To talk about this topic let's use a concrete example:
Suppose I have a one-dimensional system subjected to a linear potential, such as the hamiltonian of the system is: $$H=\frac{\hat{p}^2}{2m}-F\hat{x}, \qquad \hat{x}=i\hbar\frac{\partial}{\partial p},$$ then I might want to find the eigenfunctions of the hamiltonian: $$\psi _E(p)=\langle p|E\rangle,$$ where $|p\rangle$ are the eigenvectors of the momentum operator and $|E\rangle$ are the eigenvectors of the hamiltonian. After a bit of work with the TISE I came to the following expression for $\psi _E(p)$: $$\psi _E(p)=N\exp\left[-\frac{i}{\hbar F}\left(\frac{p^3}{6m}-Ep\right)\right].$$ I am almost there! The only thing missing is the normalization constant $N$. How should I move forward? I could try to apply the normalization condition directly by imposing the integral of this function equal to 1, but this seems like a lot of work. However my lecture notes suggest me to try to take advantage of the fact that the eigenvectors of the hamiltonian must be normalized: $$\langle E'|E\rangle=\delta(E-E')$$ where $\delta$ is the Dirac's Delta Function.1
However I cannot see how to use this information to derive the normalization constant $N$. Are my lecture notes right? How should I use the normalization condition of the eigenvectors of the hamiltonian then? Is it quicker to simply try to impose the integral equal to 1?

[1]: Based on my current understanding this is a generalization (not so rigorous) of the normalization condition of the eigenvectors of an observable in the discrete case: $$\langle E'|E\rangle=\delta _k \ \Rightarrow \ \langle E'|E\rangle=\delta(E-E')$$ where $\delta _k$ is the Kronecker Delta, equal to one if the eigenvectors are the same and zero otherwise.


2 Answers 2


The proposed "suggestion" should actually be called a requirement: you have to use it as a normalization condition. This is because the wavefunctions are not normalizable: what has to equal 1 is the integral of $|\psi|^2$, not of $\psi$, and $|\psi|^2$ is a constant. Just like a regular plane wave, the integral without $N$ is infinite, so no value of $N$ will make it equal to one.

One option here would be to just give up and not calculate $N$ (or say that it's equal to 1 and forget about it). This is not wrong! The functions $\psi_E$ are not physical - no actual particle can have them as a state. Physical states $\psi(p)$ are superpositions of our basis wavefunctions, built as

$$\psi(p) = \int dE\, f(E) \psi_E(p)$$

with $f(E)$ some function. This new wavefunction is physical, and it must be normalized, and $f(E)$ handles that job - you have to choose it so that the result is normalized.

But there are two reasons we decide to impose $\langle E | E' \rangle = \delta(E-E')$. One is that it's useful to have some convention for our basis, so that latter calculations are easier. Having a delta function is unavoidable, since regardless of the normalization the inner product will be zero for different energies and infinite for equal energies, but we could put some (possibly $E$-dependent) coefficient in front of it - that's just up to convention.

The other reason is that if you dig a little deeper into the normalization of the $\psi(p)$ above, the delta function appears anyway. We have

$$\langle \psi | \psi \rangle = \int dp\, \int dE\, \int dE'\, f(E)^* f(E') \psi_E^*(p) \psi_{E'}(p),$$

and you can see that the inner product $\langle E | E' \rangle$ is right there, in the $E$ integral. So we have to use the fact that it is proportional to $\delta(E-E')$, and it's neater to fix the constant of proportionality beforehand.

So to recap: having $\langle E | E' \rangle \propto \delta(E-E')$ just falls out of the definition of the $\psi_E(p)$, and it's also obviously the manifestation of the fact that stationary states with different energies are orthogonal. We're just free to choose what goes in front of the delta function, which is equivalent to giving a (possibly energy dependent) value for $N$. Using $\delta(E-E')$ by itself is just the simplest choice, but sometimes other factors are used.

Now, actually calculating $N$ given this convention is pretty easy: I won't give you the answer, but notice that when you calculate the inner product of two wavefunctions with different energies (that is, the integral of $\psi_E^* \psi_{E'}$), the parts with $p^3$ in the exponential cancel, because they don't depend on the energy. What's left is a regular complex exponential, and by using the identity

$$\int_{-\infty}^\infty dx\, e^{ikx} = 2\pi \delta(k)$$

(which is rigorous enough for our purposes), you show that the whole thing must be proportional to $\delta(E'-E)$, and derive the value of $N$ from there.

  • $\begingroup$ This was helpful, but I don't get why the Dirac's delta is equal to the integral shown in your last equation. Can you expand a bit on this topic? I thought that the Dirac's delta wasn't even a function at all, and now you are telling me that it is equal to a complex integral... $\endgroup$
    – Noumeno
    Sep 4, 2020 at 9:35
  • $\begingroup$ @Noumeno See here for some proofs. Getting used to "proofs" involving the delta "function" can be pretty weird, but knowing how to manipulate it is essential for physics. Everything can be made rigorous if you wish, but it's usually faster not to. $\endgroup$
    – Javier
    Sep 4, 2020 at 12:44
  • $\begingroup$ Sorry to bother you but I just realized that I have another problem with your explanation: in the second paragraph you state that the condition on the inner product of the eigenvectors of the hamiltonian is the definition of the term "normalization" for wavefunctions; but I don't see how it can be. If the integral of the wavefunction is always divergent than seems that the function cannot be normalized, why the result of this inner product has something to do with this? I think an edit to expand on this definition might be helpful. I think that this is the core of my problem with this topic. $\endgroup$
    – Noumeno
    Sep 4, 2020 at 14:21
  • $\begingroup$ @Noumeno I've added quite a bit of detail :) $\endgroup$
    – Javier
    Sep 4, 2020 at 16:08

The is a bit of confusion here. The quantum state of a system $|\psi\rangle$ must always be normalized: $\langle\psi|\psi\rangle=1$. Since the wave function of a system is directly related to the wave function: $\psi(p)=\langle p|\psi\rangle$, it must also be normalized. Otherwise, the calculations of observables won't come out right.

Now it can happen that the eigenstates of the Hamiltonian $|E\rangle$ form a continuous spectrum, so that they would obey the orthogonality condition $\langle E|E'\rangle=\delta(E-E')$. It means that these eigenstates are not normalizable. Therefore they cannot individually serve as wave functions. Instead a wave function would be composed of a superposition os such eigenstates. $$ |\psi\rangle=\int |E\rangle F(E) dE . $$ where $F(E)$ is the coefficient function. The normalization condition then means that $$ \langle\psi|\psi\rangle=\int |F(E)|^2 dE = 1 . $$

Hope this helps.


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