1 Introduction

In practice, this isn’t such a big deal. Suppose that we have a wavefunction $\mathrm{\Psi }(𝐱,t)$ that isn’t normalised but instead obeys

If $N$ is finite we say that the wavefunction is normalisable. Clearly a wavefunction is normalisable only if $\psi (𝐱,t)\to 0$ sufficiently fast as $|𝐱|\to \mathrm{\infty }$. In this case, we can always construct a normalised wavefunction

$\psi (𝐱,t)={\displaystyle \frac{1}{\sqrt{N}}}\mathrm{\Psi }(𝐱,t)$

Quite often in these lectures, it will turn out the be useful to work with un-normalised wavefunctions $\mathrm{\Psi }$ and then remember to include the normalisation factor only at the end when computing probabilities.

From the discussion above, it should be clear that we will have little use for wavefunctions that cannot be normalised because

${\displaystyle \int d^{3}x{|\psi (𝐱,t)|}^2}=\mathrm{\infty }$

These have no probabilistic interpretation and should be discarded. They do not describe quantum states. (I should warn you that this statement is mostly true that there is a small and annoying caveat that we’ll address in Section 2.1.)

There is one other relation between wavefunctions that is important: two wavefunctions that differ by a constant, complex phase should actually be viewed as describing equivalent states

$\psi (𝐱,t)\equiv e^{i\alpha }\psi (𝐱,t)$

for any constant, real $\alpha$. Note, in particular, that this doesn’t change the probability distribution $P={|\psi |}^2$. Nor, it will turn out, does it change anything else either. (The “anything else” argument is important here. As we’ll see later, if you multiplied the wavefunction by a spatially varying phase $e^{i\alpha (𝐱)}$ then it doesn’t change the probability distribution $P={|\psi |}^2$ but it does change other observable quantities and so multiplying by such a factor does not give back the same state.)

Combining the need for normalisation, together with the phase ambiguity, is sometimes useful think of states as the collection of normalisable, complex functions with the equivalence relation

$\psi (𝐱,t)\equiv \lambda \psi (𝐱,t)$

for any complex $\lambda \in 𝐂$ with $\lambda \ne 0$. The wavefunctions $\psi$ and $\lambda \psi$ should be viewed as describing the same physical state.

The mathematics underlying this statement is simple. If ${\psi }_1$ and ${\psi }_2$ are states of the system then both must be normalisable. They may, indeed, be normalised but for now let’s just assume that

Then ${\psi }_3$ is also a possible state of the system provided that it too is normalisable. But it is straightforward to show that it has this property. First we have

${\displaystyle \int d^{3}x{\left|{\psi }_3\right|}^2}={\displaystyle \int d^{3}x{\left|\alpha {\psi }_1+\beta {\psi }_2\right|}^2}\le {\displaystyle \int d^{3}x{\left(|\alpha {\psi }_1|+|\beta {\psi }_2|\right)}^2}$

where we’ve used the triangle inequality $|z_1+z_2|\le |z_1|+|z_2|$ for any $z_1,z_2\in 𝐂$. Continuing, we have

${\displaystyle \int d^{3}x{\left|{\psi }_3\right|}^2}\le {\displaystyle \int d^{3}x\left({|\alpha {\psi }_1|}^2+{|\beta {\psi }_2|}^2+2|\alpha {\psi }_1||\beta {\psi }_2|\right)}\le {\displaystyle \int d^{3}x\left(2{|\alpha {\psi }_1|}^2+2{|\beta {\psi }_2|}^2\right)}$

where, now, in the last step we’ve used the fact that ${(|z_1|-|z_2|)}^2\ge 0$ which, rearranging, gives $2|z_1||z_2|\le {|z_1|}^2+{|z_2|}^2$. So, finally, we get the result we wanted

We learn that ${\psi }_3$ is normalisable, and hence also an allowed state of the system.

The idea that functions form a vector space might be novel. There is a simple notational trick that should help convince you that it’s not too far from things you’ve seen already. If we have some $n$-dimensional vector $\overrightarrow{y}$, then we often use subscript notation and write it as $y_i$ with $i=1,\mathrm{\dots },N$. We could equally well write it as $y(i)$ with $i=1,\mathrm{\dots },N$. A function $y(x)$ is a similar kind of object, but now with a continuous label $x\in 𝐑$ rather than the discrete label $i=1,\mathrm{\dots },N$.

