4 Momentum and Planck’s constant

On how to extract information about the momentum of particles from the wave function.

In the last section, we understood the probabilistic interpretation of the wave function ψ(x,t) for measurements of position x at time t. In this lecture we ask the question: how does the wave function ψ(x,t) encode information about measurements of momentum?

4.1 Momentum in Classical Mechanics

For motivation, we first recall the interpretation of momentum in classical mechanics as the generator of translations in space.

Recall from last term’s lectures that functions of position and momentum generate infinitesimal canonical transformations in Hamiltonian mechanics. The infinitesimal canonical transformation with parameter ϵ generated by a function A(x,p) is (4.1)xx=x+ϵ{x,A}pp=p+ϵ{p,A}. where (4.2){f,g}:=fxgpfpgx is the Poisson bracket.

In particular, the canonical transformation generated by the function (4.3)A(x,p)=p is (4.4)xx=x+ϵ{x,p}=x+ϵpp=p+ϵ{p,p}=p. Here we have used that (4.5){x,p}=1. The important lesson is that momentum p generates an infinitesimal translation of x.

4.2 Momentum in Quantum Mechanics

We now use this idea of momentum to understand momentum in quantum mechanics. Let us consider translating a wave function ψ(x) by a infinitesimal amount ϵ in the positive x direction - as shown below. As in the previous lecture, we work at some fixed moment in time.

Infinitesimal translation of a wave function to the right, \psi_{\text{trans}}(x) = \psi(x-\epsilon) (and note the sign!).

The translated wave function is of course ψ(xϵ). To first order in ϵ, the change in the wave function is found by Taylor expanding, (4.6)δϵψ(x)=ψ(xϵ)ψ(x)=ϵxψ(x)+O(ϵ2). If we want momentum to generate translations, this suggests we should identify momentum p with the derivative with respect to x.

Let us therefore define a momentum operator (4.7)p^=ix, such that (4.8)δϵψ(x)=ϵip^ψ(x), where is a constant of proportionality. The additional factors of i are introduced for convenience; we will get back to them shortly. Some comments are in order

  1. Note that while the momentum p has units of MLT1, /x has units of L1. This means must have units ML2T1, or ‘energytime’.

  2. Since the wave function is complex, we could imagine is a complex number. In a moment, we will show that must be real for momentum measurements to yield real results. This is why we introduced the extra factor of i in the definition.

Planck's constant has to be present in quantum mechanics for dimensional reasons. It has dimension of ‘energytime' or ‘action'.
. The constant is known as the (reduced) Planck constant, and pronounced ‘h-bar’.

Its value cannot be determined by mathematical arguments. It must be determined by comparing to experimental data, for example atomic spectra. In our universe, (4.9)1.05×1034kgm2s1. The smallness of this number, in units that are natural to humans, is why we do not observe quantum mechanical effects in everyday life. You will in the literature also find h=2π, which is usually called Planck’s constant (without the ‘reduced’ prefix).

4.3 A Quick Commutator

In order have position and momentum on an equal footing, we can introduce a position operator x^ that simply multiplies a wave function by x. In summary, (4.10)x^=xp^=ix. The ‘commutator’ of these operators is defined by (4.11)[x^,p^]:=x^p^p^x^. Here the operators should always be understood to act on everything to the right. Acting with this equation on a wave function ψ(x), we have (4.12)[x^,p^]ψ(x)=x^(p^ψ(x))p^(x^ψ(x))=x(ixψ(x))+ix(xψ(x))=iψ(x). where the final line follows from the product rule. Since this holds for any wave function ψ(x), we can summarise this result by (4.13)[x^,p^]=i. This is known as the canonical commutation relation.

The commutator is reminiscent of the Poisson bracket formula {x,p}=1 from classical mechanics. In fact, the commutator in quantum mechanics is found by replacing (4.14){,}i[,]. This replacement rule is known as ‘canonical quantisation’. We study it further in later lectures after introducing some more mathematical machinery.

