2.2. Relation with classical mechanics

In order to see whether we can guess the structure of the Hamiltonian for systems which have a classical analogue, we consider the time evolution of a physical quantity $Q$. We assume that $Q$ does not depend on time explicitly. However, the expectation value of $Q$ may vary in time due to the change of the wave function in the course of time. For normalised wave functions, $$\frac{d}{dt} \left( \left\langle \psi(t) \right| \hat{Q} \left |\psi(t) \right\rangle \right) = \left( \frac{\partial}{\partial t} \left\langle\psi(t) \right| \right) \hat{Q} \left| \psi(t) \right\rangle + \left\langle\psi(t) \right| \hat{Q} \left( \frac{\partial}{\partial t} \left|\psi(t) \right\rangle \right) .$$

Using the time-dependent Schrödinger equation and its Hermitian conjugate, $$-{\rm i}\hslash \frac{\partial}{\partial t} \left\langle\psi(t) \right| = \left\langle\psi(t) \right| \hat{H},$$ (note the minus sign on the left-hand side which results from the Hermitian conjugate) we obtain $${\rm i}\hslash \frac{d}{dt} \left( \left\langle \psi(t) \right| \hat{Q} \left| \psi(t) \right\rangle \right) = \left\langle \psi(t) \right| \hat{Q} \hat{H} - \hat{H} \hat{Q} \left| \psi(t) \right\rangle = \left\langle \psi(t) \right| \left[ \hat{Q}, \hat{H}\right] \left| \psi(t) \right\rangle ,$$ where $\left[ \hat{Q}, \hat{H}\right]$ is the commutator. We see that the time derivative of $\hat{Q}$ is related to the commutator between $\hat{Q}$ and $\hat{H}$. This should wake you up or ring a bell. Perhaps you have seen a derivation of the time derivative of any physical quantity $Q$ in classical mechanics, based on the Hamilton equations of classical mechanics. These equations read $$\dot{p}_j = -\frac{\partial H}{\partial q_j} ; \phantom{xxx} \dot{q}_j = \frac{\partial H}{\partial p_j} .$$ Note that $H$ is expressed as a function of the generalised coordinates $q_j$ and $p_j$.

As a side note, we provide a different form of this equation. Introducing the two-dimensional coordinates $z_j = (p_j, q_j)$, we may write the Hamilton equations in the form $$\frac{\partial z_j}{\partial t} = J \nabla_j H,$$ where $\nabla_j = (\partial/\partial p_j, \partial/ \partial q_j)$ and $$J = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right).$$

The time derivative of $Q(p_j,q_j)$ is given by $$\frac{dQ}{dt} = \sum_j \left( \frac{\partial Q}{\partial q_j} \frac{\partial H}{\partial p_j} - \frac{\partial Q}{\partial q_j} \frac{\partial H}{\partial p_j} \right) \equiv \left\{ Q, H\right\}.$$ We see that this equation is very similar to that obtained above for the time derivative of the expectation value of the operator $\hat{Q}$! The differences consist of replacing the Poisson bracket by the commutator and adding a factor $-{\rm i}\hslash$. Note that the commutator and the Poisson bracket both are anti-symmetric under exchange of the two quantities involved: $$[A, B] = - [B,A]; \phantom{xxx} \{A, B\} = - \{B,A\}.$$ It seems that classical and quantum mechanics are not that different after all. Could this perhaps be a guide to formulate quantum mechanics for systems for which we have already a classical version? This turns out to be the case. At this stage, we summarise the result by writing down the time-evolution for $\hat{Q}$ and its classical version: $${\rm i}\hslash \frac{d\left\langle\hat{Q} \right\rangle}{dt} = \left\langle\left[ \hat{Q}, \hat{H}\right] \right\rangle({\text{Quantum}});$$ $$\frac{dQ}{dt} = \left\{ Q, H\right\} \phantom{xx} ({\text{Classical}}),$$ which shows the striking similarity between the two formulations. These equations are both called Liouville equation -- the first is however usually denoted as quantum Liouville equation.

