In this section we will increase our quantum-mechanical repertoire by solving the Schrödinger equation for the one-dimensional harmonic oscillator. This system is important as a model for molecular vibrations.
Classical-Mechanical Treatment
Before looking at the wave mechanics of the harmonic oscillator, we review the classical treatment. We have a single particle of mass $m$ attracted toward the origin by a force proportional to the particle's displacement from the origin:
\(
\begin{equation}
F_{x}=-k x \tag{4.20}
\end{equation}
\)
The proportionality constant $k$ is called the force constant. $F_{x}$ is the $x$ component of the force on the particle. This is also the total force in this one-dimensional problem. Equation (4.20) is obeyed by a particle attached to a spring, provided the spring is not stretched greatly from its equilibrium position.
Newton's second law, $F=m a$, gives
\(
\begin{equation}
-k x=m \frac{d^{2} x}{d t^{2}} \tag{4.21}
\end{equation}
\)
where $t$ is the time. Equation (4.21) is the same as Eq. (4.1) with $c^{2}=k / m$; hence the solution is [Eq. (4.3) with $c=(k / m)^{1 / 2}$ ]
\(
\begin{equation}
x=A \sin (2 \pi \nu t+b) \tag{4.22}
\end{equation}
\)
where $A$ (the amplitude of the vibration) and $b$ are the integration constants, and the vibration frequency $\nu$ is
\(
\begin{equation}
\nu=\frac{1}{2 \pi}\left(\frac{k}{m}\right)^{1 / 2} \tag{4.23}
\end{equation}
\)
Since the sine function has maximum and minimum values of 1 and -1 , respectively, $x$ in (4.22) oscillates between $A$ and $-A$. The sine function repeats itself every $2 \pi$ radians, and the time needed for one complete oscillation (called the period) is the time it takes for the argument of the sine function to increase by $2 \pi$. At time $t+1 / \nu$, the argument of the sine function is $2 \pi \nu(t+1 / \nu)+b=2 \pi \nu t+2 \pi+b$, which is $2 \pi$ greater than the argument at time $t$, so the period is $1 / \nu$. The reciprocal of the period is the number of vibrations per unit time (the vibrational frequency), and so the frequency is $\nu$.
Now consider the energy. The potential energy $V$ is related to the components of force in the three-dimensional case by
\(
\begin{equation}
F{x}=-\frac{\partial V}{\partial x}, \quad F{y}=-\frac{\partial V}{\partial y}, \quad F_{z}=-\frac{\partial V}{\partial z} \tag{4.24}
\end{equation}
\)
Equation (4.24) is the definition of potential energy. Since this is a one-dimensional problem, we have [Eq. (1.12)]
\(
\begin{equation}
F_{x}=-\frac{d V}{d x}=-k x \tag{4.25}
\end{equation}
\)
Integration of (4.25) gives $V=\int k x d x=\frac{1}{2} k x^{2}+C$, where $C$ is a constant. The potential energy always has an arbitrary additive constant. Choosing $C=0$, we have [Eq. (4.23)]
\(
\begin{gather}
V=\frac{1}{2} k x^{2} \tag{4.26}\
V=2 \pi^{2} \nu^{2} m x^{2} \tag{4.27}
\end{gather}
\)
The graph of $V(x)$ is a parabola (Fig. 4.5). The kinetic energy $T$ is
\(
\begin{equation}
T=\frac{1}{2} m(d x / d t)^{2} \tag{4.28}
\end{equation}
\)
and can be found by differentiating (4.22) with respect to $t$. Adding $T$ and $V$, one finds for the total energy (Prob. 4.4)
\(
\begin{equation}
E=T+V=\frac{1}{2} k A^{2}=2 \pi^{2} \nu^{2} m A^{2} \tag{4.29}
\end{equation}
\)
where the identity $\sin ^{2} \theta+\cos ^{2} \theta=1$ was used.
