The bottom line is this: applying an operator to the quantum state of a system changes that state by multiplying each constituent eigenfunction by its eigenvalue (Eq. 4.6), and making a measurement changes the quantum state by causing the wavefunction to collapse to one of those eigenfunctions.
So operators and measurements both change the quantum state of a system, but not in the same way.
You can see examples of quantum operators in action later in this chapter and inChapter 5, but before getting to that, you may find it helpful to consider the general characteristics of quantum wavefunctions, which is the subject of the next section.
coordinate, and the second-order spatial derivative wouldn’t exist ifψ(x)or
∂ψ(x)
∂x were not continuous. An exception to this occurs in the case of infinite potential, which you can read about in the discussion of the infinite potential well inChapter 5.
Square-integrable Quantum wavefunctions must be normalizable, which means that the integral of the wavefunction’s squared magnitude cannot be infinitely large. For most functions, that means that the function itself must be finite everywhere, but you should be aware that theDirac delta function is an exception. Although the Dirac delta function is defined to have infinite height, its infinitely narrow width keeps the area under its curve finite3. Note also that some functions such as the plane-wave functionAei(kx−ωt) have infinite spatial extent and are not individually square-integrable, but it is possible to construct combinations of these functions that have limited spatial extent and meet the requirement of square-integrability.
In addition to meeting these requirements, quantum wavefunctions must also match the boundary conditions for a particular problem. As mentioned earlier, it’s necessary to know the specific potential V(x) in the region of interest in order to fully determine the relevant quantum wavefunction.
However, some important aspects of the wavefunction’s behavior may be discerned by considering the relationship of wavefunction curvature to the value of the total energyEand the potential energyVin the time-independent Schr¨odinger equation (TISE).
To understand that behavior, it may help to do a bit of rearranging of the TISE (Eq. 3.40):
d2[ψ(x)]
dx2 = −2m
¯
h2(E−V)ψ(x). (4.7) The term on the left side of this equation is just the spatial curvature, which is the change in the slope of a graph of the ψ(x) function vs. location.
According to this equation, that curvature is proportional to the wavefunction ψ(x)itself, and one of the factors of proportionality is the quantity E−V, the difference between the total energy and the potential energy at the location under consideration.
3Not technically a function due to its infinitely large value when its argument is zero, the Dirac delta function is actually a “generalized function” or “distribution,” which is the mathematical equivalent of a black box that produces a known output for a given input. In physics, the usefulness of the Dirac delta function is usually realized when it appears inside an integral, as you can see in the discussion of Fourier analysis inSection 4.4.
ψ(x)
x
If ψ(x)>0 and E–V>0, curvature is negative (slope becomes less positive or more neg- ative as x gets larger)
If ψ(x)<0 and E–V>0, curvature is positive (slope becomes less negative or more pos- itive as x gets larger)
¶
ψ(x)¶
x2
=
2–
2m2 (E–V)ψ(x)= (+)
(–) (+) (–)
¶
ψ(x)¶
x2
=
2–
2m2 (E–V)ψ(x)= (+)
(–) (–) (+)
Figure 4.1 Wavefunction curvature forE−V>0 case.
Now imagine a situation in which the total energy (E) exceeds the potential energy (V) everywhere in the region of interest (this doesn’t necessarily imply that the potential energy is constant, just that the total energy is greater than the potential energy at every location in this region). HenceE−V is positive, and the curvature has the opposite sign of the wavefunction (due to the minus sign on the right side ofEq. 4.7).
Why do the signs of the curvature and the wavefunction matter? Look at the behavior of the wavefunction as you move toward positivexin a region in which the wavefunctionψ(x)is positive (above the x-axis inFig. 4.1). Since E−Vis positive, the curvature ofψ(x)must be negative in this region (since the curvature and the wavefunction have the opposite sign ifE−Vis positive).
This means that the slope of the graph ofψ(x)becomes increasingly negative as you move to larger values of x, so the waveform must curve toward the x-axis, eventually crossing that axis. When that happens, the wavefunction ψ(x) becomes negative, and the curvature becomes positive. That means that the waveform again curves toward the x-axis, until it eventually crosses back into the positive-ψ region, where the curvature once again becomes negative.
So no matter how the potential energy function V(x)behaves, as long as the total energy exceeds the potential energy, the wavefunction ψ(x) will oscillate as a function of position. As you’ll see later in this chapter and in Chapter 5, the wavelength and amplitude of those oscillations are determined by the difference betweenEandV.
Now consider a region in which the total energy is less than the potential energy. That means thatE−V is negative, so the curvature has the same sign as the wavefunction.
If you’re new to quantum mechanics, the idea of the total energy being less than the potential energy may seem to be physically impossible. After all, if the total energy is the sum of potential plus kinetic energy, wouldn’t the kinetic energy have to be negative to make the total energy less than the potential energy? And how can the kinetic energy, which is 12mv2= 2mp2, be negative?
In classical physics, that line of reasoning is correct, which is why regions in which an object’s potential energy exceeds its total energy are called “classically forbidden” or “classically disallowed.” But in quantum mechanics, solving the Schr¨odinger equation in a region in which the potential energy exceeds the total energy leads to perfectly acceptable wavefunctions; as you’ll see later in the section, those wavefunctions decay exponentially with distance within those regions. And what happens if you measure the kinetic energy in one of these regions? The wavefunction will collapse to one of the eigenstates of the kinetic-energy operator, and the result of your measurement will be the eigenvalue of that eigenstate. Those eigenvalues are all positive, so you will definitely not measure a negative value for kinetic energy.
