The presentation of wavefunction information in different spaces or domains, such as the position and wavenumber domains discussed in the previous section, is useful in many applications of physics and engineering. In quantum mechanics, the wavefunction representations you’re likely to encounter include position and momentum, so this section is all about position and momentum wavefunctions, eigenfunctions, and operators – specifically, how to represent those functions and operators in both position space and momentum space.
You’ve already seen the connection between wavenumber (k) and momen- tum (p), which is provided by the de Broglie relation
p= ¯hk. (3.4)
This means that the Fourier-transform relationship between functions of position and wavenumber also works between functions of position and momentum. Specifically, the momentum wavefunction φ(p)˜ is the Fourier transform of the position wavefunctionψ(x):
φ(p)˜ = 1
√2πh¯ ∞
−∞ψ(x)e−i
p
¯ hx
dx, (4.38)
in whichφ(p)˜ is a function of momentum (p)10.
Additionally, the inverse Fourier transform of the momentum wavefunction φ(p)˜ gives the position wavefunctionψ(x):
ψ(x)= 1
√2πh¯ ∞
−∞
φ(p)e˜ iph¯x
dp. (4.39)
Sincek=ph¯,dk=dph¯, and substituting ph¯ forkanddph¯ fordkinEq. 4.15for the inverse Fourier transform yields
ψ(x)= 1
√2π ∞
−∞
φ(p)e˜ iph¯xdp
¯
h , (4.40)
which differs fromEq. 4.39by a factor of √1
¯
h. In some texts (including this one), that factor is absorbed into the functionφ, but several popular quantum˜ texts absorb1h¯ intoφ, omitting the factor of˜ √1
¯
hin the definitions of the Fourier transform and the inverse Fourier transform. In those texts, the factor in front of the integrals inEqs. 4.38and4.39is √1
2π.
Whichever convention you use for the constants, the relationship between position and momentum wavefunctions can help you understand one of the iconic laws of quantum mechanics. That law is theHeisenberg Uncertainty principle, which follows directly from the Fourier-transform relationship between position and momentum.
You can find an uncertainty principle for any Fourier-transform pair of conjugate wavefunctions, including the momentum-basis equivalent of the rectangular (flat-amplitude) wavenumber spectrumφ (k)and sinax(ax) position wavefunction discussed in the previous section. But it’s also instructive to
10This notation is quite common in quantum textbooks; the tilde (˜) distinguishes the momentum wavefunctionφ(p)˜ from the wavenumber wavefunctionφ (k).
consider a momentum wavefunctionφ˜ that doesn’t produce an extended lobe structure inψ(x)such as that shown inFig. 4.14b, since one of the goals of adding wavefunctions over a range of wavenumbers or momenta is to produce a spatially limited position wavefunction. So a position-space wavefunction that decreases smoothly toward zero amplitude without those extended lobes is desirable.
One way to accomplish that is to form a Gaussian wave packet. You may be wondering whether this means Gaussian in position space or Gaussian in momentum space, to which the answer is “both.” To understand why that’s true, start with the standard definition of a Gaussian function of position (x):
G(x)=Ae
−(x−x0)2 2σ2
x , (4.41)
in whichAis the amplitude (maximum value) ofG(x),x0is the center location (x-value of the maximum), andσxis the standard deviation, which is half the width of the function between the points at whichG(x)is reduced to√1
e(about 61%) of its maximum value.
Gaussian functions have several characteristics that make them instructive as quantum wavefunctions, including these two:
a) The square of a Gaussian is also a Gaussian, and b) The Fourier transform of a Gaussian is also a Gaussian.
The first of these characteristics is useful because the probability density is related to the square of the wavefunction, and the second is useful because position-space and momentum-space wavefunctions are related by the Fourier transform.
You can see one of the benefits of the smooth shape of the Gaussian in Fig. 4.19. The Fourier-transform relationship betweenψ(x)andφ(p)˜ means that smoothing the sharp corners of the rectangular momentum spectrumφ(p)˜ significantly reduces the value of the magnitude of the position wavefunction in the region of the sin(ax)/axlobe structure.