According to (4.22), the classical harmonic oscillator vibrates back and forth between $x=A$ and $x=-A$. These two points are the turning points for the motion. The particle has zero speed at these points, and the speed increases to a maximum at $x=0$, where the potential energy is zero and the energy is all kinetic energy. The classical harmonic oscillator spends more time in each of the regions near $x=A$ and $x=-A$ (where it is moving the slowest) than it does in the region near $x=0$. Problem 4.18 works out the probability density for finding the classical harmonic oscillator at various locations. (Interestingly, this probability density becomes infinite at the turning points.)
Quantum-Mechanical Treatment
The harmonic-oscillator Hamiltonian operator is [Eqs. (3.27) and (4.27)]
\(
\begin{equation}
\hat{H}=\hat{T}+\hat{V}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+2 \pi^{2} \nu^{2} m x^{2}=-\frac{\hbar^{2}}{2 m}\left(\frac{d^{2}}{d x^{2}}-\alpha^{2} x^{2}\right) \tag{4.30}
\end{equation}
\)
where, to save time in writing, $\alpha$ was defined as
\(
\begin{equation}
\alpha \equiv 2 \pi \nu m / \hbar \tag{4.31}
\end{equation}
\)
The Schrödinger equation $\hat{H} \psi=E \psi$ reads, after multiplication by $2 m / \hbar^{2}$,
\(
\begin{equation}
\frac{d^{2} \psi}{d x^{2}}+\left(2 m E \hbar^{-2}-\alpha^{2} x^{2}\right) \psi=0 \tag{4.32}
\end{equation}
\)
We might now attempt a power-series solution of (4.32). If we do now try a power series for $\psi$ of the form (4.4), we will find that it leads to a three-term recursion relation, which is harder to deal with than a two-term recursion relation like Eq. (4.14). We therefore modify the form of (4.32) so as to get a two-term recursion relation when we try a series solution. A substitution that will achieve this purpose is (see Prob. 4.22) $f(x) \equiv e^{\alpha x^{2} / 2} \psi(x)$. Thus
\(
\begin{equation}
\psi=e^{-\alpha x^{2} / 2} f(x) \tag{4.33}
\end{equation}
\)
This equation is simply the definition of a new function $f(x)$ that replaces $\psi(x)$ as the unknown function to be solved for. (We can make any substitution we please in a differential equation.) Differentiating (4.33) twice, we have
\(
\begin{equation}
\psi^{\prime \prime}=e^{-\alpha x^{2} / 2}\left(f^{\prime \prime}-2 \alpha x f^{\prime}-\alpha f+\alpha^{2} x^{2} f\right) \tag{4.34}
\end{equation}
\)
Substituting (4.33) and (4.34) into (4.32), we find
\(
\begin{equation}
f^{\prime \prime}(x)-2 \alpha x f^{\prime}(x)+\left(2 m E \hbar^{-2}-\alpha\right) f(x)=0 \tag{4.35}
\end{equation}
\)
Now we try a series solution for $f(x)$ :
\(
\begin{equation}
f(x)=\sum{n=0}^{\infty} c{n} x^{n} \tag{4.36}
\end{equation}
\)
Assuming the validity of term-by-term differentiation of (4.36), we get
\(
\begin{equation}
f^{\prime}(x)=\sum{n=1}^{\infty} n c{n} x^{n-1}=\sum{n=0}^{\infty} n c{n} x^{n-1} \tag{4.37}
\end{equation}
\)
\(
f^{\prime \prime}(x)=\sum{n=2}^{\infty} n(n-1) c{n} x^{n-2}=\sum{j=0}^{\infty}(j+2)(j+1) c{j+2} x^{j}=\sum{n=0}^{\infty}(n+2)(n+1) c{n+2} x^{n}
\)
where we made the substitution $j=n-2$ and then changed the summation index from $j$ to $n$. [See Eq. (4.9).] Substitution into (4.35) gives
\(
\begin{align}
& \sum{n=0}^{\infty}(n+2)(n+1) c{n+2} x^{n}-2 \alpha \sum{n=0}^{\infty} n c{n} x^{n}+\left(2 m E \hbar^{-2}-\alpha\right) \sum{n=0}^{\infty} c{n} x^{n}=0 \
