Zero-Point Energy

The zero-point energy of a quantum mechanical system is the energy of the system in its ground state (when the system has the lowest possible energy). Take for example, a harmonically oscillating particle in a potential well. While classically, it may be possible to suppose that the particle could have zero energy if its kinetic energy and the energy potential at its position are both zero, quantum mechanically speaking, such a system is impossible. The Heisenberg uncertainty relation forbids such a scenario where both the particle's position and momentum are known. Here, the Heisenberg uncertainty relation and the Schrodinger equation are used to derive the zero-point energy of a quantum harmonic oscillator.

Hamiltonian of a Harmonic Oscillator

The one-dimensional Hamiltonian is generally written as:

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

or stated otherwise,

\begin{equation}
\hat{H} = \frac{\hat{p}^2}{2m} + V(x)
\end{equation}

where $\hat{p}$ is the one-dimensional momentum operator. Classically, the potential energy of a harmonic oscillator is given by $\int_0^x kx_1 dx_1 = \frac{1}{2}kx^2$, where $k$ is the spring constant. Because $k=m\omega^2$, where $m$ is the mass of the oscillating body and $\omega$ is the angular frequency of the oscillation, the potential energy can also be written as $\frac{1}{2}m\omega^2x^2$. Analogously, for a quantum harmonic oscillator, $V(x)=\frac{1}{2}m\omega^2\hat{x}^2$, where $\hat{x}$ is the one-dimensional position operator.

\begin{equation}
\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2
\end{equation}

Zero-Point Energy by the Heisenberg Uncertainty Relation

The Heisenberg uncertainty relation states that:

\begin{equation}
\sigma_\hat{x} \sigma_{\hat{p}} \geq \frac{\hbar}{2}
\end{equation}

where $\sigma_\hat{x}$ and $\sigma_\hat{p}$ respectively are the uncertainties of the position operator, $\hat{x}$, and momentum operator, $\hat{p}$. The standard deviations $\sigma_\hat{x}$ and $\sigma_\hat{p}$ can also be written as the following:

\begin{align}
\sigma_\hat{x} & = \left\langle\left(\hat{x} - \left\langle\hat{x}\right\rangle\right)^2 \right\rangle^{1/2} \\
\sigma_\hat{p} & = \left\langle\left(\hat{p} - \left\langle\hat{p}\right\rangle\right)^2 \right\rangle^{1/2}
\end{align}

It can be derived therefore that:

\begin{equation}
\left\langle\left(\hat{x} - \left\langle\hat{x}\right\rangle\right)^2 \right\rangle^{1/2} \left\langle\left(\hat{p} - \left\langle\hat{p}\right\rangle\right)^2 \right\rangle^{1/2} \geq \frac{\hbar}{2}
\end{equation}

For the harmonic oscillator we are modelling, the restoring force acting on the particle is proportional to the particle's position, so the energy potential of the well must then be $0$ at $x=0$. The scenario we are trying to induce involves the particle having $0$ potential energy as well as kinetic energy, thus $\langle\hat{x}\rangle=0$ and $\langle\hat{p}\rangle=0$.

\begin{equation}
\left\langle \hat{x}^2 \right\rangle^{1/2} \left\langle \hat{p}^2 \right\rangle^{1/2} \geq \frac{\hbar}{2}
\end{equation}
\begin{equation}
\left\langle \frac{1}{2}m\omega^2\hat{x}^2 \right\rangle \left\langle \frac{\hat{p}^2}{2m} \right\rangle \geq \left(\frac{\hbar\omega}{4} \right)^2
\end{equation}

To utilize the above inequality and apply it to equation $(3)$, consider the following. Let $a$ and $b$ be arbitrary positive numbers and $c$ be a positive constant. If $ab\geq c$, how can the sum $a+b$ be related to $c$?

\begin{align*}
a+b & \geq a + \frac{c}{a} \\
a+b & \geq b + \frac{c}{b}
\end{align*}