As an example, we start by considering a one-dimensional system for which the relevant classical observables are the position $x$ and the momentum $p$. Classically, we have $$\left\{ x, p \right\} = \frac{\partial x}{\partial x} \frac{\partial p}{ \partial p} -\frac{\partial x}{\partial p} \frac{\partial p}{ \partial x} =1.$$ The second term in the expression vanished because $x$ and $p$ are to be considered as independent coordinates. From this, and from the factor ${\rm i}\hslash$ occurring in the quantum evolution equation above, we may guess the quantum version of this Poisson relation: $$\left[ x, p \right] = {\rm i}\hslash$$ which should look familiar (if it does not, please review the second year quantum mechanics course). It seems that our recipe for making quantum mechanics out of classical mechanics makes sense! Therefore we can now state the following rule:

If the Hamiltonian of some classical system is known, we can use the same form in quantum mechanics, taking into account the fact that the coordinates $q_j$ and $p_j$ become Hermitian operators and that their commutator relations are: $$[q_j, q_k] = 0; \phantom{xx} [p_j, p_k ] = 0; \phantom{xx} [q_j, p_k] = {\rm i}\hslash \delta_{jk}.$$

You can verify these extended commutation relations easily by working out the corresponding classical Poisson brackets.

In the second year, you have learned that $$\hat{p} = \frac{\hslash}{{\rm i}} \frac{d}{dx}.$$ What about this relation? It was not mentioned here so far. The striking message is that this relation can be derived from the commutation relation. In order to show this, we must discuss another object you might have missed too in the foregoing discussion: the wave function written in the form $\psi({\bf r})$ (for a particle in 3D). It is important to study the relation between this and the state $\left| \psi \right\rangle$. Consider a vector $\bf a$ in two dimensions. This vector can be represented by two numbers $a_1$ and $a_2$, which are the components of the vector $\bf a$. However, the actual values of the components depend on how we have chosen our basis vectors. The vector $\bf a$ is an arrow in a two dimensional space. In that space, $\bf a$ has a particular length and a particular orientation. By changing the basis vectors, we do not change the object $\bf a$, but we do change the numbers $a_1$ and $a_2$.

In the case of the Hilbert space of a one-dimensional particle, we can use as basis vectors the states in which the particle is localised at a particular position $x$. We call these states $\left| x \right\rangle$. They are eigenvectors of the position operator $\hat{x}$ with eigenvalue $x$: $$\hat{x} \left| x\right\rangle= x \left| x\right\rangle.$$ It is convenient to normalize the states $\left| x \right\rangle$ as follows: $$\left\langle x | x' \right\rangle = \delta (x-x'),$$ where $\delta (x-x')$ is the Dirac delta function. We now can define $\psi(x)$: $$\psi(x) = \left\langle x | \psi \right\rangle,$$ that is, $\psi(x)$ are the 'components' of the 'vector' $\left| \psi \right\rangle$ with respect to the basis $\left| x \right\rangle$. For three dimensions, we have a wave function which is expressed with respect to the basis $\left| {\bf r}\right\rangle$.

In order to derive the representation of the momentum operator, $\hat{p} = \frac{\hslash}{{\rm i}}\frac{d}{dx}$, we first calculate the matrix element of the commutator: $$\left\langle x | [\hat{x}, \hat{p}] | x' \right\rangle = \left\langle x |\hat{x} \hat{p} - \hat{p} \hat{x} | x' \right\rangle = (x-x') \left\langle x |\hat{p}| x' \right\rangle.$$ The last expression is obtained by having $\hat{x}$ in the first term act on the bra-vector $\left\langle x \right|$ on its left, and on the ket $\left| x' \right\rangle$ on the right in the second term.

On the other hand, using the commutation relation, we know that $$\left\langle x | [\hat{x}, \hat{p}] | x' \right\rangle = {\rm i}\hslash \left\langle x | x' \right\rangle.$$ This is an even function of $x-x'$, as interchanging $x$ and $x'$ does not change the matrix element on the right-hand side. Since this function is equal to $(x-x') \left\langle x |\hat{p}| x'\right\rangle$, we know that $\left\langle x | \hat{p} | x' \right\rangle$ must be an odd function of $x-x'$.