& \sum{n=0}^{\infty}\left[(n+2)(n+1) c{n+2}-2 \alpha n c{n}+\left(2 m E \hbar^{-2}-\alpha\right) c{n}\right] x^{n}=0 \tag{4.38}
\end{align}
\)
Setting the coefficient of $x^{n}$ equal to zero [for the same reason as in Eq. (4.11)], we have
\(
\begin{equation}
c{n+2}=\frac{\alpha+2 \alpha n-2 m E \hbar^{-2}}{(n+1)(n+2)} c{n} \tag{4.39}
\end{equation}
\)
which is the desired two-term recursion relation. Equation (4.39) has the same form as (4.14), in that knowing $c{n}$ we can calculate $c{n+2}$. We thus have two arbitrary constants: $c{0}$ and $c{1}$. If we set $c_{1}$ equal to zero, then we will have as a solution a power series containing only even powers of $x$, multiplied by the exponential factor:
\(
\begin{equation}
\psi=e^{-\alpha x^{2} / 2} f(x)=e^{-\alpha x^{2} / 2} \sum{n=0,2,4, \ldots}^{\infty} c{n} x^{n}=e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l} x^{2 l} \tag{4.40}
\end{equation}
\)
If we set $c_{0}$ equal to zero, we get another independent solution:
\(
\begin{equation}
\psi=e^{-\alpha x^{2} / 2} \sum{n=1,3, \ldots}^{\infty} c{n} x^{n}=e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l+1} x^{2 l+1} \tag{4.41}
\end{equation}
\)
The general solution of the Schrödinger equation is a linear combination of these two independent solutions [recall Eq. (2.4)]:
\(
\begin{equation}
\psi=A e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l+1} x^{2 l+1}+B e^{-\alpha x^{2} / 2} \sum{l=0}^{\infty} c{2 l} x^{2 l} \tag{4.42}
\end{equation}
\)
where $A$ and $B$ are arbitrary constants.
We now must see if the boundary conditions on the wave function lead to any restrictions on the solution. To see how the two infinite series behave for large $x$, we examine the ratio of successive coefficients in each series. The ratio of the coefficient of $x^{2 l+2}$ to that of $x^{2 l}$ in the second series is [set $n=2 l$ in (4.39)]
\(
\frac{c{2 l+2}}{c{2 l}}=\frac{\alpha+4 \alpha l-2 m E \hbar^{-2}}{(2 l+1)(2 l+2)}
\)
Assuming that for large values of $x$ the later terms in the series are the dominant ones, we look at this ratio for large values of $l$ :
\(
\begin{equation}
\frac{c{2 l+2}}{c{2 l}} \sim \frac{4 \alpha l}{(2 l)(2 l)}=\frac{\alpha}{l} \quad \text { for } l \text { large } \tag{4.43}
\end{equation}
\)
Setting $n=2 l+1$ in (4.39), we find that for large $l$ the ratio of successive coefficients in the first series is also $\alpha / l$. Now consider the power-series expansion for the function $e^{\alpha x^{2}}$. Using (Prob. 4.3)
\(
\begin{equation}
e^{z}=\sum_{n=0}^{\infty} \frac{z^{n}}{n!}=1+z+\frac{z^{2}}{2!}+\cdots \tag{4.44}
\end{equation}
\)
FIGURE 4.1 Lowest five energy levels for the one-dimensional harmonic oscillator.
we get
\(
e^{\alpha x^{2}}=1+\alpha x^{2}+\cdots+\frac{\alpha^{l} x^{2 l}}{l!}+\frac{\alpha^{l+1} x^{2 l+2}}{(l+1)!}+\cdots
\)
The ratio of the coefficients of $x^{2 l+2}$ and $x^{2 l}$ in this series is
\(
\frac{\alpha^{l+1}}{(l+1)!} \div \frac{\alpha^{l}}{l!}=\frac{\alpha}{l+1} \sim \frac{\alpha}{l} \quad \text { for large } l
\)
Thus the ratio of successive coefficients in each of the infinite series in the solution (4.42) is the same as in the series for $e^{\alpha x^{2}}$ for large $l$. We conclude that, for large $x$, each series behaves as $e^{\alpha x^{2}}$. [This is not a rigorous proof. A proper mathematical derivation is given in H. A. Buchdahl, Am. J. Phys., 42, 47 (1974); see also M. Bowen and J. Coster, Am. J. Phys., 48, 307 (1980).]
