5 Schrödinger’s Equation

5.1 Time Evolution in Classical Mechanics

In the Hamiltonian formulation of classical mechanics, a particle is described by a definite position and momentum \((x(t),p(t))\), which evolve in time according to Hamilton’s equations \[\dot x = + \frac{\partial H}{\partial p} \qquad \dot p = - \frac{\partial H}{\partial x} \, ,\] where \(H\) is the Hamiltonian of the system.

It is useful to reformulate time-evolution as a canonical transformation. First, we note that Hamilton’s equations can be expressed as \[\dot x = \{ x,H\} \qquad \dot p = \{ p,H\} \, ,\] where \[\{ A , B \} := \frac{\partial A}{\partial x}\frac{\partial B}{\partial p}-\frac{\partial A}{\partial p}\frac{\partial B}{\partial x}\] is the Poisson bracket. This shows that an infinitesimal time evolution \(t \mapsto t +\epsilon\) of position and momentum can be expressed \[\begin{aligned} x & \mapsto x+\epsilon \dot x = x + \epsilon\{ x,H\} \\ p & \mapsto p+\epsilon \dot p = p + \epsilon\{ p,H\} \, , \end{aligned}\] which is an infinitesimal canonical transformation generated by the Hamiltonian \(H\). In other words, the Hamiltonian is the generator of time translations.

This is similar to the statement that momentum is the generator of translations in space from lecture \(\ref{lec3}\). We will follow the same logic here to understand time-evolution in quantum mechanics.

5.2 Schrödinger’s Equation

We now use the idea of the Hamiltonian as the generator of time translations to understand time-evolution in quantum mechanics.

Let us consider a wave function \(\psi(x,t)\). The small change in the wave function due to an infinitesimal time translation \(t \mapsto t+\epsilon\) is \[\begin{aligned} \delta_\epsilon \psi(x,t) & = \psi(x,t+\epsilon) - \psi(x,t) \\ & = \epsilon \frac{\partial \psi(x,t)}{\partial t} + {\cal O}(\epsilon^2) \, . \end{aligned}\] We want this transformation to be “generated" by the Hamiltonian operator \(\hat H\). This means the change in the wave function is proportional to the action of the Hamiltonian \(H\) on the wave function, \[\begin{aligned} \delta_\epsilon \psi(x,t) & = \frac{\epsilon}{\alpha} \hat H \, \psi(x,t) \, . \end{aligned}\] where \(\alpha \in \mathbb{C}\) is an unknown constant of proportionality.

Equating the two expressions for the change in the wave function under the infinitesimal time translation \(t \mapsto t + \epsilon\), we find \[\alpha \frac{\partial \psi(x,t)}{\partial t} = \hat H \, \psi(x,t) \, .\] How can we determine the constant of proportionality \(\alpha \in \mathbb{C}\)?

1. Dimensional Analysis. The hamiltonian has units of energy \(ML^2T^{-2}\), while the derivative \(\partial_t\) has units \(T^{-1}\). The constant of proportionality \(\alpha\) therefore has units of angular momentum \(ML^2T^{-1}\). Note that these are the same units as \(\hbar\).

2. Conservation of Total Probability. The probability to find the particle anywhere in space should be \(1\) at any time. This requires that the normalisation \(\langle \psi,\psi\rangle\) is constant in time for any square-normalisable wave function, \[\begin{aligned} 0 & = \partial_t \langle \psi,\psi\rangle \\ & = \langle \partial_t \psi , \psi\rangle + \langle \psi, \partial_t\psi \rangle \\ & = \langle \frac{1}{\alpha} \hat H \, \psi, \psi\rangle + \langle \psi, \frac{1}{\alpha} \hat H\, \psi \rangle \\ & = \left( \frac{1}{\bar \alpha} + \frac{1}{\alpha} \right) \langle \psi,\hat H \, \psi\rangle \\ & = \frac{\alpha+\bar\alpha}{|\alpha|^2} \langle H \rangle \, . \end{aligned}\] In passing to the fourth line, we have used the fact that we know the form of \(\hat H\) in terms of derivatives with respect to \(x\). We can integrate by parts with respect to those, and hence show that you can ‘move’ the operator \(\hat H\) from the \(\bar\psi\) factor to the \(\psi\) factor. We thus conclude that \(\alpha = -\bar\alpha\) and therefore \(\alpha\) is imaginary.

These observations suggest the constant of proportionality is \(\alpha = i \hbar\) where \(\hbar\) is Planck’s constant - this is indeed the value chosen by nature!

In summary we have \[i\hbar \frac{\partial \psi(x,t)}{\partial t} = \hat H \, \psi(x,t)\] or written out in full \[i \hbar \frac{\partial \psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2\psi(x,t)}{\partial x^2} + V(x) \psi(x,t) \, .\] This is “Schrödinger’s equation". It is a linear partial differential equation for the wave function \(\psi(x,t)\). This is probably the most profound equation you will come across in your degree: it is the most fundamental description of nature we have at short distances.

Note that there is also something called the “time-independent Schrödinger equation”, which is essentially what you get when \(\partial\psi(x,t)/\partial t\) evaluates to a constant times \(\psi(x,t)\). We will return to that special case in the chapter on “stationary states”.

