degree of freedom: the complex phase.

    Note: Another, more technical reason for this is that if the wavefunction magnitude squared is to represent probability, then the wavefunction's time-evolution must be unitary. This means that the total probability must remain equal to 1 as the wavefunction evolves in time. If the wavefunction is to produce any stationary probability distribution (which we know it must, because in nature there are stationary states), then either there must be no time evolution at all (a trivial theory), or the wavefunction must have some internal degree of freedom which can evolve with time while leaving the magnitude squared constant. The simplest choice is to give the wavefunction a complex phase.
     
    Last edited: Feb 11, 2010
  11. Feb 15, 2010 #10
    Hi Ben,

    That's definitely the most convincing argument I've heard so far.
     
  12. Feb 15, 2010 #11
    I think complex numbers are a bit more fundamental in QM than they are in E&M because of the interpretation of |psi|^2.
     
  13. Feb 16, 2010 #12

    SpectraCat

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    I don't think I said anything about how fundamental they are .. I just was trying to give you an analogy that worked. I would agree that complex numbers are more fundamental in QM ... after all [tex]i[/tex] appears in the fundamental equation of motion for Q.M.:

    [tex]i\hbar\frac{\partial}{\partial t}\Psi\left(\vec{r};t\right) = \hat{H}\Psi\left(\vec{r};t\right)[/tex]

    Or if you prefer the Heisenberg version:

    [tex]\frac{d}{dt}\hat{A}\left(t\right) = \frac{i}{\hbar}\left[\hat{H},\hat{A}\right] + \left(\frac{\partial A}{\partial t}\right)[/tex]

    Either way, the 'complex' nature of the problem is there from the beginning. I have often thought of this as a hand-waving post-justification of why it is often so hard to picture what happens in a QM experiment. We have evolved in the classical world, and all of our intuitive tools are based on real solutions ... thus the "imaginary" world of QM dynamics seems quite bizarre to us.
     
  14. Feb 16, 2010 #13
    jdstokes -> I will just add one small remark here. All these "derivation" are necessarily heuristic in origin. The reason is that to derive something you must start with something more fundamental. And there is nothing more fundamental than the Schroedinger (or Heisenberg) equation. In other words, you postulate the Schroedinger equation and start deriving other stuff. You postulate Schroedinger's equation and see if it's of any use for the real world. As it happens, it is. But you cannot properly justify it. But you sure can use various arguments, like the ones you've been using yourself, to give a bit of a feeling for the Schroedinger equation.
     
  15. Feb 16, 2010 #14
    I would rather say that the fundamental QM postulate is the one where we define the quantum state vector (or the wave-function, as you prefer). As a matter of fact, in all exposures of QM that I have seen, this is always the first postulate. We should start deriving the other postulates from this first one.

    This can be done in different ways. I personally prefer the derivation where we describe the rotation of the unitary vector |psi> that represents a particle (an arrow). The time evolution is given by the vector difference d|psi> of two infinitesimally close orientations of the vector |psi>. We can therefore write an equation of the form :

    d|psi> = i . omega . |psi> . dt,

    where:
    * omega is the angular velocity of the rotating arrow,
    * the imaginary "i" indicates that |psi> and d|psi> are at 90° with each other.

    The hbar in the Schrödinger equation is needed as a conversion factor between omega and the energy eigenvalues of the Hamiltonian.
     
  16. Feb 16, 2010 #15

    SpectraCat

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    You cannot derive postulates ... the definition of a postulate is something that you assume to be true from the start. Postulates can only be rationalized, never derived.

    It is true that there are different versions of the postulates of Q.M., which seems a bit strange to me. The ones I learned first are:

    1) definition of the wavefunction
    2) definition of observables in Q.M.
    3) measurement postulate: single measurements yield eigenvalues of corresponding observables
    4) defintion of expectation value (often conflated with 3)
    5) the Born interpretation of [tex]\left|\Psi\left(\vec{r},t\right)\right|^{2}d\tau[/tex] as probability density
    6) the time-dependent Schrodinger equation
    7) the existence of "spin"
     
  17. Feb 16, 2010 #16
    Yes, postulates are not derived. Yet to me, all the current confusion about the quantum postulates is a hint that we are missing a consistent interpretation in which most of the postulates are just corollaries of the first postulate (about the state vector). The derivation I gave in my previous post of the time evolution of a state vector was aiming towards such an interpretation.
     
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