If each series behaves as $e^{\alpha x^{2}}$, then (4.42) shows that $\psi$ will behave as $e^{\alpha x^{2} / 2}$ for large $x$. The wave function will become infinite as $x$ goes to infinity and will not be quadratically integrable. If we could somehow break off the series after a finite number of terms, then the factor $e^{-\alpha x^{2} / 2}$ would ensure that $\psi$ went to zero as $x$ became infinite. (Using l'Hôpital's rule, it is easy to show that $x^{p} e^{-\alpha x^{2} / 2}$ goes to zero as $x \rightarrow \infty$, where $p$ is any finite power.) To have one of the series break off after a finite number of terms, the coefficient of $c{n}$ in the recursion relation (4.39) must become zero for some value of $n$, say for $n=v$. This makes $c{v+2}, c{v+4}, \ldots$ all equal to zero, and one of the series in (4.42) will have a finite number of terms. In the recursion relation (4.39), there is one quantity whose value is not yet fixed, but can be adjusted to make the coefficient of $c{v}$ vanish. This quantity is the energy $E$. Setting the coefficient of $c_{v}$ equal to zero in (4.39) and using (4.31) for $\alpha$, we get
\(
\begin{gather}
\alpha+2 \alpha v-2 m E \hbar^{-2}=0 \
2 m E \hbar^{-2}=(2 v+1) 2 \pi \nu m \hbar^{-1} \
E=\left(v+\frac{1}{2}\right) h \nu, \quad v=0,1,2, \ldots \tag{4.45}
\end{gather}
\)
The harmonic-oscillator stationary-state energy levels (4.45) are equally spaced (Fig. 4.1). Do not confuse the quantum number $v$ (vee) with the vibrational frequency $\nu$ (nu).
Substitution of (4.45) into the recursion relation (4.39) gives
\(
\begin{equation}
c{n+2}=\frac{2 \alpha(n-v)}{(n+1)(n+2)} c{n} \tag{4.46}
\end{equation}
\)
By quantizing the energy according to (4.45), we have made one of the series break off after a finite number of terms. To get rid of the other infinite series in (4.42), we must set the arbitrary constant that multiplies it equal to zero. This leaves us with a wave function that is $e^{-\alpha x^{2} / 2}$ times a finite power series containing only even or only odd powers of $x$, depending on whether $v$ is even or odd, respectively. The highest power of $x$ in this power series is $x^{v}$, since we chose $E$ to make $c{v+2}, c{v+4}, \ldots$ all vanish. The wave functions (4.42) are thus
\(
\psi{v}=\left{\begin{array}{lr}
e^{-\alpha x^{2} / 2}\left(c{0}+c{2} x^{2}+\cdots+c{v} x^{v}\right) & \text { for } v \text { even } \tag{4.47}\
e^{-\alpha x^{2} / 2}\left(c{1} x+c{3} x^{3}+\cdots+c_{v} x^{v}\right) & \text { for } v \text { odd }
\end{array}\right.
\)
where the arbitrary constants $A$ and $B$ in (4.42) can be absorbed into $c{1}$ and $c{0}$, respectively, and can therefore be omitted. The coefficients after $c{0}$ and $c{1}$ are found from the recursion relation (4.46). Since the quantum number $v$ occurs in the recursion relation, we get a different set of coefficients $c{i}$ for each different $v$. For example, $c{2}$ in $\psi{4}$ differs from $c{2}$ in $\psi_{2}$.
As in the particle in a box, the requirement that the wave functions be well-behaved forces us to quantize the energy. For values of $E$ that differ from (4.45), $\psi$ is not quadratically integrable. For example, Fig. 4.2 plots $\psi$ of Eq. (4.40) for the values $E / h \nu=0.499,0.500$, and 0.501 , where the recursion relation (4.39) is used to calculate the coefficients $c_{n}$ (see also Prob. 4.23). Figure 4.3 gives an enlarged view of these curves in the region near $\alpha^{1 / 2} x=3$.