5.3 Properties

The following properties of Schrödinger’s equation are very important in the development of quantum mechanics.

1. 1st Order in Time. Schrödinger’s equation has only a first order time derivative. This means that if we know the initial wave function \(\psi(x,0)\), Schrödinger’s equation uniquely determines the wave function \(\psi(x,t)\) for \(t\geq 0\).

2. Linearity. Schrödinger’s equation is linear PDE for the wave function \(\psi(x,t)\). This means that given two solutions \(\psi_1(x,t)\) and \(\psi_2(x,t)\), any linear combination \[a_1 \psi_1(x,t) + a_2 \psi_2(x,t)\] with constants \(a_1,a_2\in \mathbb{C}\) is another solution. This is known as the “principle of superposition".

The combination of these two properties, together with the decomposition of a wave function \(\psi(x,t)\) into orthonormal eigenstates of the Hamiltonian \(\hat H\) provides a powerful and systematic method of solving Schrödinger’s equation. The next lecture is dedicated to explaining this method. For now we look at some simple examples.

5.4 Example: Plane Waves

Consider a wave function of the form \[\psi_p(x,t) = \frac{1}{\sqrt{2\pi \hbar}} e^{ipx/\hbar}e^{-iE(p) t / \hbar}\, ,\] where \(E(p)\) is some function of momentum \(p \in \mathbb{R}\). Then, \[\begin{aligned} i\hbar \frac{\partial \psi(x,t)}{\partial t} = E(p)\psi(x,t) \\ -\frac{\hbar^2}{2m} \frac{\partial^2\psi(x,t)}{\partial x^2} = \frac{p^2}{2m} \psi(x,t) \, . \end{aligned}\] If we choose \(E(p) = p^2/2m\) then \[i\hbar \frac{\partial \psi(x,t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2\psi(x,t)}{\partial x^2} \, .\] We have therefore found a solution of Schrödinger’s equation with \(V(x)=0\). This corresponds to plane wave solution for a free particle of mass \(m\).

5.5 Example: Ground State in Infinite Potential Well

Now consider the wave function \[\begin{aligned} \psi(x,t) & = \phi_1(x) e^{-iE_1t/\hbar} \\ & = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L} \right)\exp\left( -i \frac{\hbar\pi^2}{2mL^2} t \right) \end{aligned}\] in an infinite potential well \(0 < x < L\).

We find \[\begin{aligned} i\hbar \frac{\partial \psi(x,t)}{\partial t} & = \frac{\hbar^2\pi^2}{2mL^2} \psi(x,t) \\ -\frac{\hbar^2}{2m} \frac{\partial^2\psi(x,t)}{\partial x^2} & = \frac{\hbar^2}{2m}\left(\frac{\pi}{L}\right)^2 \psi(x,t) \end{aligned}\] and therefore \[i\hbar \frac{\partial \psi(x,t)}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2\psi(x,t)}{\partial x^2} \, .\] so \(\psi(x,t)\) is a solution of Schrödinger’s equation in the infinite potential well. Note that \(\psi(x,0) = \phi_1(x)\) is the ground state Hamiltonian eigenfunction with lowest energy \[E_1 = \frac{\hbar^2}{2m}\left(\frac{\pi}{L}\right)^2 = \frac{\hbar\pi^2}{2mL^2} \, .\] The wave function \(\psi(x,t)\) is therefore the unique time evolution of \(\phi_1(x)\).

These examples have something in common: they are both of the form \[\label{e:first_stationary_state} \psi(x,t) = \phi(x) e^{-iEt/\hbar}\,,\] where \(\phi(x)\) is an eigenfunction of the Hamiltonian \(H\) with eigenvalue \(E\). We will see later (chapter \(\ref{sec:energy}\)) why this is always a solution of Schrödinger’s equation and that any solution can be expressed as a linear combination of such solutions.

5.6 Wave function continuity and differentiability

When we first introduced the wave function, and discussed its form in an infinite potential well, we mentioned in passing that the wave fuction is always continuous, and differentiable except at points where \(V(x)\) is not finite. We are now in a position to back up this claim.

First look at differentiability. You can prove this formally by writing down the Schrödinger equation for a state of the form \(\eqref{e:first_stationary_state}\). Re-arranging a bit, this gives \[\label{e:first_time_indep_schroedinger} -\frac{\hbar^2}{2m} \frac{{\rm d}^2}{{\rm d}x^2} \phi(x) = (E-V(x))\phi(x)\,.\] Now integrate both sides over a small interval interval around a point of interest. If \(E-V(x)\) remains finite, the right-hand side goes to zero in the limit of infinitesimally small interval, so the left-hand side should too. That means that the 2nd derivative of \(\phi(x)\) has to remain finite, which can only be true if the 1st derivative is continuous.

You have seen a counterexample of this when we discussed the infinite potential well, where \(V(x)\) is infinite outside the range \(0<x<L\). Another counterexample which we will see later is the delta-function potential.