4.4 Momentum Expectation Values

Just like position, in quantum mechanics we can only compute the probabilities of the outcomes of momentum measurements. For now, we satisfy ourselves with computing expectation values of functions of momentum.

First, recall from the last lecture that the expectation value x can be written (4.15)x=x|ψ(x,t)|2dx=ψ(x,t)x^ψ(x,t)dx. We now propose, similarly, that the expectation value of momentum is (4.16)p=ψ(x,t)p^ψ(x,t)dx=iψ(x,t)xψ(x,t)dx. As for the position expectation value, we emphasise that p is interpreted as the average of momentum measurements on an ensemble of particles with the same wave function ψ(x,t).

Let us now return to explain why must be real. Since the outcomes of momentum measurements are real numbers, we require pR. Let us imagine for a second that is complex and compute the complex conjugate of p, (4.17)p=i¯dxψ(x,t)xψ(x,t)=i¯dxxψ(x,t)ψ(x,t)+i[|ψ(x,t)|2]=i¯dxψ(x,t)xψ(x,t)=¯p. In the passing second line, we have integrated by parts. In passing to the third line, we discarded the boundary term because |ψ(x,t)|2 must vanish as x± if the wave function is square normalisable. Therefore, the momentum expectation value is real if and only if is real.

In a similar way, we can compute more general expectation values (4.18)f(p)=dxψ(x,t)f(p^)ψ(x,t)=dxψ(x,t)f(ix)ψ(x,t). Of particular importance is the momentum uncertainty (4.19)Δp=p2p2, which gives a measure of the spread of momentum measurements around p made on an ensemble of particles with identical wave function ψ(x,t).

Example: Gaussian Wave function

Let us again consider the wave function (4.20)ψ(x)=Cex2/4Δ2, with normalisation C=(2πΔ2)1/4. In the last lecture, we showed that the position expectation values are given by x=0 and x2=Δ2, and therefore the uncertainty in position is Δx=Δ.

Gaussian wave function with width \Delta.

The action of the momentum operator on this wave function is (4.21)p^ψ(x)=i2Δ2xψ(x)p^2ψ(x)=22Δ2ψ(x)24Δ4x2ψ(x). Note that the result is always a polynomial in x times the original wave function. This means we can recycle our results for position expectation values to compute momentum expectation values. For example, (4.22)p=i2Δ2x|ψ(x)|2=i2Δ2x=0 and similarly (4.23)p2=22Δ2|ψ(x)|224Δ4x2|ψ(x)|2=22Δ224Δ4x2=24Δ2. The momentum uncertainty is therefore Δp=2Δ.

Note that the product of position and momentum uncertainties is independent of Δ, (4.24)ΔxΔp=2. This means that if we attempt to localise the particle in space by making Δx smaller, the the uncertainly in momentum Δp necessarily increases, and vice verse.

4.5 Heisenberg’s Uncertainty Principle

This example illustrates an important result known as Heisenberg’s uncertainty principle. This states that for any normalised wave function, (4.25)ΔxΔp2. We will prove this result later in the course. It shows that there is a fundamental limit in quantum mechanics on the degree we can simultaneously reduce the uncertainty in position and momentum. The Gaussian wave function saturates this limit: it is a ‘minimal uncertainty’ wave function.

Remember that is an extremely small number in human units. So while we cannot arrange for both Δx and Δp to vanish, both uncertainties can be simultaneously small in human units. This goes some way to explaining why in everyday life, objects appear to have a definite position and momentum.

4.6 Problems

  1. PROBLEMS CLASS 1
    Momentum expectation value The expectation value of momentum measurements on a particle with wave function ψ(x) is (4.26)p=idxψ(x)xψ(x). You may assume the wave function is normalised, dx|ψ(x)|2=1.

    1. Show that p is real.

    2. Show that if ψ(x) is real then p=0.

    3. Suppose a wave function ϕ(x) has momentum expectation value pϕ=P.
      Compute the expectation value pψ for (4.27)ψ(x)=eixp0/ϕ(x).