The harmonic-oscillator ground-state energy is nonzero. This energy, $\frac{1}{2} h \nu$, is called the zero-point energy. This would be the vibrational energy of a harmonic oscillator in a collection of harmonic oscillators at a temperature of absolute zero. The zero-point energy can be understood from the uncertainty principle. If the lowest state had an energy of zero, both its potential and kinetic energies (which are nonnegative) would have to be zero. Zero kinetic energy would mean that the momentum was exactly zero, and $\Delta p{x}$ would be zero. Zero potential energy would mean that the particle was always located at the origin, and $\Delta x$ would be zero. But we cannot have both $\Delta x$ and $\Delta p{x}$ equal to zero. Hence the need for a nonzero ground-state energy. Similar ideas apply for the particle in a box. The definition of the zero-point energy (ZPE) is $E{\mathrm{ZPE}}=E{\mathrm{gs}}-V{\min }$, where $E{\mathrm{gs}}$ and $V_{\min }$ are the groundstate energy and the minimum value of the potential-energy function.
FIGURE 4.2 Plots of the harmonic-oscillator Schrödinger-equation solution containing only even powers of $x$ for $E=0.499 h \nu, E=0.500 h \nu$, and $E=0.501 h \nu$. In the region around $x=0$ the three curves nearly coincide. For $\left|\alpha^{1 / 2} x\right|>3$ the $E=0.500 h \nu$ curve nearly coincides with the $x$ axis.
FIGURE 4.3 Enlargement of Fig. 4.2 in the region $\alpha^{1 / 2} x=3$
Even and Odd Functions
Before considering the wave functions in detail, we define even and odd functions. If $f(x)$ satisfies
\(
\begin{equation}
f(-x)=f(x) \tag{4.48}
\end{equation}
\)
then $f$ is an even function of $x$. Thus $x^{2}$ and $e^{-b x^{2}}$ are both even functions of $x$ since $(-x)^{2}=x^{2}$ and $e^{-b(-x)^{2}}=e^{-b x^{2}}$. The graph of an even function is symmetric about the $y$ axis (for example, see Fig. 4.4a). Therefore
\(
\begin{equation}
\int{-a}^{+a} f(x) d x=2 \int{0}^{a} f(x) d x \text { for } f(x) \text { even } \tag{4.49}
\end{equation}
\)
If $g(x)$ satisfies
\(
\begin{equation}
g(-x)=-g(x) \tag{4.50}
\end{equation}
\)
then $g$ is an odd function of $x$. Examples are $x, 1 / x$, and $x^{3} e^{x^{2}}$. Setting $x=0$ in (4.50), we see that an odd function must be zero at $x=0$, provided $g(0)$ is defined and singlevalued. The graph of an odd function has the general appearance of Fig. 4.4b. Because positive contributions on one side of the $y$ axis are canceled by corresponding negative contributions on the other side, we have
\(
\begin{equation}
\int_{-a}^{+a} g(x) d x=0 \quad \text { for } g(x) \text { odd } \tag{4.51}
\end{equation}
\)
It is easy to show that the product of two even functions or of two odd functions is an even function, while the product of an even and an odd function is an odd function.
The Harmonic-Oscillator Wave Functions
The exponential factor $e^{-\alpha x^{2} / 2}$ in (4.47) is an even function of $x$. If $v$ is an even number, the polynomial factor contains only even powers of $x$, which makes $\psi{v}$ an even function. If $v$ is odd, the polynomial factor contains only odd powers of $x$, and $\psi{v}$, being the product of an even function and an odd function, is an odd function. Each harmonic-oscillator stationary state $\psi$ is either an even or odd function according to whether the quantum number $v$ is even or odd. In Section 7.5, we shall see that, when the potential energy $V$ is an even function, the wave functions of nondegenerate levels must be either even or odd functions.
