Chapter 5

The Hamiltonian and the Schrödinger Equation

The law of evolution we have derived contains a matrix H that we do not yet know. Here we determine it, by requiring agreement with Newton’s equation and using a single measurement.

Determination of the Hamiltonian matrix.

To determine the matrix H we will exploit the information we know from Classical Mechanics. We know that Newton's laws are very accurate over a vast range of phenomena, so we will choose the matrix H in such a way that the time-evolution law of Quantum Mechanics and Newton's law $f=ma$ give the same results over this range of phenomena. We postulate the principle of agreement with Newton: under the conditions in which Newton’s mechanics is confirmed by experience — bodies sufficiently large and heavy — the theory must give the same predictions.

For simplicity we consider a not very complicated case: we determine the Hamiltonian matrix for a charged particle in a one-dimensional space, in the presence of a time-independent electric field.

Newton's law can be written in the form

\begin{aligned} m\frac{d^2x}{dt^2} & =qE(x) \\ & =-q\frac{dV(x)}{dx} \end{aligned}

The law of Quantum Mechanics is certainly very different

i\frac{d}{dt}|\psi t\rangle=H(t)|\psi t\rangle

The difference between the two equations is due to the way we represent the evolution of the system. From the classical point of view we have a material particle and must determine the equation of motion $x(t)$ . From the quantum point of view, instead, we have a system with its observable quantity x, that is, the position, and must determine the distribution of probability amplitudes $\psi(x,t)$ as time varies.

To make a comparison we can derive, from the Schrödinger equation for the amplitudes $\psi(x,t)$ , an equation for the mean value $\langle x\rangle$ of the variable x. In this way we will have an equation derived from the quantum law that will be easily comparable with Newton's law.

The mean value of the variable x can be written in the following way

\begin{aligned} \langle x\rangle & =\int_{-\infty}^{+\infty}p(x)x\:dx \\ & =\int_{-\infty}^{+\infty}\psi^*(x)x\psi(x)\:dx \\ & =\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\psi^*(x)X(x,x')\psi(x)\:dx'\:dx \\ & =\langle\psi|X|\psi\rangle \end{aligned}

where we have introduced the operator X represented by the function $X(x,x')$

X(x,x')=x\delta(x-x')

The operator X thus defined is called the position operator.

Let us now compute the derivative $d\langle x\rangle/dt$

\begin{aligned} \frac{d}{dt}\langle x\rangle & =\frac{d}{dt}\langle\psi t|X|\psi t\rangle \\ & =\left(\frac{d}{dt}\langle\psi t|\right)X|\psi t\rangle+\langle\psi t|X\left(\frac{d}{dt}|\psi t\rangle\right) \end{aligned}

On the basis of the Schrödinger equation we can write

\begin{aligned} \frac{d}{dt}|\psi t\rangle & =-iH|\psi t\rangle, \\ \frac{d}{dt}\langle\psi t| & =i\langle\psi t|H^+ \\ & =i\langle\psi t|H \end{aligned}

Substituting, we have

\begin{aligned} \frac{d}{dt}\langle x\rangle & =i\langle\psi t|HX|\psi t\rangle-i\langle\psi t|XH|\psi t\rangle \\ & =\langle\psi t|i(HX-XH)|\psi t\rangle \end{aligned}

Defining the velocity operator $\dot{X}=i(HX-XH)$ we can write

\frac{d}{dt}\langle x\rangle=\langle\psi t|\dot{X}|\psi t\rangle

with identical steps we compute the second derivative

\begin{aligned} \frac{d^2}{dt^2}\langle x\rangle & =\frac{d}{dt}\frac{d}{dt}\langle x\rangle \\ & =\frac{d}{dt}\langle\psi t|\dot{X}|\psi t\rangle \\ & =\dots \\ & =\langle\psi t|i(H\dot{X}-\dot{X}H)|\psi t\rangle \end{aligned}

defining the acceleration operator $\ddot{X}=i(H\dot{X}-\dot{X}H)$ we can write

\frac{d^2}{dt^2}\langle x\rangle=\langle\psi t|\ddot{X}|\psi t\rangle

For convenience one defines the commutation brackets $[,]$ with the following meaning: $[A,B]=AB-BA$ .