Of course, the big difference is that we’re now dealing with an infinite dimensional vector space rather than a finite dimensional vector space, and I would be lying if I told you that this doesn’t bring new issues to the table. Indeed, we’ve already met one of them above: we don’t consider any old function $\psi (𝐱)$ to be in our vector space, but rather only normalisable functions. There are many further subtleties in store but we will be best served by simply ignoring them at this point. We’ll return to some issues in Section 3.

The principle of superposition has profound physical consequences. Suppose that you have a particle that, at some time $t_0$, you know is localised somewhere near the point $𝐗$. For example, we could describe it by the Gaussian wavefunction

$\psi (𝐱)={\displaystyle \frac{1}{\sqrt{N}}}{e}^{-a{(𝐱-𝐗)}^2}$

for some choice of $a$ that tells you how spread out the wavefunction is. Here $N$ is a normalisation factor that won’t concern us. (For what it’s worth, we should take $N={(\pi /2a)}^{3/2}$ if we want to ensure that $\psi$ is normalised.) This is a state in which you might imagine that the particle still retains something of its classical particle nature, at least in the sense that the probability distribution is not spread out over long distances. However, the superposition principle tells us that we should also entertain the idea that the particle can sit in a state

$\psi (𝐱)={\displaystyle \frac{1}{\sqrt{N^{\prime }}}}\left(e^{-a{(𝐱-{𝐗}_1)}^2}+e^{-a{(𝐱-{𝐗}_2)}^2}\right)$

for arbitrary positions ${𝐗}_1$ and ${𝐗}_2$. But now the cat is really out of the bag! The interpretation of this state is that the particle has somehow split and now sits both near ${𝐗}_1$ and near ${𝐗}_2$. Indeed, we’ll shortly see clear experimental consequences of states like the one above, where elementary particles – which are, as far as we can tell, indivisible – are coaxed into travelling along two or more different paths simultaneously. Taken to the logical extreme, it is states like the one above that lead us to seemingly paradoxical situations with cats that are both alive and dead.

A few comments are in order about the Schrödinger equation and Hamiltonian. First, the Schrödinger equation replaces the venerable $𝐅=m𝐚$ in classical mechanics. Importantly, the Schrödinger equation is first order in time, rather than second order. This, ultimately, is why the state of a quantum system is described by $\psi (𝐱)$ and not both $\psi (𝐱)$ and $\dot{\psi }(𝐱)$.

Next, the “Hamiltonian” is not something entirely novel to quantum physics. It is actually a concept that arises in more sophisticated presentations of Newtonian physics. You can read about this approach in the lectures on Classical Dynamics. In the classical world, the Hamiltonian is closely related to the energy of the system which, for us, is

$E={\displaystyle \frac{1}{2m}}{𝐩}^2+V(𝐱)$

(1.3)

where $𝐩=m\dot{𝐱}$ is the momentum of the particle. As these lectures proceed, we will also see the connection between the operator Hamiltonian $\widehat{H}$ and energy. For now, we can just note that there is clearly a similarity in structure between the equations (1.2) and (1.3). If we wanted to push this analogy further, we might postulate some relationship in which momentum is replaced by a derivative operator,

$𝐩\to \pm i\mathrm{\hslash }\nabla \mathit{\hspace{1em}\hspace{1em} }\mathrm{?}$

where it’s not clear whether we should take $+$ or $-$ since the energy depends only on ${𝐩}^2$. More importantly, it’s not at all clear what a relationship of this kind would mean! How could the momentum of a particle – which, after all, is just a number — possibly be related to something as abstract as taking a derivative? As these lectures proceed, we’ll see the connection between momentum and derivatives more clearly. (We’ll also see that we should actually take the minus sign!)