For the continuity of \(\phi(x)\) itself, consider what happens if you have the simplest type of discontinuity, namely a step, \[\phi(x) = \begin{cases}a & \text{for $x<0$,}\\[1ex] b & \text{for $x>0$.}\end{cases}\] Then \(\phi'(x) = (b-a) \delta(x)\), which you may still be happy with, but the second derivative is then undefined, and you will struggle to make sense of \(\eqref{e:first_time_indep_schroedinger}\) above. For this reason, we take the wave function to be continuous.

5.7 Problems

1. Conservation of the inner product:

   Consider two square normalizable solutions \(\psi_1(x,t)\) and \(\psi_2(x,t)\) of Schrödinger’s equation.

   1. Show that the inner product of these wavefunctions is conserved, \[\partial_t\langle \psi_1,\psi_2\rangle = 0 \, .\]

   2. Hence explain why

      1. a normalized wavefunction remains normalized,

      2. a pair of orthogonal wavefunctions remain orthogonal.

   Solution ▶

   1. Schrödinger’s equation is \[i\hbar \partial_t \psi = \hat H \cdot \psi\] and therefore \[\partial_t \psi = - \frac{i}{\hbar} \hat H \cdot \psi \, .\] Computing the time derivative of the inner product, \[\begin{aligned} \partial_t\langle \psi_1,\psi_2\rangle & = \langle \partial_t\psi_1,\psi_2\rangle + \langle \psi_1,\partial_t \psi_2\rangle \\ \nonumber & = \langle -\frac{i}{\hbar} \hat H \cdot \psi_1,\psi_2\rangle + \langle \psi_1,-\frac{i}{\hbar} \hat H \cdot \psi_2\rangle \\ & = \frac{i}{\hbar}\left( \langle \hat H \cdot \psi_1,\psi_2\rangle - \langle \psi_1, \hat H \cdot \psi_2\rangle\right) = 0 \label{e:firstHhermitian} \end{aligned}\] To see that this last line is true, write out what the pointy bracket notation means: \[\begin{aligned} \langle \hat{H}\cdot \psi_1 ,\psi_2\rangle &= \int_{-\infty}^{\infty} \left(-\frac{\hbar}{2m}\frac{\partial^2}{\partial x^2} \psi_1(x) + V(x)\psi_1(x)\right)^* \psi_2(x) \\[1ex] &= \text{bdy. terms} + \int_{-\infty}^{\infty} \psi^*_1(x) \left(-\frac{\hbar}{2m}\frac{\partial^2}{\partial x^2} \psi_2(x) + V(x)\right)\psi_2(x) \\[1ex] &= \langle \psi_1, \hat{H}\cdot\psi_2\rangle\,. \end{aligned}\] In going to the pre-last line, we have integrated by parts twice to move the derivative to the \(\psi_2(x)\) factor, and we have used that \(V(x)\) is real.

      The cancellation of these two terms in \(\ref{e:firstHhermitian}\) is a manifestation of the fact that \(\hat H\) is a so-called Hermitian operator. We will discuss this in more detail in chapter 7 & 8.

   2. We conclude that \(\langle \psi_1,\psi_2\rangle\) is constant in time for any square-normalizable solutions \(\psi_1\) and \(\psi_1\) of Schrödinger’s equation. The results follow immediately.

2. TUTORIAL 1
   Heat equation:

   Consider the Schrödinger equation with, for simplicity, \(\hbar=1\) and \(m=1\), and vanishing potential, \[i\frac{\partial \psi}{\partial t} = -\frac{1}{2}\frac{\partial \psi}{\partial x^2}\,.\]

   1. Make the change of variable \(t = -i\tilde{t}\) and show that this leads to the heat equation with diffusivity \(1/2\) (look up these terms if you do not know them).

   2. The heat equation has the well-known solution \[\psi(x,\tilde{t}) = \frac{1}{\sqrt{2\pi \tilde{t}}} \exp\Big( -\frac{x^2}{2\tilde{t}}\Big)\,,\] What does this solution describe? (make some plots at various fixed values of \(\tilde{t}\)).

   3. Do the inverse transformation \(\tilde{t}=it\) on this solution, to obtain your first time-dependent solution to the Schrödinger equation. Make plots or sketches at a fixed value of \(t\) for the real and imaginary part, as well as the norm \(|\psi(x,t|^2\).

   Solution ▶

   1. Under this transformation, and using \[\frac{\partial}{\partial t} = \frac{\partial \tilde{t}}{\partial t} \frac{\partial}{\partial \tilde{t}}\,,\] we get \[\frac{\partial\psi(x,\tilde{t})}{\partial \tilde{t}} = \frac{1}{2}\frac{\partial^2 \psi(x,\tilde{t})}{\partial x^2}\,,\] the heat equation with coefficient \(1/2\).

   2. The solution looks like a Gaussian blob which spreads out in time,

      

   3. The inverse transform yields \[\psi(x,t) = \frac{1}{\sqrt{2\pi i t}}\exp\left( \frac{i x^2}{2t} \right)\,,\] which for \(t=0.1\) produces the plot below.

      

      We will discuss the use of this solution later when we discuss the free particle in quantum mechanics.