The Probability Density

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Consider a calculation that Max Planck could have carried out in 1901, after he had published his famous paper on blackbody radiation with its weird little hypothesis that bodies emit electromagnetic radiation in quanta (which we now call photons). Suppose that we want to calculate the energy density in a pulse of light. In accordance with the quantum hypothesis, we have

(Eq=n 1)

in which n represents the number of quanta in a unit volume, h represents Planck=s constant, and v represents the frequency of the light. In accordance with James C. Maxwell=s electromagnetic theory and its wave equation describing the propagation of electromagnetic waves, we have

(Eq=n 2)

which we derive from the electric field strength in the wave,

(Eq=n 3)

in which E0 represents the amplitude of the wave and the exponential describes the wave=s moving sinusoidal envelope. Because the exponential yields a complex number, when we square the field strength in order to calculate the energy density we must multiply Equation 3 by its complex conjugate, which gives us

(Eq=n 4)

Now if we take the time derivative of the electric field and multiply it by the complex conjugate of the electric field, we get

(Eq=n 5)

Comparing that result with Equation 1 leads us to infer that if we define an operator

(Eq=n 6)

and apply it to Equation 3, it will extract from the description of the wave=s electric field the value of the energy in one of the wave=s photons. More precisely, it extracts that information from the wave function itself; the amplitude of the electric field (E0) contributes nothing in that calculation, so it looks like we can remove that factor without affecting the calculation. Of course, it=s not that simple. The electric-field amplitude E0 is not properly constant: it conforms to an amplitude modulation that starts at zero, rises to some maximum value, and then declines back to zero over the distance occupied by the pulse. That amplitude modulation gives the pulse a location around the point x=ct as the wave function cannot (the exponential has values for all values of x-λvt=x-ct and not only zero).

If I had wanted to keep E0 as a constant, I should have written Equation 3 as

(Eq=n 7)

in which A=A(x,t) represents the amplitude modulation of the electric field. Now I can divide out E0 and obtain the bare wave function itself;

(Eq=n 8)

which describes only the shape and the propagation of the pulse. Now I can use the operator of Equation 6 and recalculate Equation 5 as

(Eq=n 9)

In the electromagnetic version of that calculation the factor A2 describes the distribution of energy over the volume of space occupied by the pulse, so in the quantum version it must also give us a kind of density. We can= t smear out a photon in the way an electric field spreads out, so the density A2 must represent a more fundamental kind of density, one that applies to both photons and fields. If we were to put a large number of photons into the pulse, then Equation 9 would turn into Equation 1; that is, A2 determines what fraction of N photons goes into a given minuscule element of volume (dV) in the pulse. That means that for a single photon A2 represents the probability density for the photon to exist in that element of volume,

(Eq=n 10)

subject to the proviso that

(Eq=n 11)

when we carry out the integration over all space.

Thus, in 1901 Max Planck could have parlayed his weird little postulate into a more or less complete quantum theory of light. The reason that he did not do so may relate to the fact that he expressed his quantum hypothesis, not as properties inherent in the quanta, but rather as properties of the putative oscillators in the blackbody that absorb and re-radiate the radiation falling on them. However, as I showed at the end of the essay on AThe Kirchhof-Clausius Law@ , Planck (or someone else) could have deduced the quantum nature of light from the law of entropy and would have understood that the quantization is an inherent property of the light. Had Planck conceived the quantum hypothesis in that way, as referring to an inherent property of light, he might very well have deduced the statement that the square of the wavefunction represents a probability density.

In reality that statement was not deduced, but was presented as a postulate in July 1926 by Max Born (1882 Dec 11 - 1970 Jan 05). It was Born=s paper, by the way, that inspired Albert Einstein to write to Born a letter in which he included the comment that, AThe Old One does not play dice.@

Oh, but It does. Our Universe would not exist if It didn=t.