It turns out that not all classical theories can be written using a Hamiltonian. Roughly speaking, only those theories that have conservation of energy can be formulated in this way. Importantly, the same is true of quantum mechanics. This means, in particular, that there is no Schrödinger equation for theories with friction. In many ways that’s no big loss. The friction forces that dominate our world are not fundamental, but the result of interactions between many (say ${10}^{23}$) atoms. There are no such friction forces in the atomic and subatomic worlds and, moreover, the formalism of quantum mechanics does not allow us to easily incorporate such forces.

(If, for some reason, you really do want to consider quantum friction then you’re obliged to include the direct coupling to the ${10}^{23}$ other atoms and track what happens. There are interesting applications of this, not least an idea known as decoherence.)

One final comment: as we proceed to more advanced physical theories, whether relativistic quantum field theory or even more speculative endeavours like string theory, the Schrödinger equation (1.1) remains the correct description for the evolution of the quantum state. All that changes is the Hamiltonian (1.2), which is replaced by something more complicated, as too are the wavefunctions on which the Hamiltonian acts.

To this end, let’s look at how the probability density changes in time. We have $P={|\psi |}^2$ and so,

${\displaystyle \frac{\partial P}{\partial t}}={\psi }^{\star }{\displaystyle \frac{\partial \psi }{\partial t}}+{\displaystyle \frac{\partial {\psi }^{\star }}{\partial t}}\psi$

(1.4)

The Schrödinger equation (1.1) and its complex conjugate give

${\displaystyle \frac{\partial \psi }{\partial t}}=-{\displaystyle \frac{i}{\mathrm{\hslash }}}\left(-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}{\nabla }^2\psi +V\psi \right)\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }{\displaystyle \frac{\partial {\psi }^{\star }}{\partial t}}={\displaystyle \frac{i}{\mathrm{\hslash }}}\left(-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}{\nabla }^2{\psi }^{\star }+V{\psi }^{\star }\right)$

We see that the terms with the potential $V$ cancel out when inserted in (1.4) and we’re left with

	${\displaystyle \frac{\partial P}{\partial t}}$	$=$	${\displaystyle \frac{i\mathrm{\hslash }}{2m}}\left({\psi }^{\star }{\nabla }^2\psi -\psi {\nabla }^2{\psi }^{\star }\right)$
		$=$	${\displaystyle \frac{i\mathrm{\hslash }}{2m}}\nabla \cdot \left({\psi }^{\star }\nabla \psi -\psi \nabla {\psi }^{\star }\right)$

We write this as

${\displaystyle \frac{\partial P}{\partial t}}+\nabla \cdot 𝐉=0$

(1.5)

where $𝐉$ is known as the probability current and is given by

$𝐉=-{\displaystyle \frac{i\mathrm{\hslash }}{2m}}\left({\psi }^{\star }\nabla \psi -\psi \nabla {\psi }^{\star }\right)$

(1.6)

Equations of the form (1.5) have a special place in the laws of physics. They are known as continuity equations and arises whenever there is a conserved quantity. (For example, the same equation appears in Electromagnetism where it relates the charge density to the electric current and ensures the conservation of electric charge.)

To see that the continuity equation implies a conservation law, consider some region $V\subset {𝐑}^3$ with boundary $S$. We then integrate to get the probability $P_V$ that the particle lies somewhere in this region.

$P_V(t)={\displaystyle {\int }_V}{d}^{3}xP(𝐱,t)$

The continuity equation tells us that this probability changes as

${\displaystyle \frac{\partial P_V}{\partial t}}=-{\displaystyle {\int }_V}{d}^{3}x\nabla \cdot 𝐉=-{\displaystyle {\int }_S}𝑑𝐒\cdot 𝐉$

(1.7)

where, in the final step, we have used the divergence theorem. We see that that the probability that the particle lies in $V$ can change, but only if there is a flow of probability through the surface $S$ that bounds $V$. If we know for sure that $𝐉=0$ everywhere on the surface $S$, then the probability that the particle is in the region $V$ doesn’t change: $\partial P_V/\partial t=0$.

The importance of the continuity equation is that it doesn’t just tell us that some quantity is conserved: it tells us that the object is conserved locally. In the present context, this quantity is probability. The probability density can’t just vanish in one region of space, to reappear in some far flung region of the universe. Instead, the evolution of the probability density is local. If it does change in some region of space then it’s because it has moved into a neighbouring region. The current $𝐉$ tells you how it does this.