In position space, the phrase “Gaussian wave packet” means that the envelope of a sinusoidally varying function has a Gaussian shape. Such a packet can be formed by multiplying the Gaussian function G(x) by the functionei
p0
¯ hx
for a plane wave with momentump0:
ψ(x)=Ae
−(x−x0)2 2σ2
x ei
p0
¯ hx
, (4.42)
Ф(p) Magnitude of ψ(x)
p x
p
0Rectangular function has sharp corners in Ф(p), so sidelobes in ψ(x) are large
Gaussian Ф(p) means
Gaussian ψ(x)
͠ ͠ ͠
Figure 4.19 Improved spatial localization of position wavefunctionψ(x)using Gaussian rather than rectangular function in momentum space.
in which the plane-wave amplitude has been absorbed into the constant A.
When you’re dealing with such a Gaussian wavefunction, it’s important to realize that the quantityσxrepresents the standard deviation of the wavefunc- tionψ(x), which isnotthe same as the standard deviation of the probability distribution that results from this wavefunction. That probability distribution is also a Gaussian, but with a different standard deviation, as you’ll see later in this section.
When you’re dealing with a quantum wavefunction, it’s always a good idea to make sure that the wavefunction is normalized. Here’s how that works for ψ(x):
1= ∞
−∞ψ∗ψdx= ∞
−∞
⎡
⎣Ae
−(x−x0)2 2σ2
x ei
p0
¯ hx
⎤
⎦
∗⎡
⎣Ae
−(x−x0)2 2σ2
x ei
p0
¯ hx
⎤
⎦dx
= ∞
−∞|A|2
⎡
⎣e−(x−x0)
2 σ2
x
⎤
⎦e
(−p0+p0)x
¯
h dx=|A|2 ∞
−∞e
−(x2−2x0x+x2 0) σ2
x dx.
This definite integral can be evaluated using ∞
−∞e−(ax2+bx+c)dx= 3π
aeb
2−4ac
4a . (4.43)
In this casea= σ12
x,b=−σ2x20
x , andc= σx202 x, so
1= |A|2 π
1 σx2
e
(−2x0 σ2
x )2−4 1
σ2 x
x2 0 σ2 x 4 1σ2
x = |A|2
σx2πe
4x2 0−4x2
0 4σ2
x = |A|2σx√ π. Solving forAyields
A= 1
(σx√
π )1/2 (4.44)
and the normalized position wavefunction is ψ(x)= 1
(σx√ π )1/2e
−(x−x0)2 2σ2
x ei
p0
¯ hx
. (4.45)
To find the momentum wavefunctionφ(p)˜ corresponding to this normalized position wavefunction, take the Fourier transform ofψ(x). To simplify the notation, you can take the origin of coordinates to be atx0, sox0 = 0. That makes the Fourier transform look like this:
φ(p)˜ = 1
√2πh¯ ∞
−∞ψ(x)e−iph¯xdx= 1
√2πh¯ ∞
−∞
1 (σx√
π )1/2e
−x2 2σ2
xe−ip−ph¯0xdx
= 1
√2πh¯ 1 (σx√
π )1/2 ∞
−∞e
−x2 2σ2
x−ip−p¯h0x
dx.
Using the same definite integral given earlier in this section witha = 2σ12 x, b= −ip−h¯p0, andc=0 gives
φ(p)˜ = 1
√2πh¯ 1 (σx√
π )1/2 3π
aeb
2−4ac
4a = 1
√2πh¯
2π σx2 (σx√
π )1/2e
−(p−p0)2σ2 x 2h¯2
=
"
σx2 πh¯2
#14 e
−(p−p0)2σ2 x 2h¯2 .
This is also a Gaussian, since it can be written φ(p)˜ =
"
σx2 πh¯2
#1
4
e
−(p−p0)2 2σ2
p , (4.46)
in which thestandard deviationof the momentum wavefunction is given by σp=σh¯x.
Multiplying the standard deviations of these Gaussian position and momen- tum wavefunctions gives
σxσp=σx
h¯ σx
= ¯h. (4.47)
It takes just one more step to get to the Heisenberg Uncertainty principle.
To make that step, note that the “uncertainty” in the Heisenberg Uncertainty principle is defined with respect to the width of the probability distribution, which is narrower than the width of the Gaussian wavefunctionψ(x).
