Ch 3. The Schrodinger Equation

    3.1 Creating the Schrodinger Equation

Now it’s time to make a slight modification to the Hamilton Jacobi Equation. Thus far, we’ve been describing particle movement in terms of velocity fields. The particles form an imaginary fluid and the solutions of the HJE describe streamlines. Real fluids behave differently though: the particles of the fluid interact with each other. The interaction is usually a function of the density of the fluid. If we wanted to alter the HJE to describe real fluids, we would need to account for an additional force on each particle (or point in phase space, more generally). The most natural way to do this would be to add another term to the Potential Energy.

(1)   \begin{equation*} V_{Total}=V+V_Q \end{equation*}

We’re going to call the new term, V_Q, the “Quantum Potential”. But what form do we want to try for it? How about this:

(2)   \begin{equation*} V_Q=-\sum _{i=1}^n \frac{\hbar ^2}{2m_i}\frac{(\nabla _i{}^2R)}{R}=-\sum _{j=1}^{3n} \frac{\hbar ^2}{2M_j}\frac{(\frac{\partial ^2R}{\partial q_j{}^2})}{R} \end{equation*}

Where “R” is the square root of the density of the fluid, and \hbar is the so-called “reduced Planck constant”. This is sort of arbitrary looking, so let’s take a closer look at it. First, we’ll simplify to the case of 1 particle in 1 dimension.

(3)   \begin{equation*} V_Q=-\frac{\hbar^{2}}{2m} \frac{\frac{\partial^{2} R}{\partial x^{2}}}{R} \end{equation*}

To get a feel for how this thing behaves, let’s look at density functions that have a uniform Quantum Potential. Letting V_Q be a constant and rearranging terms:

(4)   \begin{equation*} -\frac{2m V_Q}{\hbar ^2}R=\frac{d^2R}{dx^2} \end{equation*}

Anyone who has studied Differential Equations will immediately recognize the solution:

(5)   \begin{equation*} R(x)=A Sin(\sqrt{2m V_Q}x/\hbar +\alpha ) \end{equation*}

Where A and \alpha are arbitrary constants. But remember that R is only the square root of the density. The actual density (after using some Trig identities) is:

(6)   \begin{equation*} R^2=\frac{A}{2}( 1-Cos(2\sqrt{2m V_Q}x/\hbar +\alpha )) \end{equation*}

This fact is the root of “wave-like” properties of Quantum Mechanics. The Classical Potential, ‘V’, is usually a given, then the rest of the system’s Energy is split between particle movement, and these ripples in the density field.

Figure 3.1.1: An example of a particle density function that has a constant Quantum Potential

Figure 3.1.1: An example of a particle density function that has a constant Quantum Potential

This is a good time to switch to a convenient set of units: Atomic Units. This will set \hbar equal to 1 and simplify everything from here on out. Inserting equation (1) into the HJE (see last chapter for details), gives us what we might call the “Quantum Hamilton Jacobi Equation”:

    \[ \frac{\partial S}{\partial t}+\sum _{j=1}^n \frac{1}{2m_j}(\nabla _jS)^2+V_{Total}=\frac{\partial S}{\partial t}+\sum _{i=1}^n \frac{1}{2m_i}(\nabla _iS)^2+V \]

(7)   \begin{equation*} -\sum _{i=1}^n \frac{1}{2m_i}\frac{(\nabla _i{}^2R)}{R}=0 \end{equation*}

or, using our “extended mass variables”

(8)   \begin{equation*} \frac{\partial S}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}(\frac{\partial S}{\partial q_i})^2+V-\sum _{j=1}^{3n} \frac{1}{2M_j}\frac{(\frac{\partial ^2R}{\partial q_j{}^2})}{R}=0 \end{equation*}

Like the standard HJE, this equation expresses how the energy of each particle in our fluid is related to the all important generating function, S. To make things absolutely explicit:

    \[ \frac{\partial S}{\partial t}+\sum _{j=1}^n \frac{1}{2m_j}(p_j)^2+V-\sum _{i=1}^n \frac{1}{2m_i}\frac{(\nabla _i{}^2R)}{R}=\frac{\partial S}{\partial t}+T+V+V_Q \]

(9)   \begin{equation*} =\frac{\partial S}{\partial t}+E=0 \end{equation*}

Another property that we will demand of our fluid is that no “amount” of it is either created or destroyed, meaning that within a given volume and for a given amount of time, the increase in the amount of fluid equals how much fluid entered the volume minus how much fluid left the volume.

