For short, define a new variable¹ \(|\bar \psi\rangle \equiv H|\psi\rangle\). We can express the state \(|\bar\psi\rangle\) as a function of \(x\) because we know how to express \(H\) and \(\psi\) in terms of \(x\): \(H\) is the differential operator \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\), and \(\psi(x)\) is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that \(\bar\psi(x)\) is constant.

Having introduced \(|\bar\psi\rangle\), we can calculate the average energy-squared of \(|\psi\rangle\) in the following way.

\(\begin{eqnarray} \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H \psi\rangle\\ &=& \langle \psi H | H\psi \rangle\\ &=& \langle \bar\psi | \bar\psi \rangle\\ &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ &=& \frac{A^2\hbar^4 a}{m^2}\\ \end{eqnarray}\)

For future reference, observe that this value is nonzero (which makes sense).

We can also calculate the average energy-squared of \(|\psi\rangle\) in the following way.

\begin{eqnarray} \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ &=& \langle \psi |H \bar\psi \rangle\\ &=&\int_0^a Ax(x-a) \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\ &=& 0\quad (!)\\ \end{eqnarray}

The second-to-last term must be zero because the second derivative of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.

Physical states are represented as wavefunctions in quantum mechanics. Just as we disallow certain physically nonsensical states in classical mechanics (for example, we consider it to be nonphysical for an object to spontaneously disappear from one place and reappear in another), we also disallow certain wavefunctions in quantum mechanics.

For example, since wavefunctions are supposed to correspond to probability amplitudes, we require wavefunctions to be normalized \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow wavefunctions that do not satisfy this property (although there are some exceptions²).

As another example, we generally expect probability to vary smoothly—if a particle is very likely or very unlikely to be found at a particular location, it should also be somewhat likely or somewhat unlikely to be found near that location. In more precise terms, we expect that for physically meaningful wavefunctions, the probability \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of \(x\) and, again, we disallow wavefunctions that do not satisfy this property because we consider them to be physically nonsensical.

So, physical wavefunctions must satisfy certain properties like the two just described. Wavefunctions that do not satisfy these properties are rejected for being physically nonsensical: even though we can perform calculations with them, the mathematical results we obtain do not mean anything physically.

Now, in quantum mechanics, an operator is a function that converts states into other states. Some operators correspond to physical quantities such as energy, momentum, or position, and as a result, the mathematical properties of these operators correspond to physical properties of the system. Such operators are called hermitian operators; one important property of hermitian operators is this rule:

Hermitian operator rule: A hermitian operator must only operate on the wavefunctions we have deemed physical, and must only produce physical wavefunctions³.

As you can see, this rule comes in two pieces. The first part is a constraint on you, the physicist: you must never feed a nonphysical state into a Hermitian operator, as it may produce nonsense. The second part is a constraint on the operator: the operator is guaranteed only to produce physical wavefunctions.

In fact, this rule for hermitian operators is the source of our problem, as we unknowingly violated it when applying our second method!

You'll remember that in the second method we had wavefunctions within the well

\( \begin{eqnarray} \psi(x) &=& A\;x(x-a)\\ \bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\ \end{eqnarray} \)

Using this, we wrote

\(\begin{eqnarray} \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ &=& \langle \psi |H \bar\psi \rangle\\ & \vdots&\\ &=& 0\\ \end{eqnarray}\)

However, there are two issues here. First, we were not allowed to operate with \(H\) on the wavefunction \(|\bar \psi\rangle\), because \(H\) is supposed to be a physical operator and \(|\bar \psi\rangle\) is a nonphysical state. Indeed, the constant function \(|\bar\psi\rangle\) does not approach zero at the edges of the well. By feeding \(H\) a nonphysical wavefunction, we obtained nonsensical results.

Second, and more importantly, we were wrong to claim that \(H\) was a physical operator—that \(H\) was hermitian. According to the rule, a hermitian operator must convert physical states into other physical states. But \(|\psi\rangle\) is a physical state, as we said when we first introduced it —it is a normalized, continuous function which approaches zero at the edges of the well and doesn't exist outside it. On the other hand, \(|\bar\psi\rangle\) is nonphysical because it does not go to zero at the edges of the well. It is therefore impermissible for \(H\) to transform the physical state \(|\psi\rangle\) into the nonphysical state \(|\bar\psi\rangle\). Because \(H\) converts some physical states into nonphysical states, it cannot be a hermitian operator as we assumed.

It may surprise you (and it certainly surprised me) to find that the Hamiltonian is not hermitian. One of the fundamental principles of quantum mechanics is that hermitian operators correspond to physically observable quantities; for this reason, surely the Hamiltonian—the operator corresponding to energy—ought to be hermitian?

But we must understand the correspondence between physically observable quantities and hermitian operators: every hermitian operator corresponds to a physically observable quantity, but not every quantity that intuitively “ought” to be observable will correspond to a hermitian operator⁴. The true definition of a hermitian operator imply that the Hamiltonian stops being hermitian in the infinitely deep well. Here we arrive at a crucial point:

Operators do not change form between problems: the one-dimensional Hamiltonian is always \(H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V\), the one-dimensional canonical momentum is always \(P=i\hbar\frac{d}{dx}\), and so on.

However, operators do change in this respect: hermitian operators must only take in physical states, and must only produce physical states; because in different problems we do change the requirements for being a physical state, we also change what it takes for an operator to be called hermitian. As a result, an operator that is hermitian in one setting may fail to be hermitian in another.

Having seen how boundary conditions can affect hermiticity, we ought to be extra careful about which conditions we impose on our wavefunctions.

We have said already that physicists require wavefunctions to satisfy certain properties in order to be deemed “physical”. To be specific, wavefunctions in the infinitely deep well

• Must be normalizable, because they correspond to probability amplitudes.
• Must have smoothly-varying probability, because if a particle is very likely to be at a location, it ought to be likely to be near it as well.
• Must not exist outside the well, because it would take an infinite amount of energy to do so.

These conditions are surely reasonable. However, physicists sometimes assert that in order to satisfy the second and third conditions, physical wavefunctions

• (?) Must smoothly approach zero towards the edges of the well.

This final constraint is our reason for rejecting \(|\bar\psi\rangle\) as nonphysical and is consequently the reason why \(H\) is not hermitian. If we can convince ourselves that the final constraint is unnecessary, \(H\) may again be hermitian. This will satisfy our intuitions that the energy operator ought to be hermitian.

But in fact, we have the following mathematical observation to save us: a function \(f\) does not need to be continuous in order for the integral \(\int^x f\) to be continuous. As a particularly relevant example, you may now notice that the function \(\bar\psi(x)\) is not itself continuous, although the integral \(\int_0^x \bar\psi\) is continuous. Evidently, it doesn't matter that the wavefunction \(\bar\psi\) itself is not continuous; the probability corresponding to \(\bar\psi\) does manage to vary continuously anyways. Because the probability corresponding to \(\bar\psi\) is the only aspect of \(\bar\psi\) which we can detect physically, we can safely omit the final constraint while keeping the other three.