The harmonic oscillator (HO) potential
- Any smoothly varying potential energy function looks like the HO potential at small distances from equilibrium,
- The quantum HO has an exact solution,
- There's an elegant solution involving "ladder operators" which has wider use in QM and more advanced field theories.
The classical HO
...or a mass on a spring. You know how to solve this classically (with no friction). The restoring force is $F=-kx$ and so...
$$\begineq F&=&ma\\
-kx&=&m\frac{d^2x}{dt^2}\endeq$$
One way of writing the general solution is $x(t)=A\sin \omega t+B\cos \omega t$, where
$$\omega \equiv \sqrt{\frac{k}{m}}.$$
The potential function for the HO ($F(x)=-\grad V(x)$) is $$V(x)=\frac{1}{2}kx^2.$$ This is a parabolic function. The curvature at the bottom of the potential is $\frac{d^2V(x)}{dx^2}=k$.
[Actually, the curvature does not depend on $x$ for this potential, so the curvature is $k$ everywhere.]
Small oscillations: the pendulum
$$V=mgh = mg(L-L\cos\theta)=mgL(1-cos\theta)$$
For small angles, the (Taylor) power series for cosine is: [See a 'handout' on CoCalc for how to use the taylor(...) method in Sagemath] $$\cos(\theta)=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+....$$ So, to lowest order in $\theta$, $$1-\cos\theta \approx \theta^2/2.$$ Putting this into $V$... $$V\approx mgL\theta^2/2=\frac{mg}{2L}x^2$$ where the arclength $x=L\theta$.
This potential looks like $V(x)=\frac{1}{2}kx^2$ for $k=mg/L$, so we take advantage of our spring formula to write the frequency for oscillations (for small $\theta$):
$$\omega = \sqrt\frac{k}{m}=\sqrt\frac{g}{L}.$$
More generally, any smoothly varying potential can be written as a Taylor series wrt $x_0$, the position of the bottom of the potential:
$$V(x)=V(x_0)+V'(x_0)\times[x-x_0]+\frac{1}{2}V''(x_0)\times[x-x_0]^2 +\frac{1}{3!}V'''(x_0)\times[x-x_0]^3+...$$
- $V(x_0)$ is the vertical offset of the bottom of the well. Changing that $\to 0$ amounts to shifting the graph down (or up) but doesn't affect the force that a particle in the well feels ($F=-\grad V=-dV/dx$).
- $V'(x_0)$ is the slope of the well at the bottom of the well. That is $V'(x_0)=0$
So, to lowest order in $[x-x_0]^n$, $$V(x)\approx \frac{1}{2}V''(x_0)[x-x_0]^2.$$ ...and we saw that $k= V''$ at the bottom of the well of the HO.
So, any of these valleys could be fit with a parabola of some particular curvature $k$ at the bottom, and the frequency of small oscillations of a particle of mass $m$ would be
$$\omega=\sqrt\frac{k}{m}.$$
Sketches
Before we find the answer analytically, try your hand at sketching the first three stationary states--with energies corresponding to the colored lines in this sketch of $V(x)$:
Quantum Harmonic Oscillator
We want to solve the stationary Schrödinger equation (using $\omega=\sqrt\frac{k}{m}$), $$\begineq \hat H \psi= \left(\frac{1}{2m}\hat p^2+V(\hat x)\right)\psi&=&E\psi\\ \left(\frac{1}{2m}\hat p^2+\frac{1}{2}k\hat x^2\right)\psi&=&\\ \left(\frac{1}{2m}\hat p^2+\frac{1}{2}m\omega^2\hat x^2\right)\psi&=&\\ \frac{1}{2m}\[\hat p^2+(m\omega\hat x)^2\]\psi&=&E\psi.\endeq$$
Now, consider these two operators which I'll define with one statement:
$$\hat a_\pm\equiv \frac{1}{\sqrt{2\hbar m\omega}} \left(\mp i\hat p+m\omega\hat x\right)$$
The product of these two operators is almost proportional to $\hat p^2 +(m\omega \hat x)^2$: $$\begineq \hat a_-\hat a_+ &=&\frac{1}{2\hbar m\omega}(i\hat p+m\omega \hat x)(-i\hat p + m\omega \hat x)\\ &=&\frac{1}{2\hbar m\omega}\[\hat p^2+(m\omega \hat x)^2-im\omega (\hat x\hat p-\hat p\hat x)\] \label{commproduct}\endeq$$
That last factor involves a difference of operators called the commutator of $\hat x$ and $\hat p$:
$$\[\hat x, \hat p\]\equiv \hat x \hat p - \hat p \hat x$$.
