The time-independent Schrödinger equation

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 2.

Once we have the Schrödinger equation, most of non-relativistic quantum mechanics is devoted to finding solutions to this equation for various potential functions. Since it is a second-order partial differential equation, the number of cases for which an exact solution may be found is depressingly small. As a result there are a lot of approximation techniques that allow solutions in various cases, and of course there is always the option of a numerical solution using a computer.

An important feature of the Schrödinger equation is that it is linear, meaning that the function {\Psi} and its derivatives occur to the first power only, and there are no products or other non-linear functions of {\Psi} to be found. This has the important consequence that if we find two different solutions of the Schrödinger equation, then any linear combination of these two solutions is also a solution. A linear combination of two solutions {\Psi_{1}} and {\Psi_{2}} has the form

\displaystyle   \Psi \displaystyle  = \displaystyle  a\Psi_{1}+b\Psi_{2} \ \ \ \ \ (1)

where {a} and {b} are complex constants.

There are, however, several important potentials for which exact solutions may be found, and many of these potentials are time-independent, meaning that they depend only on position. In such a case, the Schrödinger equation may be simplified by the trick of separation of variables. To see how this works, let’s start with the general Schrödinger equation:

\displaystyle  -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V(x,t)\Psi=i\hbar\frac{\partial\Psi}{\partial t} \ \ \ \ \ (2)

The only place in this equation where an explicit time dependence can occur is in the potential function {V(x,t)} (of course the wave function itself will have an explicit time dependence, but that’s what we’re trying to solve for!). In the situation where the potential depends only on space, the Schrödinger equation becomes

\displaystyle  -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V(x)\Psi=i\hbar\frac{\partial\Psi}{\partial t} \ \ \ \ \ (3)

Not much of an improvement, you might think. But suppose we propose a solution for the wave function that is the product of one function {\psi(x)} that depends on space ({x}) only and another function {\Xi(t)} (the Greek capital ‘xi’) that depends only on time ({t}). That is

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \psi(x)\Xi(t) \ \ \ \ \ (4)

The partial derivatives become a little simpler:

\displaystyle   \frac{\partial^{2}\Psi}{\partial x^{2}} \displaystyle  = \displaystyle  \frac{\partial^{2}\psi}{\partial x^{2}}\Xi(t)\ \ \ \ \ (5)
\displaystyle  \frac{\partial\Psi}{\partial t} \displaystyle  = \displaystyle  \psi(x)\frac{\partial\Xi}{\partial t} \ \ \ \ \ (6)

Substituting this back into the Schrödinger equation we get

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}\Xi(t)+V(x)\psi(x)\Xi(t) \displaystyle  = \displaystyle  i\hbar\frac{\partial\Xi}{\partial t}\psi(x)\ \ \ \ \ (7)
\displaystyle  -\frac{\hbar^{2}}{2m}\frac{1}{\psi(x)}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x) \displaystyle  = \displaystyle  i\hbar\frac{1}{\Xi(t)}\frac{\partial\Xi}{\partial t} \ \ \ \ \ (8)

where in the last line we have divided the first line through by {\psi(x)\Xi(t)}.

Notice that a magical thing has happened here: the left side of the equation depends only on {x} and the right side depends only on {t}. We have therefore separated the two independent variables in the equation. How does that help us? Well, since {x} and {t} are independent variables, we can vary either of them without changing the other. If we varied {x} for example, then in principle the left side of the equation would change while the right side, which does not depend on {x} wouldn’t. So it looks like we’ve ended up with an impossible situation.

Not quite. The equation can still be satisfied if both sides of the equation are equal to the same constant, which we’ll call {E}. That is, we must have

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{1}{\psi(x)}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x) \displaystyle  = \displaystyle  E\ \ \ \ \ (9)
\displaystyle  i\hbar\frac{1}{\Xi(t)}\frac{\partial\Xi}{\partial t} \displaystyle  = \displaystyle  E \ \ \ \ \ (10)

which we can rewrite as two separate ordinary (not partial) differential equations, since each equation now has only one independent variable:

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x) \displaystyle  = \displaystyle  E\psi(x)\ \ \ \ \ (11)
\displaystyle  i\hbar\frac{d\Xi(t)}{dt} \displaystyle  = \displaystyle  E\Xi(t) \ \ \ \ \ (12)

The clever thing about this separation of variables is that the potential function has disappeared from the second equation, so we can solve for the time part of the equation in general, and get

\displaystyle   \int\frac{d\Xi}{\Xi} \displaystyle  = \displaystyle  -\frac{i}{\hbar}E\int dt\ \ \ \ \ (13)
\displaystyle  \ln\Xi \displaystyle  = \displaystyle  -\frac{i}{\hbar}Et+\ln C\ \ \ \ \ (14)
\displaystyle  \Xi(t) \displaystyle  = \displaystyle  Ce^{-iEt/\hbar} \ \ \ \ \ (15)

where {C} is a constant of integration, and will be determined by the normalization of the wave function.

But what is this constant {E}? To get an idea of what {E} represents, we can rewrite the {\psi}equation like this:

\displaystyle   \left[-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x)\right]\psi(x) \displaystyle  = \displaystyle  E\psi(x)\ \ \ \ \ (16)
\displaystyle  \left[\frac{1}{2m}\left(\frac{\hbar}{i}\frac{d}{dx}\right)^{2}+V(x)\right]\psi(x) \displaystyle  = \displaystyle  E\psi(x) \ \ \ \ \ (17)

Now if we remember the expression of the momentum as an operator:

\displaystyle   p \displaystyle  = \displaystyle  \frac{\hbar}{i}\frac{d}{dx} \ \ \ \ \ (18)

(whether we use a total or partial derivative doesn’t matter when the function being operated on depends on {x} only), we can see that the {\psi}equation can be written as:

\displaystyle   \left[\frac{p^{2}}{2m}+V(x)\right]\psi(x) \displaystyle  = \displaystyle  E\psi(x) \ \ \ \ \ (19)

The terms in square brackets are the kinetic plus the potential energy, so we can view this as an operator equation, where the operator in square brackets operates on the spatial part of the wave function with the result of giving the same wave function back again, but multiplied by the constant {E} which can therefore be interpreted as the total energy in the state.

In more advanced language, the function {\psi(x)} is an eigenfunction of the operator {\frac{p^{2}}{2m}+V(x)} with eigenvalue {E}. (If you’re just starting out in quantum mechanics and these terms are unfamiliar, don’t worry about them right now.)

All this may look a bit arbitrary, since from the appearance of the {\psi} equation, it looks like we can just pick any old value (including complex numbers) for {E}. However, the magic of this equation is that when we solve it for particular potential functions, we discover that only certain values of {E} (and they are all real, too) are allowed. This equation, therefore, puts the ‘quantum’ in ‘quantum mechanics’, since it shows that energies can be only certain discrete values. More on this when we discuss some of the individual potentials.

For now, it is worthwhile summarizing the chain of logic that got us this far. Starting with the Schrödinger equation (which can be accepted as a postulate), we considered the special case of a time-independent potential. This led to a solution using the mathematical technique of separation of variables, which in turn led to the interpretation of the spatial part of the solution as an equation that gives the allowed energy levels in the system. Note that the prediction of the energy levels is an entirely mathematical consequence of the assumption that the Schrödinger equation is the correct equation for describing nature.

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