To determine the relationship between these two different widths, remember that the probability density is proportional toψ∗ψ. That means that the width xof the probability distribution can be found from
e−
x2 2(x)2 =
"
e−
x2 2σ2
x
#∗"
e−
x2 2σ2
x
#
=e−
x2 σ2
x. (4.48)
So 2(x)2 = σx2, or σx = √
2x. The same argument applies to the momentum-space wavefunction φ(p), so it’s also true that˜ σp = √
2p, in whichp represents the width of the probability distribution in momentum space.
This is the reason that many instructors and authors define the exponential term in the position wavefunctionψ(x)ase
−(x−x0)2 4σ2
x . In that case, theσx that they write in the exponential ofψ(x)is the standard deviation of the probability distribution rather than the standard deviation of the wavefunction.
WritingEq. 4.47in terms of the widths of the probability distributions in position (x) and momentum (p) gives
σxσp=(√ 2x)(√
2p)= ¯h (4.49)
or
xp= ¯h
2. (4.50)
This is the uncertainty relation for Gaussian wavefunctions. For any other functions, the product of the standard deviations gives a value greater than this, so the general uncertainty relation between conjugate variables such as
position and momentum (or any other two variables related by the Fourier transform) is
xp≥ ¯h
2. (4.51)
This is the usual form of the Heisenberg Uncertainty principle. It says that for this pair of conjugate or “incompatible” observables, there is a fundamental limit to the precision with which both may be known. So precise knowledge of position (smallx) is incompatible with precise knowledge of momentum (small p), since the product of their probability-distribution uncertainties (xp) must be equal to or larger than half of the reduced Planck constanth.¯
Another important aspect of incompatible observables concerns the opera- tors associated with those observables. Specifically, the operators of incom- patible observables do not commute, which means that the order in which those operators are applied matters. To see why that’s true for the position and momentum operators, it helps to have a good understanding of the form and behavior of these operators in both position and momentum space.
Students learning quantum mechanics often express confusion about quan- tum operators and their eigenfunctions, and that confusion is frequently embodied in questions such as:
– Why is the result of operating on a position wavefunction with the position operatorXequal to that wavefunction multiplied byx?
– Why are position eigenfunctions given by delta functions δ(x− x0) in position space?
– Why is the result of operating on a momentum wavefunction with the momentum operator P equal to the spatial derivative of that function multiplied by−ih?¯
– Why are momentum eigenfunctions given by √1
2πh¯eiph¯xin position space?
To answer these questions, start by considering how an operator and its eigenfunctions are related to the expectation value of the observable associated with the operator. As explained in Section 2.5, the expectation value for a continuous observable such as positionxis given by
x = ∞
−∞xP(x)dx, (4.52)
in whichP(x)represents the probability density as a function of positionx.
For normalized quantum wavefunctions ψ(x), the probability density is given by the square magnitude of the wavefunction|ψ(x)|2 = ψ(x)∗ψ(x), so the expectation value may be written as
x = ∞
−∞x|ψ(x)|2dx= ∞
−∞[ψ(x)]∗x[ψ(x)]dx. (4.53) Compare this to the expression fromSection 2.5for the expectation value of an observablexassociated with operatorXusing the inner product:
x = ψ|X|ψ = ∞
−∞[ψ(x)]∗X[ψ(x)]dx. (2.60) For these expressions to be equal, the result of the operatorXacting on the wavefunctionψ(x)must be to multiplyψ(x)byx. And why does it do that?
Because an operator’s job is to pull out the eigenvalues (that is, the possible results of observations) from the eigenfunctions of that operator, as described inSection 4.2. In the case of the position observations, the possible results of measurement are every positionx, so that’s what the position operatorXpulls out from its eigenfunctions.
And what are those eigenfunctions of the X operator? To answer that question, consider how those eigenfunctions behave. The eigenvalue equation for the position operator acting on the first of the eigenfunctions (ψ1(x)) is
Xψ1(x)=x1ψ1(x), (4.54)
in whichx1represents the eigenvalue associated with eigenfunctionψ1. But since the action of the position operator is to multiply the function upon which it’s operating byx, it must also be true that
Xψ1(x)=xψ1(x). (4.55)
Setting the right sides ofEqs. 4.54and4.55equal to one another gives
xψ1(x)=x1ψ1(x). (4.56)
Think about what this equation means: the variablextimes the first eigenfunc- tion ψ1 is equal to the single eigenvaluex1 times that same function. Since xvaries over all possible positions whilex1represents only a single position, how can this statement be true?