If that commutator only vanished, we'd have the Hamiltonian. Does it?
To evaluate the commutator we have to let it operate on some function of $x$, and see what happens: $$\begineq\[\hat x, \hat p\]f(x) =x \frac{\hbar}{i}\frac{d}{dx}f(x)-\frac{\hbar}{i}\frac{d}{dx}(xf(x))\\ = \frac{\hbar}{i}\left(xf'-f -xf')\right)=i\hbar f(x). \endeq$$
So, now we can take away $f(x)$ above and write this as...
This result is known as the canonical commutation relation: $$[\hat x, \hat p]=i\hbar$$
Using the same technique, it can be shown that
- Any operator $\hat Q(\hat x, \hat p)$ commutes with itself: $$[\hat Q, \hat Q]=0,$$
- And any complex number $z$ commutes with any operator: $$[\hat Q, z]=0.$$
Using the canonical conjucation relation, we can re-write (\ref{commproduct}) as: $$\begineq \hat a_-\hat a_+ &=&\frac{1}{2\hbar m\omega}\[\hat p^2+(m\omega \hat x)^2-im\omega (\hat x\hat p-\hat p\hat x)\]\\ &=&\frac{1}{2\hbar m\omega}\[\hat p^2+(m\omega \hat x)^2+\hbar m\omega\]\\ &=&\frac{\hat H}{\hbar \omega}+\frac{1}{2} \endeq$$
Running through the same line of argument, with $\hat a_+\hat a_-$ instead of $\hat a_-\hat a_+$, it turns out that: $$ \hat a_+\hat a_- = \frac{\hat H}{\hbar \omega}-\frac{1}{2}$$
Problem: Use these last two results to calculate the commutator $[\hat a_-,\hat a_+]$.
Depending on which of these equations I choose to solve for $\hat H$, I can write the stationary Schrödinger equation two ways: $$\hat H \psi = \hbar\omega\left(\hat a_\pm \hat a_\mp \pm \frac 1 2\right)\psi=E\psi.$$
Up and down the ladder...
We've got a commutation relation $$[\hat a_-,\hat a_+]=1$$ and the time independent Schrödinger equation: $$\hat H \psi = \hbar\omega\left(\hat a_\pm a_\mp \pm \frac 1 2\right)\psi=E\psi.$$
Now, imagine that we live in some happier world,
where we have already discovered one solution $\psi$ that has an energy $E$....
We wouldn't be so lucky as to find that $\hat a_+ \psi$ is also a solution...would we??
Try putting this into the Sch equation and let's see if it's a solution too, that is $$\hat H(\hat a_+\psi)\stackrel ? = E'(\hat a_+\psi).$$
$$\begineq \hat H(\hat a_+\psi)&=& \hbar\omega\left(\hat a_+ \hat a_- + \frac 1 2\right)(\hat a_+\psi)\\ &=&\hbar\omega\left(\hat a_+ \hat a_- \hat a_+ + \frac 1 2 \hat a_+ \right)\psi\\ &=&\hbar\omega\hat a_+ \left(\hat a_- \hat a_+ + \frac 1 2 \right)\psi\\ &=&\hbar\omega\hat a_+ \left(1+\hat a_+ \hat a_- + \frac 1 2 \right)\psi\\ &=&\hat a_+ \left(\hbar\omega+\hat H \right)\psi\\ &=&\hat a_+ \left(\hbar\omega\psi+\hat H\psi\right)\\ &=&\hat a_+ \left(\hbar\omega\psi+E\psi\right)\\ &=&\hat a_+ \left(\hbar\omega+E\right)\psi\\ \hat H(\hat a_+ \psi)&=& \left(\hbar\omega+E\right)\hat a_+\psi\label{goup}\\ \endeq$$ This new solution $\hat a_+\psi$, apparently has an energy $E'=E+\hbar\omega$.
Using the same kinds of manipulations, we can show that $$\hat H(\hat a_-\psi)=(E-\hbar\omega)(\hat a_-\psi).$$ (BTW, there's no guarantee that $\hat a_-\psi$ is a normalized stationary state.)
