Dirac’s delta function and Schrödinger’s equation in three dimensions

Feynman’s rather informal derivation of Schrödinger’s equation – following Schrödinger’s own logic when he published his famous paper on it back in 1926 – is wonderfully simple but, as I mentioned in my post on it, does lack some mathematical rigor here and there. Hence, Feynman hastens to dot all of the i‘s and cross all of the t‘s in the subsequent Lectures. We’ll look at two things here:

  1. Dirac’s delta function, which ensures proper ‘normalization’. In fact, as you’ll see in a moment, it’s more about ‘orthogonalization’ than normalization. 🙂
  2. The generalization of Schrödinger’s equation to three dimensions (in space) and also including the presence of external force fields (as opposed to the usual ‘free space’ assumption).

The second topic is the most interesting, of course, and also the easiest, really. However, let’s first use our energy to grind through the first topic. 🙂

Dirac’s delta function

When working with a finite set of discrete states, a fundamental condition is that the base states be ‘orthogonal’, i.e. they must satisfy the following equation:

ij 〉 = δij, with δij = 1 if i = j and δij = 0 if ij

Needless to say, the base states and j are rather special vectors in a rather special mathematical space (a so-called Hilbert space) and so it’s rather tricky to interpret their ‘orthogonality’ in any geometric way, although such geometric interpretation is often actually possible in simple quantum-mechanical systems: you’ll just notice a ‘right’ angle may actually be 45°, or 180° angles, or whatever. 🙂 In any case, that’s not the point here. The question is: if we move an infinite number of base states – like we did when we introduced the ψ(x) and φ(p) wavefunctions – what happens to that condition?

Your first reaction is going to be: nothing. Because… Well… Remember that, for a two-state system, in which we have two base states only, we’d fully describe some state | φ 〉 as a linear combination of the base states, so we’d write:

| φ 〉 =| I 〉 CI + | II 〉 CII 

Now, while saying we were talking a Hilbert space here, I did add we could use the same expression to define the base states themselves, so I wrote the following triviality:

M1Trivial but sensible. So we’d associate the base state | I 〉 with the base vector (1, 0) and, likewise, base state | II 〉 with the base vector (0, 1). When explaining this, I added that we could easily extend to an N-state system and so there’s a perfect analogy between the 〈 i | j 〉 bra-ket expression in quantum math and the ei·ej product in the run-of-the-mill coordinate spaces that you’re used to. So why can’t we just extend the concept to an infinite-state system and move to base vectors with an infinite number of elements, which we could write as ei =(…, 0, ei = 1, 0, 0,,…) and ej =(…, 0, 0, ej = 1, 0,…), thereby ensuring 〈 i | j 〉 = ei·ej = δijalways! The ‘orthogonality’ condition looks simple enough indeed, and so we could re-write it as:

xx’ 〉 = δxx’, with δxx’ = 1 if x = x’ and δxx’ = 0 if if x ≠ x’

However, when moving from a space with a finite number of dimensions to a space with an infinite number of dimensions, there are some issues. They pop up, for example, when we insert that 〈 xx’ 〉 = δxx’ function (note that we’re talking some function here of x and x’, indeed, so we’ll write it as f(x, x’) in the next step) in that 〈φ|ψ〉 = ∫〈φ|x〉〈x|ψ〉dx integral.

Huh? What integral? Relax: that 〈φ|ψ〉 = ∫〈φ|x〉〈x|ψ〉dx integral just generalizes our 〈φ|ψ〉 = ∑〈φ|x〉〈x|ψ〉 expression for discrete settings for the continuous case. Just look at it. When substituting φ for x’, we get:

x’|ψ〉 = ψ(x’) = ∫ 〈x’|x〉 〈x|ψ〉 dx ⇔ ψ(x’) = ∫ 〈x’|x〉 ψ(x) dx

You’ll say: what’s the problem? Well… From a mathematical point of view, it’s a bit difficult to find a function 〈x’|x〉 = f(x, x’) which, when multiplied with a wavefunction ψ(x), and integrated over all x, will just give us ψ(x’). A bit difficult? Well… It’s worse than that: it’s actually impossible!

Huh? Yes. Feynman illustrates the difficulty for x’ = 0, but he could have picked whatever value, really. In any case, if x’ = 0, we can write f(x, 0) = f(x), and our integral now reduces to:

ψ(0) = ∫ f(x) ψ(0) dx

This is a weird expression: the value of the integral (i.e. the right-hand side of the expression) does not depend on x: it is just some non-zero value ψ(0). However, we know that the f(x) in the integrand is zero for all x ≠ 0. Hence, this integral will be zero. So we have an impossible situation: we wish a function to be zero everywhere but for one point, and, at the same time, we also want it to give us a finite integral when using it in that integral above.

