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Predicting energy levels and probabilities: The Schrödinger equation

In the last lecture, we saw that the Bohr model is able to predict the allowed energies of any single-electron atom or cation. However, the Bohr model is, by no means, a general approach and, in any case, it relies heavily on classical ideas, clumsily grafting quantization onto an essentially classical picture, and therefore, provides no real insights into the true quantum nature of the atom.



Any rule that might be capable of predicting the allowed energies of a quantum system must also account for the particle-wave duality and include a wave-like description for particles. In 1926, the Austrian physicist Erwin Schrödinger posited an equation that predicts both the allowed energies of a system as well as the probability of finding a particle in a given region of space. The equation, known as the Schrödinger wave equation, does not yield the probability directly, in fact, but rather the probability amplitude alluded to in the previous lecture. This amplitude function is, in general, a complex function denoted $\psi(x)$ (for a single particle in one spatial dimension) and is referred to as the wave function. It is related to the probability as follows:

The probability that a single quantum particle moving in one spatial dimension will be found in a region $x\in [a,b]$ if a measurement of its location is performed is

\begin{displaymath}
P(x\in[a,b]) = \int_a^b \vert\psi(x)\vert^2 dx
\end{displaymath}

The square of the wave function $\vert\psi(x)\vert^2$ is known as the probability density $p(x)$. In general, the probability that a quantum particle will be found in a very small region $dx$ about the point $x$ is

\begin{displaymath}
p(x)dx = \vert\psi(x)\vert^2 dx
\end{displaymath}

Since particles can exhibit wave-like behavior, the amplitude or wave function $\psi(x)$ should have a wave-like form.



The Schrödinger equation cannot be derived from any more fundamental principle. However, in order to motivate it, let us use the assumption that $\psi(x)$ should have a wave-like form. Thus, consider a free particle of mass $m$ and momentum $p$. Recall the de Broglie hypothesis stating that the particle has a wavelength $\lambda$ given by

\begin{displaymath}
\lambda = {h \over p}
\end{displaymath}

or

\begin{displaymath}
p = {h \over \lambda}
\end{displaymath}

If the particle is a free particle, its potential energy $V(x)=0$, so that its energy is purely kinetic

\begin{displaymath}
E = {p^2 \over 2m} = {h^2 \over 2m\lambda^2}
\end{displaymath}

If the amplitude $\psi(x)$ describes a wave, then it should take the mathematical form

\begin{displaymath}
\psi(x) = A\cos\left({2\pi x \over \lambda}\right)\;\;\;\;\;...
...}
\;\;\;\;\;\;\;\;\;\;B\sin\left({2\pi x \over \lambda}\right)
\end{displaymath}

(we are considering a wave that is not changing in time here). Consider the cosine form (the same will hold for a sine for as well) and consider the first two derivatives of $\psi(x)$:
$\displaystyle {d\psi \over dx}$ $\textstyle =$ $\displaystyle -{2\pi \over \lambda}A\sin\left({2\pi x \over
\lambda}\right)$  
$\displaystyle {d^2\psi \over dx^2}$ $\textstyle =$ $\displaystyle -{4\pi^2 \over \lambda^2}A\cos\left({2\pi x \over
\lambda}\right)$  

We see, therefore, that $\psi(x)$ and $d^2\psi(x)/dx^2$ are related by

\begin{displaymath}
{d^2 \psi \over dx^2} = -{4\pi^2 \over \lambda^2}\psi(x)
\end{displaymath}

or
$\displaystyle -{h^2 \over 8\pi^2 m}{d^2 \psi \over dx^2}$ $\textstyle =$ $\displaystyle {h^2 \over 2m
\lambda^2}\psi(x)$  
       
$\displaystyle -{\hbar^2 \over 2m}{d^2 \psi \over dx^2}$ $\textstyle =$ $\displaystyle E\psi(x)$  

The last line of the above expression is, in fact, the Schrödinger equation for a free particle moving along the $x$-axis.