Equation 11 looks like the expression of a conservation law for probability. We thus expect that the probability density will conform to a continuity equation. So let= s calculate the time derivative of the probability density,

(Eq=n 12)

The state function (Ψ) must conform to the requirements of Schrödinger=s Equation,

(Eq=n 13)

so we can use it and its complex conjugate to replace the time derivatives in Equation 12. We thus obtain

(Eq=n 14)

Note that the potential energy terms canceled each other out: that happened because the potential energy appears as a simple multiplication by a real number and not as an operator acting on the state function. Next we apply the vector-operator identity

(Eq=n 15)

noting that the dot products of the paired gradients cancel out. So we have

(Eq=n 16)

If we rewrite that equation as

(Eq=n 17)

then we have a continuity equation () in which the probability current density conforms to

(Eq=n 18)

In the standard quantum theory we use the operator

(Eq=n 19)

to extract a description of a particle=s linear momentum from the particle=s state function. In that light we see that Equation 18 becomes

(Eq=n 20)

which makes the probability current density proportional to the product of the probability density and the particle=s velocity. We expect that result of a conserved quantity.

We could have obtained Equation 14 by multiplying Schrödinger=s Equation by Ψ*, multiplying the complex conjugate of Schrödinger=s Equation by Ψ, and subtracting the latter product from the former,

(Eq=n 21)

If we apply that procedure to the relativistic Schrödinger Equation (the Klein-Gordon Equation),

(Eq=n 22)

we get

(Eq=n 23)

We can rearrange that equation easily to get

(Eq=n 24)

Multiplying that equation by iS /2m0 and dividing it by the square of lightspeed gives us

(Eq=n 25)

The second major term on the left side of that equation gives us the divergence of the probability current density that we see in Equation 17, but the first major term does not look like the time derivative of the probability density. However, this is a relativistic calculation, so we cannot truly claim that we made a mistake in devising it until we see what it does in the limit as the particle= s total energy declines to m0c2.

The formula inside the parentheses of the first term looks like a preparation to calculate the expectation value of the particle=s total energy. We use the same operators in the relativistic quantum theory as we use in the non-relativistic version, so we accept that interpretation. In this case we have no forces acting on the particle, so applying the operator of Equation 6 to the state function is the same as multiplying by the number e (and its negative for the complex conjugate). Multiplication by a constant commutes with all of the operations in the first term, so we can rewrite Equation 25 as

(Eq=n 26)

We know, of course, that

(Eq=n 27)

so we can see that we achieve the limit described above by making the particle=s velocity take values that approach zero (or at least negligible relative to the speed of light). In that limit Equation 26 becomes Equation 17, as we require.

Thus, for a fully relativistic description of a quantum system we have the probability density as

(Eq=n 28)

If we want to calculate the expectation value of some property, represented by the operator Q, of the particle, we use

(Eq=n 29)

integrated over all space. As an example let=s calculate the expectation value of a particle=s position in the x-direction. For a proper Hamiltonian theory we must have perfect symmetry in operator representations, so given Equation 19, we must also have

(Eq=n 30)

Thus for the expectation value we have

(Eq=n 31)

We get that result because we know that

(Eq=n 32)

that is, the probability density (Equation 28) integrated over the volume of all space must yield the number of certainty. We have also made the tacit assumption that the state function has the form of Equation 8 as

(Eq=n 33)

with

(Eq=n 34)

in this case.

Even though this is a relativistic calculation, the Lorentz factor does not appear explicitly. Until we know precisely how the Lorentz factor appears in a relativistic wave function we must leave it implicit in the variables that our operators extract from the wave function.

Next let=s calculate the expectation value of the force acting on a particle immersed in a forcefield in which it has potential energy U (we simply add U to Equation 34 for this). We correlate force, via Newton=s second law, with the time derivative of the particle= s linear momentum, so we have

(Eq=n 35)

As we expect, the force equals the negative gradient of the potential energy, just as in classical dynamics.

And finally let= s calculate the expectation value of the time at which an event involving the particle occurs. Because we have

(Eq=n 36)

our Hamiltonian symmetry requires that we also have an operator

(Eq=n 37)

Noting that Equation 34 gives us

(Eq=n 38)

we thus calculate

(Eq=n 39)

That result resembles the fourth equation of the Lorentz Transformation, again with the Lorentz factor left implicit, which we should expect of the quantum theory if it encodes Lorentz invariance. I don=t want to pursue that thought here; rather, I want to wait until I have worked out the explicit description of the single-particle state function.

habg

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