Using these brackets we can write

\begin{aligned} \ddot{X} & =i[H,\dot{X}] \\ & =-[H,[H,X]] \end{aligned}

So in the end we have:

\begin{aligned} \frac{d^2}{dt^2}\langle x\rangle & =\langle\psi_t|\ddot{X}|\psi_t\rangle \\ & =-\langle\psi_t|[H,[H,X]]|\psi_t\rangle \end{aligned}

To have agreement with Newton's equation we expect that

$d^2\langle x\rangle/dt^2=q\langle E(x)\rangle/m$ which is equivalent to $d^2x/dt^2=qE(x)/m$

Note 06 Here we require agreement on the mean of the field, not on the field at the mean position: the caveat →

The mean value $\langle E(x)\rangle$ can be written in the form $\langle\psi t|E(X)|\psi t\rangle$ , where $E(X)$ is an operator built by applying the function $E(x)$ to the position operator X. Since X maps $\psi(x)$ to $x\psi(x)$ , each of its powers $X^n$ maps $\psi(x)$ to $x^n\psi(x)$ : applying a function to the position operator amounts to multiplying by that function.

Now we compare the two equations for $d^2\langle x\rangle/dt^2$ , the one obtained from the Schrödinger equation and the one suggested by Newton's equation:

\frac{d^2}{dt^2}\langle x\rangle=-\langle\psi_t|[H,[H,X]]|\psi_t\rangle

\frac{d^2}{dt^2}\langle x\rangle=\frac{q}{m}\langle\psi t|E(X)|\psi t\rangle

For agreement to hold, the equality between the right-hand sides must be valid

-\langle\psi t|[H,[H,X]]\psi t\rangle=\frac{q}{m}\langle\psi t|E(X)\psi t\rangle

This last one must be true for every choice of $|\psi t\rangle$ , so we can simplify it

-[H,[H,X]]=\frac{q}{m}E(X)

At this point we must find a matrix H that satisfies this equation. Let us narrow the search to matrices of the form $H=f(X)+g(K)$ , with f and g functions to be determined. Narrowing the search is not an assumption about the result, it is a choice of where to look: if this family contains a solution, it is a solution in its own right, because in the end we will test it against the original equation. Here X is the position operator, which maps $\psi(x)$ to the function $x\psi(x)$ , while $K=iD$ is the derivative operator, which maps $\psi(x)$ to the function $i\,d\psi(x)/dx$ .

To find the solution we need some very useful mathematical results:

1a $[K,f(X)]=i\,df(X)/dX$

2a $[X,f(K)]=-i\,df(K)/dK$

3a $[X,f(X)]=0$

4a $[K,f(K)]=0$

These formulas are proved by induction; here we take them as established and go on to solve the equation for the unknown H.

Note 10 The proofs by induction of the four formulas: the calculation →

Let us now look for the solution of the equation

-[H,[H,X]]=\frac{q}{m}E(X)

As we said, we look for H in the form $H=f(X)+g(K)$ ; substituting we have

\begin{aligned} -\left[f(X)+g(K),\left[f(X)+g(K),X\right]\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ & \quad\text{By formula 3a} \\ -\left[f(X)+g(K),\left[g(K),X\right]\right] & =\frac{q}{m}E(X) \end{aligned}

To move forward with the calculation we narrow the field further: we look for g among the functions for which $[g(K),X]=icK$ , with c constant. With this choice we have

\begin{aligned} -\left[f(X)+g(K),icK\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ -\left[f(X),icK\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ \left[icK,f(X)\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ & \quad\text{By formula 1a} \\ -c\frac{df(X)}{dX} & =\frac{q}{m}E(X) \end{aligned}

Recall that the operator $E(X)$ can be written in derivative form $E(X)=-dV(X)/dX$ , where $V(x)$ is the potential function. Substituting we have

-c\frac{df(X)}{dX}=-\frac{q}{m}\frac{dV(X)}{dX}

This equation is solved if one chooses $f(X)=qV(X)/cm$ .

Let us recall the condition we imposed on g, $[g(K),X]=icK$ : this equation is equivalent to $i\,dg(K)/dK=icK$ , which is satisfied if one chooses $g(K)=cK^2/2$ .

So for the matrix H we have

\begin{aligned} H & =f(X)+g(K) \\ & =\frac{q}{cm}V(X)+\frac{1}{2}cK^2 \end{aligned}

The reasoning has led us to a candidate; now it must be put to the test. Let us verify that this matrix satisfies the original equation

\begin{aligned} -\left[\frac{q}{cm}V(X)+\frac{1}{2}cK^2,\left[\frac{q}{cm}V(X)+\frac{1}{2}cK^2,X\right]\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ -\left[\frac{q}{cm}V(X)+\frac{1}{2}cK^2,\left[\frac{1}{2}cK^2,X\right]\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ -\left[\frac{q}{cm}V(X)+\frac{1}{2}cK^2,icK\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ -\left[\frac{q}{cm}V(X),icK\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ ic\frac{q}{cm}\left[K,V(X)\right] & =\frac{q}{m}E(X)\Leftrightarrow \\ -\frac{q}{m}\frac{dV(X)}{dX} & =\frac{q}{m}E(X)\quad\text{Verified.} \end{aligned}

At this point we can say we have determined the Schrödinger equation for a charged particle in an electric potential $V(x)$

\begin{aligned} i\frac{d}{dt}|\psi t\rangle & =H(t)|\psi t\rangle \\ & =\left(\frac{q}{cm}V(X)+\frac{1}{2}cK^2\right)|\psi t\rangle\Leftrightarrow \\ i\frac{d}{dt}\psi(x,t) & =\frac{q}{cm}V(x)\psi(x,t)-\frac{1}{2}c\frac{d^2}{dx^2}\psi(x,t) \end{aligned}

This equation was derived so as to be in agreement with Newton's equation, in the cases where the latter is applicable.

