Schrödinger Equation with Potential

Question

I have a question concerning the Schrödinger equation with a potential, \begin{equation} i \partial_t \Psi = - \Delta \Psi + V \Psi. \end{equation} Usually, when dealing with differential equations with a potential, we are given a potential function $V:\mathbb{R}^n \to \mathbb{R}$ and we consider the ODE \begin{equation} \partial_t \Phi = -\nabla V(\Phi). \end{equation} But I guess the potential in the Schrödinger equation is a different potential from what I know as a potential. Why is the $V$ in the Schrödinger equation called "potential"?

Best, Luke

Answer 1

I will only give an informal (heuristic) treatment.

In classical mechanics, consider the reformulation of Newton's equation in Hamiltonian mechanics \begin{align} \dot x =&\ p\\ \dot p =&\ -\nabla V(x) \end{align} then we see that \begin{align} H = \frac{p^2}{2}+V(x) = \text{energy}. \end{align}

If you are given an initial point $(x_0, p_0)$, called a state, in the phase space, then solving Hamilton's equations yields a trajectory that lies on the level curve of $H$. But, sometimes you are not given the initial state. Instead, you are given an approximation of the actual location (e.g. when you point to a location in space, you are not actually giving the exact location but a region in space). Hence your starting point is not a single state, but a probability distribution of the location of the initial state in phase space, $\rho_0$. Studying the evolution of $\rho_0$ with respect to Hamilton's equation yields Liouville's equation \begin{align} \partial_t \rho = \{H, \rho\} \end{align} where $\{\cdot, \cdot\}$ is the Poisson bracket.

Back to quantum mechanics. If you accept the idea that $\rho(t, x)=|\Psi(t, x)|^2 = \operatorname{Tr}(\hat \rho)$ is the probability density that the particle is at $x$ and $\hat\rho =$ is the density operator (integral operator with kernel $\Psi(t, x)\Psi^\ast(t, y)$, then the quantum analog of Liouville's equation is given by \begin{align} \partial_t \hat\rho = \frac{1}{i\hbar}[\hat H, \hat\rho] \end{align} where $[\cdot, \cdot]$ is the operator commutator. Note, I have use the correspondence principle \begin{align} \frac{1}{i\hbar}[\cdot, \cdot] \rightarrow \{\cdot, \cdot\}. \end{align}

Writing everything out explicitly in the kernel yields \begin{align} \partial_t (\Psi(t, x)\Psi^\ast(t, y)) = \frac{1}{i\hbar} \left((-\hbar^2\Delta_x+V(x))\Psi(t, x)\Psi^\ast(t, y)-(-\hbar^2\Delta_y +V(y))\Psi(t, x)\Psi^\ast(t, y) \right). \end{align} Massaging the above mess a little yields \begin{align} (i\hbar\partial_t\Psi(t, x))\Psi^\ast(t, y) +\Psi(t, x) (i\hbar\partial_t\Psi^\ast(t, y)) = (H\Psi(t, x))\Psi^\ast(t, y)-\Psi(t, x)(H\Psi^\ast(t, y)) \end{align} or, equivalently, \begin{align} (i\hbar\partial_t\Psi(t, x)-H\Psi(t, x))\Psi^\ast(t, y) -\Psi(t, x) (i\hbar\partial_t\Psi(t, y)-H\Psi(t, y))^\ast=0. \end{align}

In short, the potential in the Schrodinger equation is the "same" potential in the classical case, if you accept the correspondence principle.

Answer 2

Quantum theory and classical theory are two theories at different scales, but the potential represents the same physical object. In the quantum theory, particles are replaced by wave functions, which is different from the classical density of particles!

Newton laws tells you that the correct form is the Hamiltonian system $$ \begin{align*} \dot x &= v \\ \dot v &= -\nabla V(x) \end{align*} $$ The associate PDE is called the Liouville or Vlasov equation. Then, yes, if you zoom-out, you can (at least formally) get your second equation. This is sometimes called an hydrodynamic limit. In the same way, if you zoom out from Schrödinger equation, you will find the Newton equations (You should look at the words semiclassical limit and mean field limit).

So these are different models representing the same physical background at different scales.

Answer 3

Your idea that a conservative potential $V$, one that only depends on the positions of the interacting particles, not their velocities, is a mapping $\mathbb{R}^n \rightarrow \mathbb{R}$ is entirely correct. And indeed, the $V$ in Schrödinger's Equation is that exact type of $V$. In particular, if you have $N$ particles interacting in the 3D physical space, then $V:\mathbb{R}^{3N}\rightarrow\mathbb{R}$, since the potential has a scalar value for every possible particle configuration in 3D space.

I think the confusion you are having is when we consider orbits of particles. An orbit is a continuous path, a sequence of points in space. We will denote them $\gamma(t)$. An orbit is a slice, a subset of the space on which $V$ is defined. Technically the orbit is a slice of a larger space, the phase space, which also includes the velocities of the particle in the orbit, not only its position but I digress.

In classical mechanics we are all about finding orbits and paths in $\mathbb{R}^{3N}$. One such way of doing that is by using Newton's Second Law. So, I believe the equation you have shown us is Newton's second Law.

$$ \tag{1} \label{eq:newton-2nd-law} \frac{d^2 \vec x}{dt^2} = -\nabla V(\vec x) $$

where $\vec x \in \mathbb{R}^{3N}$. In your post you equated the first time derivative to the inverse gradient of the potential, when it should have been the second time derivative, the acceleration. That is, if indeed the equation is Newton's Second Law.

So when you have a conservative potential $V$, and some initial velocities for your particles, you plug $V$ in Equation \eqref{eq:newton-2nd-law} and what you get are the forces acting on your particles, which you can use to integrate Newton's Equations and find the path that your particles trace in space as they move in time, which corresponds to an orbit $\gamma(t)$ in your position space, a subset of the entire space. That is how classical mechanics use the potential $V$

Quantum mechanics is different. In quantum mechanics you are not trying to find the subset of the phase space that is accessible to your system, the subset that matters for the description of its mechanic behavior. In quantum mechanics your are describing the dynamics of $\Psi$, and $\Psi$ is a mapping from $\mathbb{R}^{3N}$, like $V$, only it maps $\mathbb{R}^{3N}$ to $\mathbb{C}$. That is what the Schrödinger Equation is, it describes the dynamics of $\Psi$, how it changes in time and what are its stationary states.

$$ \tag{2} \label{eq:schrodinger-equation} i \partial_t \Psi = - \frac{1}{2} \nabla^2 \Psi + V\Psi $$

notice the $\partial_t$ in the left-hand side of the equation.

That is why both equations are so different. Whereas in Equation \eqref{eq:newton-2nd-law} we are trying to find subsets of the phase space, in \eqref{eq:schrodinger-equation} we are trying to find an entire function $\Psi$ of the same space as $V$. So the potential $V$ is the same, each equation just uses it differently.