As a special case, if we take the region $V$ to be all of space, so $V={𝐑}^3$, then

$P_{\mathrm{anywhere}}={\displaystyle \int d^{3}x{|\psi (𝐱,t)|}^2}$

As we’ve already seen, $P_{\mathrm{anywhere}}=1$. The change in the probability is, using (1.7),

${\displaystyle \frac{\partial P_{\mathrm{anywhere}}}{\partial t}}=-{\displaystyle {\int }_{S_{\mathrm{\infty }}^2}}𝑑𝐒\cdot 𝐉$

where $S_{\mathrm{\infty }}^2$ is the asymptotic 2-sphere at the infinity of ${𝐑}^3$. But any normalised wavefunction must have $\psi \to 0$ as $|𝐱|\to \mathrm{\infty }$ and, from the definition (1.6), we must then have $𝐉\to 0$ as $|𝐱|\to \mathrm{\infty }$. Furthermore, you can check that $𝐉$ falls off quickly enough so that

${\displaystyle \frac{\partial P_{\mathrm{anywhere}}}{\partial t}}=0$

This is the statement that if the wavefunction is normalised at some time, then it remains normalised for all time.

As an example, suppose that at time $t_0$ we have a particle sitting in a superposition of two different positions far well-separated positions, ${𝐗}_1$ and ${𝐗}_2$, something like

$\psi (𝐱,t_0)={\displaystyle \frac{1}{\sqrt{N^{\prime }}}}\left(e^{-a{(𝐱-{𝐗}_1)}^2}+e^{-a{(𝐱-{𝐗}_2)}^2}\right)$

Left to its own devices, this state will evolve through the Schrödinger equation. But suppose instead that we chose to do a measurement of the particle. For example, we can put a detector in the vicinity of ${𝐗}_1$ and see if it registers.

Suppose that the detector does indeed see the particle. Then we know for sure that the particle is sitting near ${𝐗}_1$ and that means that it can’t possibly be elsewhere. The probability that it sits near ${𝐗}_2$ must then vanish! Correspondingly, after the measurement, the wavefunction can’t have any support near ${𝐗}_2$. Instead, at time $t_{+}$ immediately after the measurement the wavefunction must jump to something localised near ${𝐗}_1$, like

$\psi (𝐱,t_{+})={\displaystyle \frac{1}{\sqrt{N}}}{e}^{-a{(𝐱-{𝐗}_1)}^2}$

This is known as the collapse of the wavefunction.

There is something more than a little disconcerting about this aspect of quantum mechanics. Just a few pages ago, I told you that the probability in the wavefunction was not due to our ignorance, but instead to an inherent randomness of the system. Yet now we see that once we gain additional knowledge about the system – like where the particle actually is – the wavefunction necessarily changes. Which makes it sound like the wavefunction is describing some aspect of what we know after all!

The collapse of the wavefunction also demonstrates that our gain in knowledge is far from innocent. There is no way to learn something about a quantum system without perturbing that system, sometimes violently so. In the discussion above, it may appear that the collapse of the wavefunction only increases knowledge since we now have a better understanding of the particle’s position. However, the wavefunction contains other, more subtle information and this too has been disturbed by the collapse which, typically, will lead to an increase of uncertainty in other properties.

Furthermore, the collapse of the wavefunction doesn’t happen locally. Now the probability density really does just vanish from one region of space and re-emerge in the region where we detected the particle. You might worry that this will lead to contradiction with causality or with special relativity. It turns out that it doesn’t but we won’t get to fully understand why until later lectures. (See, for example, the section on Bell’s inequality in the lectures on Topics in Quantum Mechanics.)

The fact that the wavefunction can evolve in two very different ways – the first through the smooth development of the Schrödinger equation, the second from the abrupt collapse after measurement – is one of the most unsettling aspects of quantum mechanics. It has led to much handwringing and philosophising and gnashing of teeth and mumbling of words like “ontic” and “epistemic”, none of which seems to be quite enough to calm the nerves. We’ll say more about these issues in Section 3.5 when we discuss the different interpretations of quantum mechanics.