We now find the explicit forms of the wave functions of the lowest three levels. For the $v=0$ ground state, Eq. (4.47) gives
\(
\begin{equation}
\psi{0}=c{0} e^{-\alpha x^{2} / 2} \tag{4.52}
\end{equation}
\)
where the subscript on $\psi$ gives the value of $v$. We fix $c_{0}$ by normalization:
\(
1=\int{-\infty}^{\infty}\left|c{0}\right|^{2} e^{-\alpha x^{2}} d x=2\left|c{0}\right|^{2} \int{0}^{\infty} e^{-\alpha x^{2}} d x
\)
where Eq. (4.49) has been used. Using the integral (A.9) in the Appendix, we find $\left|c_{0}\right|=(\alpha / \pi)^{1 / 4}$. Therefore,
\(
\begin{equation}
\psi_{0}=(\alpha / \pi)^{1 / 4} e^{-\alpha x^{2} / 2} \tag{4.53}
\end{equation}
\)
if we choose the phase of the normalization constant to be zero. The wave function (4.53) is a Gaussian function (Fig. 4.4a).
For the $v=1$ state, Eq. (4.47) gives
\(
\begin{equation}
\psi{1}=c{1} x e^{-\alpha x^{2} / 2} \tag{4.54}
\end{equation}
\)
After normalization using the integral in Eq. (A.10), we have
\(
\begin{equation}
\psi_{1}=\left(4 \alpha^{3} / \pi\right)^{1 / 4} x e^{-\alpha x^{2} / 2} \tag{4.55}
\end{equation}
\)
Figure 4.4 b shows $\psi_{1}$.
For $v=2$, Eq. (4.47) gives
\(
\psi{2}=\left(c{0}+c_{2} x^{2}\right) e^{-\alpha x^{2} / 2}
\)
The recursion relation (4.46) with $v=2$, gives
\(
c{2}=\frac{2 \alpha(-2)}{1 \cdot 2} c{0}=-2 \alpha c_{0}
\)
Therefore
\(
\begin{equation}
\psi{2}=c{0}\left(1-2 \alpha x^{2}\right) e^{-\alpha x^{2} / 2} \tag{4.56}
\end{equation}
\)
Evaluating $c_{0}$ by normalization, we find (Prob. 4.10)
\(
\begin{equation}
\psi_{2}=(\alpha / 4 \pi)^{1 / 4}\left(2 \alpha x^{2}-1\right) e^{-\alpha x^{2} / 2} \tag{4.57}
\end{equation}
\)
Note that $c{0}$ in $\psi{2}$ is not the same as $c{0}$ in $\psi{0}$.
The number of nodes in the wave function equals the quantum number $v$. It can be proved (see Messiah, pages 109-110) that for the bound stationary states of a onedimensional problem, the number of nodes interior to the boundary points is zero for the ground-state $\psi$ and increases by one for each successive excited state. The boundary points for the harmonic oscillator are $\pm \infty$. Moreover, one can show that a one-dimensional wave function must change sign as it goes through a node (see Prob. 4.45).
FIGURE 4.4 Harmonicoscillator wave functions. The same scale is used for all graphs. The points marked on the $x$ axes are for $\alpha^{1 / 2} x= \pm 2$.
FIGURE 4.5 The classically
allowed $(-a \leq x \leq a)$ and forbidden ( $x<-a$ and $x>a)$ regions for the harmonic oscillator.
The polynomial factors in the harmonic-oscillator wave functions are well known in mathematics and are called Hermite polynomials, after a French mathematician. (See Prob. 4.21.)
According to the quantum-mechanical solution, there is some probability of finding the particle at any point on the $x$ axis (except at the nodes). Classically, $E=T+V$ and the kinetic energy $T$ cannot be negative: $T \geq 0$. Therefore, $E-V=T \geq 0$ and $V \leq E$. The potential energy $V$ is a function of position, and a classical particle is confined to the region of space where $V \leq E$; that is, where the potential energy does not exceed the total energy. In Fig. 4.5, the horizontal line labeled $E$ gives the energy of a harmonic oscillator, and the parabolic curve gives the potential energy $\frac{1}{2} k x^{2}$. For the regions $x<-a$ and $x>a$, we have $V>E$, and these regions are classically forbidden. The classically allowed region $-a \leq x \leq a$ in Fig. 4.5 is where $V \leq E$.