    Solution

    1. We compute the conjugate (4.28)p=+idxψ(x)xψ(x)=idxxψ(x)ψ(x)+i[|ψ(x)|2]=idxψ(x)xψ(x)=p. In passing to the second line we have integrated by parts. The boundary term vanishes provided the probability density |ψ(x)|2 vanishes at x±. This is a necessary condition for the wave function to be normalizable, |ψ(x)|2<.

    2. Assuming the wave function is real ψ(x)=ψ(x), then (4.29)p=idxψ(x)xψ(x)=i2dxx(ψ(x)2)=i2[ψ(x)2], which vanishes for the same reason as the boundary term in part (a).

    3. By direct computation (4.30)pψ=idxψ(x)xψ(x)=idxeixp0/ϕ(x)x(eixp0/ϕ(x))=idxϕ(x)(xϕ(x)+ip0ϕ(x))=idxϕ(x)xϕ(x)+p0dx|ϕ(x)|2=P+p0 where in the last step we used that the wave function is normalized.

  2. Gaussian wave function:

    Reconsider the Gaussian wave function which you have seen in one of the problems in the previous chapter.

    1. Explain why the momentum expectation value is zero, p=0.

    2. Compute the momentum uncertainty Δp.

    3. Show that the wave function saturates Heisenberg’s uncertainty principle, (4.31)ΔxΔp2.

    4. What happens to Δp when the particle is localised in space?

    5. What changes if you multiply the wave function by eixp0/?

    Solution

    1. The wave function is real.

    2. First compute the action of the momentum operator p=ix on the wave function, (4.32)pψ(x)=ix(Ce(xx0)2/4Δ2)=i2Δ2(xx0)Ce(xx0)2/4Δ2=i2Δ2(xx0)ψ(x). Acting again, we find (4.33)p2ψ(x)=24Δ4((xx0)22Δ2)ψ(x). We can now re-use the position expectation values x=x0 and x2=x02+Δ2 computed in question 3 (together with the normalisation condition 1=1) to compute the momentum expectation values, (4.34)p=i2Δ2xx0=0p2=24Δ2(xx0)22Δ2=24Δ2(x22x0x+x022Δ2)=24Δ2. If you are uncomfortable with the above manipulations, you can first compute the momentum expectation values using the integral definitions (4.35)p=idxψ(x)xψ(x)p2=2dxψ(x)x2ψ(x). You will find that the computation is equivalent to the one above! Finally, Δp=p2p2=/2Δ.

    3. Recalling that Δx=Δ, we have ΔxΔp=/2.

    4. The uncertainty in momentum increases.

    5. If you multiply the wave function by eixp0/, the momentum expectation value changes to p=p0.

  3. TUTORIAL 1
    Non-normalisable wave functions:

    Consider the wave function (4.36)ψ1(x)=Ceikx, for some constant C and k. The total integrated probability is not defined, because the integral (4.37)|ψ(x)|2dx does not exist for any non-zero value of C. However, we can make sense of it by ‘cutting off’ the integral so that L<x<L (‘putting the particle in a box’), doing all computations at finite L, and then taking the limit L. If you use this ‘regularisation’, what is the expectation value p^ when the system is described by the wave function above? Also compute p2.

    Solution If we cut off the integrals, (4.38)C2=LLdx=2L,C=1/2L, and the expectation value of the momentum becomes (4.39)p=C2LL(i)(ik)dx=k. This is independent of L. However, manipulations like these are on a shoddy mathematical ground (we e.g. did not discuss the boundary conditions on the wave function), and we will see some examples where naive reasoning like this goes flat on its face.

  4. Momentum in a box:

    Consider a particle in an infinite potential well, so that L/2<x<L/2. If you want to show that p is real (see above), you will encounter boundary terms which need to vanish. Do they? Why?

    Solution The boundary term you encounter when computing pp reads (4.40)i[ψ(x)ψ¯(x)]L/2L/2. For generic wave functions ψ(x) this would not vanish. It will, however, vanish if the wave function vanishes at the edge of the box, which is precisely the condition we have used in the chapter where we first introduced the wave function.