If we let v_j represent the j^{th} component of the fluid velocity, then the differential form of this statement (AKA the “Continuity Equation“), is:

(10)   \begin{equation*} \frac{\partial R^2}{\partial t}+\sum _{j=1}^{3n} \frac{\partial }{\partial q_j}(R^2v_j)=\frac{\partial R^2}{\partial t}+\sum _{j=1}^{3n} \frac{\partial }{\partial q_j}(R^2\frac{\frac{\partial S}{\partial q_i}}{M_j}) \end{equation*}

Expanding out the derivatives:

(11)   \begin{equation*} \frac{\partial R}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}(2\frac{\partial R}{\partial q_j}\frac{\partial S}{\partial q_j}+R\frac{\partial ^2S}{\partial q_j{}^2})=0 \end{equation*}

So now we have 2 equations that describe our fluid (equations (8) and 11). Is there a nice way to combine them? Let’s try this: (i(equation (11))-R(equation (8)))e^{iS}=0:

    \[ (i(\frac{\partial R}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}(2\frac{\partial R}{\partial q_j}\frac{\partial S}{\partial q_j}+R\frac{\partial ^2S}{\partial q_j{}^2})) \]

(12)   \begin{equation*} -R(\frac{\partial S}{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}(\frac{\partial S}{\partial q_i})^2+V-\sum _{j=1}^{3n} \frac{1}{2M_j}\frac{(\frac{\partial ^2R}{\partial q_j{}^2})}{R}))e^{iS}=0 \end{equation*}

After some Algebra:

(13)   \begin{equation*} i\frac{\partial }{\partial t}(R e^{iS})+\sum _{j=1}^{3n} \frac{1}{2M_j}\frac{\partial ^2}{\partial q_i{}^2}(R e^{iS})-V(R e^{iS})=0 \end{equation*}

Understand that nothing magical has happened to arrive at equation (13). The introduction of complex numbers has no “mysterious” connotations whatsoever. Equation (13) is completely equivalent to equations (8) and (11) taken together. Now, let’s introduce

(14)   \begin{equation*} \Psi =R e^{iS} \end{equation*}

Then we will have,

(15)   \begin{equation*} i\frac{\partial \Psi }{\partial t}+\sum _{j=1}^{3n} \frac{1}{2M_j}\frac{\partial ^2\Psi }{\partial q_i{}^2}-V \Psi =0 \end{equation*}

Or, in another form:

(16)   \begin{equation*} i\frac{\partial \Psi }{\partial t}+\sum _{i=1}^n \frac{1}{2m_i}\nabla _i{}^2\Psi -V \Psi =0 \end{equation*}

And at last we have the Schrodinger Equation in it’s usual form. To recap, in the last chapter we reformulated Classical Mechanics in terms of the velocity field of an imaginary fluid. Then we altered this fluid by letting it’s constituents interact with each other by adding a term to the Potential Energy that depends on the fluid’s density at each point. We also imposed the rule that the “amount” of fluid never changes. This led directly to the Schrodinger Equation! Therefore, Classical Physics is precisely the limiting case of Quantum Physics when the Quantum Potential becomes small compared with the other Energies in a system. This is usually the case for systems of particles with lots of mass (because mass is in the denominator of the Quantum Potential).

So, if you were “living inside” of a solution of the Schrodinger Equation (i.e. you were part of a system of particles that moved along a particular streamline in the “Quantum Fluid”, \Psi), and you were equipped with sensory organs that could only detect the gross movement of large sets of particles, you would observe a universe that appeared to follow the laws of Classical Physics!) Furthermore, you would see collections of particles arranged in stable structures (Atoms, Molecules, Chairs, etc). We’ll study the classical Limit in a forthcoming chapter.

This is one side of the coin. The other side is this: Why would someone “living inside” of \Psi find that small particles behaved in the probabilistic manner described by “standard” Quantum Mechanics? This is a slightly more subtle question that will have to wait for now.

    3.2 States of Uniform Energy

In this section, I’d like to examine the most typical procedure for solving the Schrodinger Equation., “Separation of Variables”. My approach is different enough from the usual one that I thought it would be worthwhile. If you are familiar enough with QM and understand how Energy is related to the Separable Solutions, you can probably skip this section.

We start by seeking solutions that are “separable”, meaning they can be written as a product of 2 functions, one of position only, and one of time only:

(17)   \begin{equation*} \Psi (\vec{Q},t)=\psi (\vec{Q})f(t) \end{equation*}

We are also going to keep in mind that \Psi can be written as

(18)   \begin{equation*} \Psi (\vec{Q},t)=R e^{i S} \end{equation*}

and we’ll carry out identical analysis with both forms.