In reality we still have to determine the constant c. To determine this constant we must refer to an experience that lies outside the domain of Classical Mechanics, because we have already exploited all the information we could draw from that theory. Indeed, Newton's equation contains all the information of the classical theory.

To determine c we will use the De Broglie relation $p\lambda=h$ , which we verified with the electron-diffraction experiment.

Consider the Schrödinger equation written for the case in which the electric potential is zero $V(x)=0$

i\frac{d}{dt}\psi(x,t)=-\frac{1}{2}c\frac{d^2}{dx^2}\psi(x,t)

This equation admits solutions of complex-exponential type

\begin{aligned} \psi(x,t) & =e^{i(kx-\omega t)} \\ & =e^{i\left(\frac{2\pi}{\lambda}x-\omega t\right)} \end{aligned}

Substituting, we have

i\frac{d}{dt}e^{i\left(\frac{2\pi}{\lambda}x-\omega t\right)}=-\frac{1}{2}c\frac{d^2}{dx^2}e^{i\left(\frac{2\pi}{\lambda}x-\omega t\right)}\Leftrightarrow

\Leftrightarrow\omega e^{i\left(\frac{2\pi}{\lambda}x-\omega t\right)}=\frac{1}{2}c\frac{4\pi^2}{\lambda^2}e^{i\left(\frac{2\pi}{\lambda}x-\omega t\right)}

which is satisfied if $\omega=2\pi^2c/\lambda^2$ . So we have the solutions

\psi(x,t)=e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}

These distributions of probability amplitudes are not acceptable from a physical point of view, because they give a probability of finding the particle that is constant over the whole space, from $-\infty$ to $+\infty$

\begin{aligned} p(x,t) & \propto|\psi(x,t)|^2 \\ & =\psi^*(x,t)\psi(x,t) \\ & =e^{-i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)} \\ & =e^0 \\ & =1\ \text{Constant}. \end{aligned}

However, the Schrödinger equation is linear, and so one can build other solutions by superposing the exponential ones. In this way wave packets are realised that have a limited spatial extension and are physically acceptable

\psi(x,t)=\int_{-\infty}^{+\infty}c(\lambda')e^{i\left(\frac{2\pi}{\lambda'}x-\frac{2\pi^2c}{\lambda^{'2}}t\right)}\:d\lambda'

If, for example, we choose $c(\lambda')=\delta(\lambda'-\lambda)$ we recover the exponential function

\begin{aligned} \psi(x,t) & =\int_{-\infty}^{+\infty}\delta(\lambda'-\lambda)e^{i\left(\frac{2\pi}{\lambda'}x-\frac{2\pi^2c}{\lambda^{'2}}t\right)}d\lambda' \\ & =e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)} \end{aligned}

This means that the exponential distribution represents a wave packet that has infinite spatial extension but a well-defined wavelength. A finite wave packet, on the other hand, cannot have such a precise wavelength because it is made of the superposition of several exponential functions, each with a different wavelength. So the exponential solution, even if it is not physically acceptable, is very convenient for representing the limiting cases of monochromatic beams, that is, with a very precise wavelength.

In general we know that every distribution of probability amplitudes must be normalised, that is, divided by its own modulus. This is not possible for an exponential solution because its modulus is infinite

\begin{aligned} {\left|e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}\right|}^2 & =\int_{-\infty}^{+\infty}e^{-i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}\:dx \\ & =\int_{-\infty}^{+\infty}e^0\:dx \\ & =\int_{-\infty}^{+\infty}1dx \\ & =\infty \end{aligned}

However, in order to save the exponential solutions, we will use an artifice: whenever in the calculations we introduce an exponential distribution, we will write at the denominator the integral that expresses its modulus, without ever computing it, waiting for an equal integral to appear at the numerator with which it can be simplified. We will apply this artifice now, as we compute the momentum of a particle in a state associated with an exponential distribution.