The $a$ value that gives the least $a+c/a$ can be found by setting to $0$ the derivative of $a+c/a$ with respect to $a$. This minimum is thus found to occur for $1 - c/a^2 = 0$ when $a = \sqrt{c}$. (This is a minimum since the derivative of $a+c/a$ is negative for $0 < a < \sqrt{c}$ and positive for $a > \sqrt{c}$.) Similarly, the minimum of $b + c/b$ occurs when $b = \sqrt{c}$. Therefore, if $ab \geq c$, then the sum of the positive numbers $a$ and $b$ can be related to the positive constant $c$ by the following inequality:

\begin{equation*}
a+b \geq 2\sqrt{c}
\end{equation*}

Continuing the derivation of the zero-point energy, if we consider $\left\langle m\omega^2\hat{x}^2/2 \right\rangle$ as $a$, $\left\langle \hat{p}^2/(2m) \right\rangle$ as $b$, and $(\hbar\omega/4)^2$ as $c$, then this can be said:

\begin{equation}
\left\langle \frac{1}{2}m\omega^2\hat{x}^2 \right\rangle + \left\langle \frac{\hat{p}^2}{2m} \right\rangle \geq \frac{\hbar\omega}{2}
\end{equation}

By equation $(3)$,

\begin{equation}
\left\langle \hat{H} \right\rangle = \left\langle \frac{\hat{p}^2}{2m} \right\rangle + \left\langle\frac{1}{2}m\omega^2\hat{x}^2 \right\rangle
\end{equation}

The zero-potential energy, $E_0$, according to the Heisenberg uncertainty relation, is consequently found to be non-zero.

\begin{equation}
E_0 \geq \frac{\hbar\omega}{2}
\end{equation}

Zero-Point Energy by the Schrodinger Equation

To pin down the exact value of the zero-point energy, the time-independent Schrodinger equation, $E\psi=\hat{H}\psi$, can be used. The Hamiltonian, $\hat{H}$, has already been found in equation $(3)$ and the one-dimensional momentum operator, $\hat{p}$, is given by $-i\hbar\frac{d}{dx}$.

\begin{equation}
E\psi(x) = -\frac{\hbar^2}{2m}\frac{d^2}{{dx}^2}\psi(x) + \frac{1}{2}m\omega^2\hat{x}^2\psi(x)
\end{equation}
\begin{equation}
0 = \frac{\hbar^2}{2m}\frac{d^2}{{dx}^2}\psi(x) + \left(E - \frac{1}{2}m\omega^2x^2\right)\psi(x)
\end{equation}

The above equation is a homogeneous second-order linear differential equation. Since the second derivative of $\psi(x)$ must return the product of $\psi(x)$ and the sum of a constant and a term of $x^2$, it can heuristically be deduced that a possible solution is:

\begin{equation}
\psi(x) = Ae^{Cx^2}
\end{equation}

where $A$ and $C$ are both constants yet to be found. To see whether or not this actually is a solution, substitute in $Ae^{Cx^2}$ for $\psi(x)$ in equation $(14)$.

\begin{equation}
0 = \frac{\hbar^2}{2m} \left(2ACe^{Cx^2} + 4AC^2x^2e^{Cx^2} \right)+ \left(E - \frac{1}{2}m\omega^2x^2\right)Ae^{Cx^2}
\end{equation}
\begin{equation}
0 = \frac{\hbar^2}{m} \left(C + 2C^2x^2 \right)+ \left(E - \frac{1}{2}m\omega^2x^2\right)
\end{equation}
\begin{align}
0 & = \frac{\hbar^2}{m}C + E \\
0 & = \frac{2\hbar^2}{m}C^2 x^2 - \frac{1}{2}m\omega^2x^2
\end{align}
\begin{align}
C & = -\frac{mE}{\hbar^2} \\
C & = -\frac{m\omega}{2\hbar}
\end{align}
\begin{equation}
E = \frac{\hbar\omega}{2}
\end{equation}

From this, two conclusions can be made. First, $\psi(x) = Ae^{Cx^2}$ is a solution to the time-independent Schrodinger equation when $E = \hbar\omega/2$ if $C = -mE/\hbar^2 = -m\omega/(2\hbar)$. Second, by equation $(12)$, this solution must be for the ground state of the harmonic oscillator and the zero-point energy is therefore:

\begin{equation}
E_0 = \frac{\hbar\omega}{2}
\end{equation}