Inserting (17) into (16), then (18) into (16):

(19)   \begin{equation*} i \psi  \frac{df}{dt}+\sum _{i=1}^n \frac{1}{2m_i}(\nabla _i{}^2\psi )f-V \psi  f=0 \end{equation*}

    \[ i(\frac{\partial R}{\partial t}+R(i\frac{\partial S}{\partial t}))e^{i S}+\sum _{i=1}^n \frac{1}{2m_i}(\nabla _i{}^2R-R (\nabla S)^2+ \]

(20)   \begin{equation*} i 2R\nabla R \nabla S+\nabla ^2S)e^{i S}-V R e^{i S}=0 \end{equation*}

Dividing by –\Psi:

(21)   \begin{equation*} -i \frac{df/dt}{f}-\sum _{i=1}^n \frac{1}{2m_i}\frac{(\nabla _i{}^2\psi )}{\psi }+V=0 \end{equation*}

    \[ (\frac{\partial S}{\partial t}-i\frac{\partial R/\partial t}{R})+((\sum _{i=1}^n \frac{1}{2m_i}(\nabla _iS)^2-\sum _{i=1}^n \frac{1}{2m_i}\frac{\nabla _i{}^2R}{R}+V)- \]

(22)   \begin{equation*} i \sum _{i=1}^n \frac{1}{2m_i}(2\nabla _iR \nabla _iS+\frac{\nabla _i{}^2S}{R})) =0 \end{equation*}

Now, because \psi and f can be varied independently,

(23)   \begin{equation*} -\sum _{i=1}^n \frac{1}{2m_i}\frac{(\nabla _i{}^2\psi )}{\psi }+V=i \frac{df/dt}{f}=C,\ \ which \Longrightarrow  \end{equation*}

    \[ (\sum _{i=1}^n \frac{1}{2m_i}(\nabla _iS)^2-\sum _{i=1}^n \frac{1}{2m_i}\frac{\nabla _i{}^2R}{R}+V)- \]

    \[ i \sum _{i=1}^n \frac{1}{2m_i}(2\nabla _iR \nabla _iS+\nabla _i{}^2S))= \]

(24)   \begin{equation*} -\frac{\partial S}{\partial t}+i\frac{\partial R/\partial t}{R}=C=C_{re}+i C_{im} \end{equation*}

where ‘C’ is a constant and we’ve split it into real and imaginary parts for the last equation. Let’s look at the real part of (24):

(25)   \begin{equation*} (\sum _{i=1}^n \frac{1}{2m_i}(\nabla _iS)^2-\sum _{i=1}^n \frac{1}{2m_i}\frac{\nabla _i{}^2R}{R}+V)=-\frac{\partial S}{\partial t}=C_{re} \end{equation*}

By now, you should be able to recognize the left side of the equation as the total energy, so that the real part of our separation constant is the Energy of each particle of the fluid:

(26)   \begin{equation*} E=(T+V_Q+V)=-\frac{\partial S}{\partial t}=C_{re} \end{equation*}

This means that the Energy of the fluid is uniform throughout phase space. (Remember that solutions to the classical HJE also had uniform Energy, so separable solutions to the SE are going to be a good point of comparison between the 2 equations).

What about the imaginary part of (24)?

(27)   \begin{equation*} \sum _{i=1}^n \frac{1}{m_i}(\nabla _iR \nabla _iS+\frac{\nabla _i{}^2S}{2R})=\frac{\partial R/\partial t}{R}= C_{im} \end{equation*}

Now, the Continuity Equation, (11), allows us to make a substitution which gives us:

    \[ -\frac{\partial R/\partial t}{R}=\frac{\partial R/\partial t}{R}= C_{im},   which \Longrightarrow  \frac{\partial R}{\partial t}=C_{im}=0 , \]

(28)   \begin{equation*}  which \Longrightarrow  C=E  \end{equation*}

To summarize our analysis, these 3 things are equivalent:
1. A solution to the Schrodinger Equation is separable.
2. The Energy of the fluid is uniform throughout space.
3. The Density of the fluid is constant throughout time.

Let’s go ahead and solve (23) for f:

(29)   \begin{equation*} \frac{df}{dt}=-i C f  =-i E f\Longrightarrow  f= A e^{-i E t}=e^{-i E t} \end{equation*}

(The last equation follows because we’re going to absorb the constant A into \Psi). Therefore, we can write

(30)   \begin{equation*} \Psi_E=\psi_E e^{(-i E t)} \end{equation*}

Where \Psi_E is a solution to the “Time Independent Schrodinger Equation”

(31)   \begin{equation*} -\sum _{i=1}^n \frac{1}{2m_i}\frac{(\nabla _i{}^2\psi )}{\psi }+V=E \end{equation*}

Then, because the Schrodinger Equation is linear, a general solution will have the form

(32)   \begin{equation*} \Psi =\sum _{i=1}^N c_i\Psi _{E_i} \end{equation*}

Chapter 4

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