Suppose we know the state of a particle and the corresponding normalised distribution of probability amplitudes $\psi(x,t)$ , and suppose we want to compute the momentum associated with this particle. In reality we have not yet defined the concept of momentum in Quantum Mechanics, but it is not hard to guess that what we want to compute is the quantity $\langle p\rangle=m\,d\langle x\rangle/dt$ , where $\langle x\rangle$ is the mean value of the random variable x. A few pages ago we wrote

\frac{d}{dx}\langle x\rangle=\langle\psi|\dot{X}|\psi\rangle

where $\dot{X}=i\left[H,X\right]$ is an operator we called the velocity operator. Let us see what form $\dot{X}$ takes with the H we have determined

\begin{aligned} \dot{X} & =i[H,X] \\ & =i\left[\frac{q}{cm}V(X)+\frac{1}{2}cK^2,X\right] \\ & =i\left[\frac{1}{2}cK^2,X\right] \\ & =-i\left[X,\frac{1}{2}cK^2\right] \\ & =-cK \end{aligned}

So we can write the formula for $\langle p\rangle$

\begin{aligned} \langle p\rangle & =m\frac{d}{dx}\langle x\rangle \\ & =m\langle\psi|(-cK)|\psi\rangle \\ & =\langle\psi|(-cmK)|\psi\rangle \end{aligned}

Defining the momentum operator $P=-cmK$ we have $\langle p\rangle=\langle\psi|P|\psi\rangle$ .

Let us now compute the momentum for a probability amplitude with exponential distribution

\begin{aligned} \psi(x,t) & =\frac{e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}}{\sqrt{\int_{-\infty}^{+\infty}e^{-i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}dx}} \\ & =\frac{e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}}{M} \end{aligned}

where by M we have denoted the modulus of the non-normalised exponential distribution.

\begin{aligned} \langle p\rangle & =\langle\psi|P|\psi\rangle \\ & =\langle\psi|(P|\psi\rangle) \\ & =\int_{-\infty}^{+\infty}\frac{e^{-i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}}{M}\left(-icm\frac{d}{dx}\frac{e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}}{M}\right)dx \\ & =\int_{-\infty}^{+\infty}\frac{e^{-i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}}{M}\left(-icmi\frac{2\pi}{\lambda}\frac{e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}}{M}\right)dx= \end{aligned}

\begin{aligned} & =cm\frac{2\pi}{\lambda}\frac{\int_{-\infty}^{+\infty}e^{-i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}e^{i\left(\frac{2\pi}{\lambda}x-\frac{2\pi^2c}{\lambda^2}t\right)}dx}{M^2} \\ & =cm\frac{2\pi}{\lambda}\frac{M^2}{M^2} \\ & =cm\frac{2\pi}{\lambda} \end{aligned}

So we have found $\langle p\rangle=2\pi cm/\lambda\Leftrightarrow\langle p\rangle\lambda=2\pi cm$ .

Recalling the De Broglie relation $p\lambda=h$ , we can conclude that it must be $2\pi cm=h\Leftrightarrow c=h/2\pi m$ . For convenience one defines the reduced Planck constant $\hbar=h/2\pi$ , with which we write $c=\hbar/m$ . Substituting this value we can finally give the Schrödinger equation in its definitive form

\begin{aligned} i\frac{d}{dt}\psi(x,t) & =\frac{q}{\hbar}V(x)\psi(x,t)-\frac{1}{2}\frac{\hbar}{m}\frac{d^2}{dx^2}\psi(x,t)\Leftrightarrow \\ i\hbar\frac{d}{dt}\psi(x,t) & =qV(x)\psi(x,t)-\frac{1}{2m}\hbar^2\frac{d^2}{dx^2}\psi(x,t)\Leftrightarrow \\ i\hbar\frac{d}{dt}\psi(x,t) & =\left(qV(X)+\frac{1}{2m}P^2\right)\psi(x,t)\Leftrightarrow \\ i\hbar\frac{d}{dt}|\psi\rangle & =\left(qV(X)+\frac{1}{2m}P^2\right)|\psi\rangle \end{aligned}

where we have used the definition $P=-cmK=-\hbar K=-i\hbar D$ .

In textbooks of Quantum Mechanics one generally finds the Schrödinger equation written with the $H$ on the right-hand side

i\hbar\frac{d}{dt}|\psi\rangle=H|\psi\rangle

with

H=qV(X)+\frac{1}{2m}P^2

From now on we too will use this form.

Conclusions.

In this card we have found an equation that represents the law of evolution for a charged particle according to the formalism of Quantum Mechanics. The equation we have found does not take into account the Theory of Relativity and is, moreover, greatly simplified: it does not consider magnetic or gravitational fields and is written for a one-dimensional space. Nevertheless it is a good example that achieves the aim of this card, that is, to show the fundamental features of Quantum Mechanics applied to a system of practical interest. In the following cards we will generalise this equation and apply it to study concrete problems.