Now we evaluate the matrix element $\left\langle x | \hat{p} | \psi \right\rangle$. We recall from linear algebra that, since $\left| x \right\rangle$ are the eigenstates of an Hermitian operator, they form a complete set, that is: $$\mathbb{1} = \int \left| x \right\rangle\left\langle x \right| \; dx,$$ where $\mathbb{1}$ is the unit operator. Then we can write $$\left\langle x | \hat{p} | \psi \right\rangle = \int \left\langle x | \hat{p} | x' \right\rangle \left\langle x' | \psi\right\rangle dx'.$$ Now we perform a Taylor expansion around $x$ in order to rewrite $\left\langle x' | \psi \right\rangle$: $$\left\langle x' | \psi \right\rangle = \left\langle x | \psi \right\rangle + (x'-x) \frac{d}{dx} \left\langle x | \psi \right\rangle + \frac{(x'-x)^2}{2!} \frac{d^2}{dx^2} \left\langle x | \psi \right\rangle + \cdots$$ We then obtain $$\left\langle x | \hat{p} | \psi \right\rangle = \int \left\langle x \right| p \left| x' \right\rangle \left[ \left\langle x | \psi \right\rangle + (x'-x) \frac{d}{dx} \left\langle x | \psi \right\rangle + \frac{(x'-x)^2}{2!} \frac{d^2}{dx^2} \left\langle x | \psi \right\rangle + \cdots \right] dx'.$$ The first term in the square brackets gives a zero after integration, as it is multiplied by $\left\langle x | \hat{p} | x' \right\rangle$, which was an odd function of $x-x'$. The second term gives $$\int \left\langle x | \hat{p} | x' \right\rangle (x'-x) \frac{d}{dx} \left\langle x | \psi \right\rangle dx' = -{\rm i}\hslash\frac{d}{dx} \left\langle x | \psi \right\rangle,$$ where we have used the relation $$\left\langle x | \hat{p} | x' \right\rangle (x'-x) = -{\rm i}\hslash \delta (x'-x).$$

We use the same relation for the second term. But then we obtain a term of the form $$(x'-x) \delta (x'-x)$$ in the integral over $dx'$. This obviously yields a zero. The same holds for all higher order terms, so we are left with $$\left\langle x | \hat{p} | \psi \right\rangle = \frac{\hslash}{i} \frac{d}{dx} \left\langle x | \psi \right\rangle,$$ which is the required result.

Having obtained this, we can analyse the form of the eigenstates of the momentum operator: $$\hat{p} \left| p \right\rangle= p \left| p \right\rangle.$$ The states $\left| p \right\rangle$ can be represented in the basis $\left\langle x \right|$; the components then are $\left\langle x | p \right\rangle$. We can find the form of these functions by using the eigenvalue equation and the representation of the momentum operator as a derivative: $$\begin{aligned} \left\langle x | \hat{p} | p \right\rangle&= p \left\langle x | p \right\rangle\phantom{xx} {\text{ and}} \\ \left\langle x | \hat{p} | p \right\rangle&= \frac{\hslash}{{\rm i}} \frac{d}{dx} \left\langle x | p \right\rangle. \end{aligned}$$ The first of these equation expresses the fact that $\left| p \right\rangle$ is an eigenstate of the operator $\hat{p}$, and the second one follows directly from the fact that the momentum operator acts as a derivative in the $x$-representation. Combining these two we obtain a simple differential equation $$\frac{\hslash}{{\rm i}} \frac{d}{dx} \left\langle x | p \right\rangle= p \left\langle x | p \right\rangle,$$ with a normalised solution: $$\left\langle x | p \right\rangle = \frac{1}{\sqrt{2\pi\hslash}} e^{{\rm i}px/\hslash}.$$

This allows us to find any state $\psi$ in the momentum representation, that is, the representation in which we use the states $\left| p \right\rangle$ as basis states: $$\psi(p) = \left\langle p | \psi \right\rangle = \int \left\langle p | x \right\rangle \left\langle x | \psi \right\rangle dx = \frac{1}{\sqrt{2\pi \hslash}} \int e^{-{\rm i}px/\hslash} \psi(x) dx.$$ This is nothing but the Fourier transform, the specific form of which we already encountered at the end of the previous chapter. The analysis presented here for a one-dimensional particle can be generalised to three or more dimensions in a natural way.