In quantum mechanics, the stationary-state wave functions are not eigenfunctions of $\hat{T}$ or $\hat{V}$, and we cannot assign definite values to $T$ or $V$ for a stationary state. Instead of the classical equations $E=T+V$ and $T \geq 0$, we have in quantum mechanics that $E=\langle T\rangle+\langle V\rangle$ (Prob. 6.35) and $\langle T\rangle \geq 0$ (Prob. 7.7), so $\langle V\rangle \leq E$ in quantum mechanics, but we cannot write $V \leq E$, and a particle has some probability to be found in classically forbidden regions where $V>E$.
It might seem that, by saying the particle can be found outside the classically allowed region, we are allowing it to have negative kinetic energy. Actually, there is no paradox in the quantum-mechanical view. To verify that the particle is in the classically forbidden region, we must measure its position. This measurement changes the state of the system (Sections 1.3 and 1.4). The interaction of the oscillator with the measuring apparatus transfers enough energy to the oscillator for it to be in the classically forbidden region. An accurate measurement of $x$ introduces a large uncertainty in the momentum and hence in the kinetic energy. Penetration of classically forbidden regions was previously discussed in Sections 2.4 and 2.5.
A harmonic-oscillator stationary state has $E=\left(v+\frac{1}{2}\right) h \nu$ and $V=\frac{1}{2} k x^{2}=2 \pi^{2} \nu^{2} m x^{2}$, so the classically allowed region where $V \leq E$ is where $2 \pi^{2} \nu^{2} m x^{2} \leq\left(v+\frac{1}{2}\right) h \nu$, which gives $x^{2} \leq\left(v+\frac{1}{2}\right) h / 2 \pi^{2} \nu m=(2 v+1) / \alpha$, where $\alpha \equiv 2 \pi \nu m / \hbar$ [Eq. (4.31)]. Therefore, the classically allowed region for the harmonic oscillator is where $-(2 v+1)^{1 / 2} \leq$ $\alpha^{1 / 2} x \leq(2 v+1)^{1 / 2}$.
Note from Fig. 4.4 that $\psi$ oscillates in the classically allowed region and decreases exponentially to zero in the classically forbidden region. We previously saw this behavior for the particle in a rectangular well (Section 2.4).
Figure 4.4 shows that, as we go to higher-energy states of the harmonic oscillator, $\psi$ and $|\psi|^{2}$ tend to have maxima farther and farther from the origin. Since $V=\frac{1}{2} k x^{2}$ increases as we go farther from the origin, the average potential energy $\langle V\rangle=\int{-\infty}^{\infty}|\psi|^{2} V d x$ increases as the quantum number increases. The average kinetic
energy is given by $\langle T\rangle=-\left(\hbar^{2} / 2 m\right) \int{-\infty}^{\infty} \psi^{*} \psi^{\prime \prime} d x$. Integration by parts gives (Prob. 7.7b) $\langle T\rangle=\left(\hbar^{2} / 2 m\right) \int_{-\infty}^{\infty}|d \psi / d x|^{2} d x$. The higher number of nodes in states with a higher quantum number produces a faster rate of change of $\psi$, so $\langle T\rangle$ increases as the quantum number increases.
A classical harmonic oscillator is most likely to be found in the regions near the turning points of the motion, where the oscillator is moving the slowest and $V$ is large. In contrast, for the ground state of a quantum harmonic oscillator, the most probable region is the region around the origin. For high oscillator quantum numbers, one finds that the outer peaks of $|\psi|^{2}$ are larger than the peaks near the origin, and the most probable regions become the regions near the classical turning points, where $V$ is large (see Prob. 4.18). This is an example of the correspondence principle (Section 2.2).
Some online simulations of the quantum harmonic oscillator are available at www.phy. davidson.edu/StuHome/cabellf/energy.html (shows energy levels and wave functions and shows how the wave function diverges when the energy is changed from an allowed value); www.falstad.com/qm1d/ (choose harmonic oscillator from the drop-down menu at the top; double click on one of the small circles at the bottom to show a stationary state; shows energies, wave functions, probability densities; $m$ and $k$ can be varied); demonstrations. wolfram.com/HarmonicOscillatorEigenfunctions (shows $\left|\psi{v}\right|^{2}$ ).