You’re likely to shake your head now and say: what the hell? Does it matter? It does: it is an actual problem in quantum math. Well… I should say: it was an actual problem in quantum math. Dirac solved it. He invented a new function which looks a bit less simple than our suggested generalization of Kronecker’s delta for the continuous case (i.e. that 〈 xx’ 〉 = δxx’ conjecture above). Dirac’s function is – quite logically – referred to as the Dirac delta function, and it’s actually defined by that integral above, in the sense that we impose the following two conditions on it:

  • δ(x‘) = 0 if x ≠ x’ (so that’s just like the first of our two conditions for that 〈 xx’ 〉 = δxx’ function)
  • δ(x)ψ(x) dx = ψ(x’) (so that’s not like the second of our two condition for that 〈 xx’ 〉 = δxx’ function)

Indeed, that second condition is much more sophisticated than our 〈 xx’ 〉 = 1 if x = x’ condition. In fact, one can show that the second condition amounts to finding some function satisfying this condition:

δ(x)dx = 1

We get this by equating x’ to zero once more and, additionally, by equating ψ(x) to 1. [Please do double-check yourself.] Of course, this ‘normalization’ (or ‘orthogonalization’) problem all sounds like a lot of hocus-pocus and, in many ways, it is. In fact, we’re actually talking a mathematical problem here which had been lying around for centuries (for a brief overview, see the Wikipedia article on it). So… Well… Without further ado, I’ll just give you the mathematical expression now—and please don’t stop reading now, as I’ll explain it in a moment:


I will also credit Wikipedia with the following animation, which shows that the expression above is just the normal distribution function, and which shows what happens when that a, i.e. its standard deviation, goes to zero: Dirac’s delta function is just the limit of a sequence of (zero-centered) normal distributions. That’s all. Nothing more, nothing less.


But how do we interpret it? Well… I can’t do better than Feynman as he describes what’s going on really:

“Dirac’s δ(xfunction has the property that it is zero everywhere except at x = 0 but, at the same time, it has a finite integral equal to unity. [See the δ(x)dx = 1 equation.] One should imagine that the δ(x) function has such fantastic infinity at one point that the total area comes out equal to one.”

Well… That says it all, I guess. 🙂 Don’t you love the way he puts it? It’s not an ‘ordinary’ infinity. No. It’s fantastic. Frankly, I think these guys were all fantastic. 🙂 The point is: that special function, Dirac’s delta function, solves our problem. The equivalent expression for the 〈 ij 〉 = δij condition for a finite and discrete set of base states is the following one for the continuous case:

xx’ 〉 = δ(x − x’)

The only thing left now is to generalize this result to three dimensions. Now that’s fairly straightforward. The ‘normalization’ condition above is all that’s needed in terms of modifying the equations for dealing with the continuum of base states corresponding to the points along a line. Extending the analysis to three dimensions goes as follows:

  • First, we replace the x coordinate by the vector r = (x, y, z)
  • As a result, integrals over x, become integrals over x, y and z. In other words, they become volume integrals.
  • Finally, the one-dimensional δ-function must be replaced by the product of three δ-functions: one in x, one in y and one in z. We write:

r | r 〉 = δ(x − x’) δ(y − y’)δ(z − z’)

Feynman summarizes it all together as follows:


What if we have two particles, or more? Well… Once again, I won’t bother to try to re-phrase the Grand Master as he explains it. I’ll just italicize or boldface the key points:

Suppose there are two particles, which we can call particle 1 and particle 2. What shall we use for the base states? One perfectly good set can be described by saying that particle 1 is at xand particle 2 is at x2, which we can write as | xx〉. Notice that describing the position of only one particle does not define a base state. Each base state must define the condition of the entire system, so you must not think that each particle moves independently as a wave in three dimensions. Any physical state | ψ 〉 can be defined by giving all of the amplitudes 〈 xx| ψ 〉 to find the two particles at x1 and x2. This generalized amplitude is therefore a function of the two sets of coordinates x1 and x1. You see that such a function is not a wave in the sense of an oscillation that moves along in three dimensions. Neither is it generally simply a product of two individual waves, one for each particle. It is, in general, some kind of a wave in the six dimensions defined by x1 and x1Hence, if there are two particles in Nature which are interacting, there is no way of describing what happens to one of the particles by trying to write down a wave function for it alone. The famous paradoxes that we considered in earlier chapters—where the measurements made on one particle were claimed to be able to tell what was going to happen to another particle, or were able to destroy an interference—have caused people all sorts of trouble because they have tried to think of the wave function of one particle alone, rather than the correct wave function in the coordinates of both particles. The complete description can be given correctly only in terms of functions of the coordinates of both particles.

Now we really know it all, don’t we? 🙂

Well… Almost. I promised to tackle another topic as well. So here it is:

Schrödinger’s equation in three dimensions

Let me start by jotting down what we had found already, i.e. Schrödinger’s equation when only one coordinate in space is involved. It’s written as:

schrodinger 3

Now, the extension to three dimensions is remarkably simple: we just substitute the ∂/∂xoperator by the ∇operator, i.e. ∇= ∂/∂x2  + ∂/∂y+ ∂/∂z2. We get:

schrodinger 4

Finally, we can also put forces on the particle, so now we are not looking at a particle moving in free space: we’ve got some force field working on it. It turns out the required modification is equally simple. The grand result is Schrödinger’s original equation in three dimensions:

schrodinger 5

V = V(x, y, z) is, of course, just the potential here. Remarkably simple equations but… How do we get these? Well… Sorry. The math is not too difficult, but you’re well equipped now to look at Feynman’s Lecture on it yourself now. You really are. Trust me. I really dealt with all of the ‘serious’ stuff you need to understand how he’s going about it in my previous posts so, yes, now I’ll just sit back and relax. Or go biking. Or whatever. 🙂


2 thoughts on “Dirac’s delta function and Schrödinger’s equation in three dimensions

  1. Pingback: Quantum-mechanical operators | Reading Feynman

  2. Pingback: Virtual particles | Reading Feynman

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