The answer is that the eigenfunctionψ1(x)must be zero everywhere except at the single locationx=x1. That way, when the value ofxis not equal tox1, both sides ofEq. 4.56are zero, and the equation is true. And whenx=x1, this equation saysx1ψ1(x)=x1ψ1(x), which is also true.
So what function is zero for all values of x except when x = x1? The Dirac delta functionδ(x−x1). And for the second eigenfunctionψ2(x)with eigenvaluex2, the delta functionδ(x−x2)does the trick, as doesδ(x−x3)for ψ3(x), and so forth.
Thus the eigenfunctions of the position operatorX are an infinite set of Dirac delta functionsδ(x−x), each with its own eigenvalue, and those eigen- values (represented byx) cover the entire range of positions from−∞to+∞.
You can bring this same analysis to bear on momentum operators and eigenfunctions, which behave in momentum space in the same way as position operators and eigenfunctions behave in position space.
That means that you can find the expectation value of momentum using the integral of the possible outcomesptimes the probability density
p = ∞
−∞p| ˜φ(p)|2dp= ∞
−∞[φ(p)]˜ ∗p[φ(p)]dp.˜ (4.57) You can also write the momentum expectation value using the inner product, with the momentum-space representation of the momentum operatorPpacting on the momentum-basis wavefunctionφ(p):˜
p =, φ˜ Ppφ˜
-= ∞
−∞[φ(p)]˜ ∗Pp[φ(p)]dp.˜ (4.58) In the notation Pp, the uppercase P wearing a hat tells you that this is the momentum operator, and the lowercasepin the subscript tells you that this is the momentum-basis version of the operator.
And just as in the case of the position operator, the action of the momentum operator is to multiply the function upon which it’s operating byp. Hence for the eigenfunctionφ˜1with eigenvaluep1
Ppφ˜1(p)=pφ˜1(p)=p1φ˜1(p). (4.59) For this equation to be true, the eigenfunctionφ˜1(p)must be zero everywhere except at the single locationp=p1. Thus the eigenfunctions of the momentum operatorPp in momentum space are an infinite set of Dirac delta functions δ(p−p), each with its own eigenvalue, and those eigenvalues (represented by p) cover the entire range of momenta.
Here are the important points: in any operator’s own space, the action of that operator on each of its eigenfunctions is to multiply that eigenfunction by the observable corresponding to the operator. And in the operator’s own space, those eigenfunctions are Dirac delta functions.
This explains the form and behavior of operators and their eigenfunctions in their own space. But it’s often useful to apply an operator to functions that reside in other spaces – for example, applying the momentum operatorPto position wavefunctionsψ(x).
Why might you want to do that? Perhaps you have the position-basis wavefunction and you wish to find the expectation value of momentum. You can do that using the position-space representation of the momentum operator Pxoperating on the position-basis wavefunctionψ(x)
p = ∞
−∞[ψ(x)]∗Px[ψ(x)]dx, (4.60) in which lowercasexin the subscript ofPxtells you that this is the position- basis version of the momentum operator P. This equation is the position- space equivalent to the momentum-space relation for the expectation value of pshown inEq. 4.58.
So what is the form of the momentum operatorPin position space? One way to discover that is to begin with the eigenfunctions of that operator.