We can always set up the Schrödinger equation via the following simple prescription. Start with an expression for the classical energy. In this case, for a free particle

\begin{displaymath}
{p^2 \over 2m} = E
\end{displaymath}

Now multiply by the wave function $\psi(x)$:

\begin{displaymath}
{p^2 \over 2m}\psi(x) = E\psi(x)
\end{displaymath}

Finally replace the momentum $p$ by the following derivative:

\begin{displaymath}
p\longrightarrow -i\hbar {d \over dx}
\end{displaymath}

which is equivalent to replacing $p^2$ by a second derivative:

\begin{displaymath}
p^2 \longrightarrow -\hbar^2 {d^2 \over dx^2}
\end{displaymath}

When this is done, we arrive at:

\begin{displaymath}
-{\hbar^2 \over 2m}{d^2 \psi \over dx^2} = E\psi(x)
\end{displaymath}

If the particle has a potential energy $V(x)$ affecting it, then the same prescription can be used. Start with the classical energy expression:

\begin{displaymath}
{p^2 \over 2m} + V(x) = E
\end{displaymath}

Multiply by $\psi(x)$:

\begin{displaymath}
\left[{p^2 \over 2m} + V(x)\right]\psi(x) = E\psi(x)
\end{displaymath}

Replace $p \longrightarrow -i\hbar (d/dx)$, and we arrive at the Schrödinger equation for the general case of a single quantum particle in one spatial dimension:
$\displaystyle -{\hbar^2 \over 2m}{d^2 \psi \over dx^2} + V(x) \psi(x)$ $\textstyle =$ $\displaystyle E\psi(x)$  
       
$\displaystyle \left[-{\hbar^2 \over 2m}{d^2 \over dx^2} + V(x)\right] \psi(x)$ $\textstyle =$ $\displaystyle E\psi(x)$  

The object on the left that acts on $\psi(x)$

\begin{displaymath}
-{\hbar^2 \over 2m}{d^2 \over dx^2} + V(x)
\end{displaymath}

is an example of an operator. In effect, what is says to do is ``take the second derivative of $\psi(x)$, multiply the result by $-(\hbar^2/2m)$ and then add $V(x)\psi(x)$ to the result of that.'' Quantum mechanics involves many different types of operators. This one, however, plays a special role because it appears on the left side of the Schrödinger equation. It is given a special name, therefore - it is called the Hamiltonian operator and is denoted as

\begin{displaymath}
\hat{H} = -{\hbar^2 \over 2m}{d^2 \over dx^2} + V(x)
\end{displaymath}

Therefore, the Schrödinger equation is generally written as

\begin{displaymath}
\hat{H}\psi(x) = E\psi(x)
\end{displaymath}

Note that $\hat{H}$ is derived from the classical energy $p^2/2m + V(x)$ simply by replacing $p \longrightarrow -i\hbar (d/dx)$.



Since $\vert\psi(x)\vert^2 dx$ is a probability, we require that the probability of finding the particle somewhere in space be exactly 1. That is, we require that the probability that $x\in (-\infty,\infty)$ be 1, which means

\begin{displaymath}
\int_{-\infty}^{\infty}\vert\psi(x)\vert^2 dx = 1
\end{displaymath}

This is known as the normalization condition on $\psi(x)$. Note that if we are working on a subset of the real line, then the integral in the normalization condition must be restricted to the part of the line to which we are restricted.



Finally, we need to specify how $\psi(x)$ behaves at the physical boundaries of the space we are working in. These conditions are known as boundary conditions.



Once we specify the Schrödinger equation, the boundary conditions on $\psi(x)$ and the normalization condition, we have all the information we need to calculate both the allowed energies and the wave function $\psi(x)$. We will see shortly how this prescription is applied to a few simple examples.



Before we embark on this, however, let us pause to comment on the validity of quantum mechanics. Despite its weirdness, its abstractness, and its strange view of the universe as a place of randomness and unpredictability, quantum theory has been subject to intense experimental scrutiny. It has been found to agree with experiments to better than 10$^{-10}$% for all cases studied so far. When the Schrödinger equation is combined with a quantum description of the electromagnetic field, a theory known as quantum electrodynamics, the result is one of the most accurate theories of matter that has ever been put forth. Keeping this in mind, let us forge ahead in our discussion of the quantum universe and how to apply quantum theory to both model and real situations.


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Next: A simple model of Up: lecture_6 Previous: lecture_6
Mark E. Tuckerman 2011-10-04