- $\hat a_+$ is called the raising operator and
- $\hat a_-$ is the lowering operator.
- Together they are referred to as ladder operators.
The energy must be positive, just as in the case of the infinite square well. Repeated application of $\hat a_-$ ought to get us down to the lowest, ground state $\psi_0$ with an $E_0\le\hbar \omega$. If we apply the lowering operator just one more time to $\psi_0$, we ought to get either some non-normalizable solution, or else just zero: $$\hat a_-\psi_0=0$$ Is there some solution, $\psi_0(x)$, to this equation?
$$0=\hat a_-\psi_0=\frac{1}{\sqrt{2\hbar m\omega}} \left( i\hat p+m\omega\hat x\right)\psi_0.$$ With $\hat p=\frac \hbar i \frac{d}{dx}$, this becomes $$\frac{d\psi_0}{dx}=-\frac{m\omega}{\hbar}x\psi_0.$$
Problem: Use separation of variables to find the functional form of $\psi_0$:
$$\psi_0(x)=?$$
Problem: It turns out that we can apply the stationary Schrödinger equation to find the energy without worrying about the functional form of $\psi_0$ if we recall how the ladder operators work on $\psi_0$...
$$\hat H \psi_0=\hbar\omega\left(\hat a_+\hat a_- +\frac 1 2 \right)\psi_0=E_0\psi_0$$
$$\Rightarrow E_0=?$$
Up the ladder
Once we know that $E_0=\hbar\omega/2$, equation (\ref{goup}) tells us that there is always another solution with an energy $\hbar \omega$ above any stationary state, so, $$E_n=\left(n +\frac 1 2 \right) \hbar \omega.$$
Zero-point energy: Unlike a classical HO, the QMHO ground state $E_0\ne 0$...!
[It is an unfortunate convention, that the ground state of the HO is written $\psi_0$, while the ground state of the infinite well is written $\psi_1$!]
The normalized ground state is
$$\begineq\psi_0&=&\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^2}
=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{1}{\hbar\omega}\frac{m\omega^2x^2}{2}}\\
&=&\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{1}{\hbar\omega}V(x)}.\endeq$$
Plotting $V(x)=x^2$ with $x$ in units of $\sqrt{\frac{1}{2}m\omega^2}$ and $y$ in units of $\hbar \omega$,
- A classical particle with the ground state energy of $\frac 1 2$ would never be found outside of $-\frac{1}{\sqrt 2} \lt x \lt +\frac{1}{\sqrt 2}$.
- The wave function $\psi_0$ has an amplitude outside of this range.
- But the curvature switches sign at $x\pm\frac{1}{\sqrt 2}$.
To get to the next state up in energy, apply the raising operator once (I'm just interested in the functional form, so I'm going to write a bunch of the constants as $[...]$):
$$\begineq\psi_1&=&[...]\,\hat a_+\psi_0(x)\\
&=&[...]\left(-\hbar\frac d{dx}+m\omega x
\right)e^{-\frac{m\omega}{2\hbar}x^2}\\
&=&[...]\left(-\hbar e^{-\frac{m\omega}{2\hbar}x^2}\frac d{dx}\[-\frac{m\omega}{2\hbar}x^2\]+m\omega xe^{-\frac{m\omega}{2\hbar}x^2}\right)\\
&=&[...]\left(+m\omega x e^{-\frac{m\omega}{2\hbar}x^2}+m\omega xe^{-\frac{m\omega}{2\hbar}x^2}\right)\\
&=&[...] xe^{-\frac{m\omega}{2\hbar}x^2}\\
\endeq$$
To generate higher states, we could keep on applying the raising operator... $$\psi_n=A_n(\hat a_+)^n\psi_0(x),$$ where $A_n$ is a constant that we calculate so as to normalize each solution.
The first couple of stationary states look like...
In this picture, the classical turning points are labelled as $x_{\pm n}$.
- Number of nodes is $n-1$.
- Successive states are odd, even, odd,....
Problem: In classical terms why would you expect that the probability density is higher farther away from the center of the well?
Problem: Can you think of an argument (one possibility is based on curvature) to convince yourself that we haven't missed some solutions (maybe on a parallel ladder of solutions...)?
Image credits
Anjan, George Seurat, "A Sunday Afternoon on the Island of La Grande Jatte"