Since you know that in momentum space the eigenfunctions of the momentum operator are the Dirac delta functionsδ(p−p), you can use the inverse Fourier transform to find the position-space momentum eigenfunctions:
ψ(x)= 1
√2πh¯ ∞
−∞φ(p)e˜ iph¯x
dp= 1
√2πh¯ ∞
−∞δ(p−p)eiph¯x
dp
= 1
√2πh¯ei
p
¯ hx
,
in which p is the continuous variable representing all possible values of momentum. Naming that variable p instead of p makes the position repre- sentation of the momentum eigenfunctions
ψp(x)= 1
√2πh¯ei
p
¯ hx
, (4.61)
in which the subscript “p” is a reminder that these are the momentum eigenfunctions represented in the position basis. You can use this position- space representation of the momentum eigenfunctions to find the position- space representationPx of the momentum operatorP. To do that, remember that the action of the momentum operator on its eigenfunctions is to multiply those eigenfunctions byp:
Pxψp(x)=pψp(x). (4.62)
Plugging in the position-space representations of the momentum eigenfunc- tions forψp(x)makes this
Px
1
√2πh¯eip¯hx
=p 1
√2πh¯eiph¯x
. (4.63)
Thepthat the operator must pull out of the eigenfunction is in the exponential, which suggests that a spatial derivative may be useful:
∂
∂x 1
√2πh¯eiph¯x
=ip
¯ h
1
√2πh¯eiph¯x
.
So the spatial derivative does bring out a factor ofp, but two constants come along with it. You can deal with those by multiplying both sides byh¯i:
¯ h i
∂
∂x 1
√2πh¯eiph¯x
= ¯h i
ip
¯ h
√1 2πh¯eiph¯x
=p 1
√2πh¯eiph¯x
exactly as needed. So position-space representation of the momentum opera- torPis
Px= ¯h i
∂
∂x = −ih¯ ∂
∂x. (4.64)
This is the form of the momentum operatorPin position space, and you can usePxto operate on position-basis wavefunctionsψ(x).
The same approach can be used to determine the form of the position operatorXand its eigenfunctions in momentum space. This leads to position eigenfunctions in momentum space:
φ˜x(p)= 1
√2πh¯e−i
p
¯ hx
(4.65) and the momentum-space representationXpof the position operatorX:
Xp=ih¯ ∂
∂p. (4.66)
If you need help getting to these expressions, check out the problems at the end of this chapter and the online solutions.
Given these position-basis representations of the position and momentum operators, you can determine an important quantity in quantum mechanics.
That quantity is the commutator [ X,P]:
[ X,P]=X P−P X=x(−ih)¯ d
dx −(−ih)¯ d
dxx. (4.67)
Trying to analyze this expression in this form leads many students astray.
To correctly determine the commutator, you shouldalwaysprovide a function
on which the operators can operate, like this:
[X,P]ψ=( XP−PX)ψ =
x(−ih)¯ d
dx−(−ih)¯ d dxx
ψ
=x(−ih)¯ dψ
dx −(−ih)¯ d(xψ) dx .
You can see the reason for inserting the functionψin the last term – it reminds you that the spatial derivatived/dxmust be applied not only tox, but to the productxψ:
[ X,P]ψ=x(−ih)¯ dψ
dx −(−ih)¯ d(xψ)
dx =(−ih)x¯ dψ
dx −(−ih)¯ d(x)
dx ψ−(−ih)¯ dψ dxx
=(−ih)x¯ dψ
dx −(−ih)(1)ψ¯ −(−ih)¯ dψ dxx
=ihψ¯ .
Now that the wavefunctionψ has done its job of helping you take all the required derivatives, you can remove it and write the commutator of the position and momentum operators as
[ X,P]=ih.¯ (4.68)
Using the momentum-space representation of the operatorsX andPleads to the same result, as you can see in the chapter-end problems and online solutions.
This nonzero value of the commutator [ X,P] (called the “canonical commu- tation relation”) has extremely important implications, since it shows that the order in which certain operators are applied matters. Operators such asXandP are “non-commuting,” which means they don’t share the same eigenfunctions.
Remember that the process of making a position measurement of a quantum observable for a particle or system in a given state causes the wavefunction to collapse to an eigenfunction of the position operator. But since the position and momentum operators don’t commute, that position eigenfunction is not an eigenfunction of momentum. So if you then make a momentum measurement, the wavefunction (not being in a momentum eigenfunction) collapses to a momentum eigenfunction. That means the system is now in a different state, so your position measurement is no longer relevant. This is the essence of quantum indeterminacy.
In the next chapter, the quantum wavefunctions for three specific potentials are derived and explored. Before getting to that, here are some problems to help you apply the concepts discussed in this chapter.