How is that result consistent with the potential energy exceeding the total energy in this region? The answer is hidden in the phrase “in this region.” Since position and momentum are incompatible observables, and kinetic energy depends on the square of momentum, the uncertainty principle says that you cannot simultaneously measure both position and kinetic energy with arbitrar- ily great precision. Specifically, the more precisely you measure kinetic energy, the larger the uncertainty in position. So when you measure kinetic energy and get a positive value, the possible positions of the quantum particle or system always include a region in which the total energy exceeds the potential energy.
With that understanding, take a look at the behavior of the wavefunction ψ(x) as you move toward positive x in a region in which ψ(x) is initially positive (above the x-axis) in Fig. 4.2. IfE −V is negative in this region, then the curvature must be positive, since the curvature and the wavefunction must have the same sign ifE−Vis negative andψ(x)is positive. This means that the slope of the graph ofψ(x)becomes increasingly positive as you move
ψ (x)
x
If ψ(x)>0 and E – V<0, curvature is positive (slope becomes more positive as x gets larger)
If ψ(x)<0 and E–V<0, curvature is negative (slope becomes more negative as x gets larger)
¶
ψ(x)¶
x2
=
2–
2m2 (E–V)ψ(x)= (–)
(–) (+) (+)
¶
ψ(x)¶
x2
=
2–
2m2 (E–V)ψ(x)= (–)
(–) (–) (–)
Initial slope positive
Initial slope slightly negative
Initial slope very negative
Figure 4.2 Wavefunction curvature forE−V<0 case.
to larger values ofx, which means that the waveform must curveawayfrom the x-axis. And if the slope ofψ(x)is positive (or zero) at the position shown, the wavefunction will eventually become infinitely large.
Now think about what happens if the slope of ψ(x) is negative at the position shown. That depends on exactly how negative that slope is. As shown in the figure, even if the slope is initially slightly negative, the positive curvature will cause the slope to become positive, and the graph ofψ(x)will turn upward before crossing the x-axis, which means the value ofψ(x)will eventually become infinitely large.
But if the slope at the position shown is sufficiently negative,ψ(x) will cross the x-axis and become negative. And whenψ(x)becomes negative, the curvature will also become negative, sinceE−V is negative. With negative curvature below the axis,ψ(x)will curve away from the x-axis, eventually becoming infinitely large in the negative direction.
So for each of the initial slopes shown in Fig. 4.2, the value of the wavefunction ψ(x) will eventually reach either +∞ or −∞. And since wavefunctions with infinitely large amplitude are not physically realizable, the slope of the wavefunction cannot have any of these values at the position
ψ (x)
x
Initial slope just right to cause ψ(x) to approach zero as x→
∞
Initial slope not negative enough to prevent ψ(x) from becoming infinite as x→
∞
Initial slope too negative to prevent ψ(x) from becoming infinite as x→
∞
Figure 4.3 Effect of initial slope onψ(x)forE−V<0 case.
shown. Instead, the curvature of the wavefunction must be such that the amplitude of ψ(x)remains finite at all locations so that the wavefunction is normalizable. This means that the integral of the square magnitude of ψ(x) must converge to a finite value, which means that the value ofψ(x)must tend toward zero asxapproaches±∞. For that to happen, the slope ∂ψ∂x must have just the right value to causeψ(x)to approach the x-axis asymptotically, never turning away from the axis, but also never crossing below the axis. In that case, ψ(x)will approach zero asxapproaches∞, as shown inFig. 4.3.
What conclusion can you reach about the behavior of the wavefunction ψ(x)in the regions in whichE−V is negative? Just this: oscillations are not possible in such regions, because the slope of the wavefunction at any position must have the value that will cause the wavefunction to decay toward zero asx approaches±∞.
So just by considering the Schr¨odinger equation’s relationship of wavefunc- tion curvature to the value ofE−V, you can determine thatψ(x)oscillates in regions in which the total energyEexceeds the potential energyVand decays in regions in whichEis less thanV. More details of that behavior can be found by solving the Schr¨odinger equation for specific potentials, as you can see in Chapter 5.
To get a better sense of how the wavefunction behaves, consider the cases in which the potentialV(x)is constant over the region of interest and the total energyEis either greater than or less than the potential energy.
First taking the case in whichE−V is positive, the TISE (Eq. 4.7) can be written as
d2[ψ(x)]
dx2 = −2m
¯
h2(E−V)ψ(x)= −k2ψ(x), (4.8) in which the constantkis given by
k=
2m
¯
h2(E−V). (4.9)
The general solution to this equation is
ψ(x)=Aeikx+Be−ikx, (4.10) in whichAandBare constants to be determined by the boundary conditions.4
Even without knowing those boundary conditions, you can see that quantum wavefunctions oscillate sinusoidally in regions in whichE is greater thanV (classically allowed regions), sincee±ikx=coskx±isinkxby Euler’s relation.
This fits with the curvature analysis presented earlier.
And here’s another conclusion you can draw from the form of the solution inEq. 4.10:krepresents the wavenumber in this region, which determines the wavelength of the quantum wavefunction through the relationk=2π/λ. The wavenumber determines how “fast” the wavefunction oscillates with distance (cycles per meter rather than cycles per second), andEq. 4.9tells you that large E−Vmeans largek, and large wavenumber means short wavelength. Thus the larger the difference between the total energyE and the potential energy V of a quantum particle, the greater the curvature and the faster the particle’s wavefunction will oscillate withx(higher number of cycles per meter).
Enforcing the boundary conditions of continuous ψ(x) and continuous slope (∂ψ(x)∂x ) at the boundary between two classically allowed regions with different potentials can help you understand the relative amplitude of the wavefunction in the two regions. To see how that works, useEq. 4.10to write the wavefunction and its first spatial derivative on both sides of the boundary.
Since taking that derivative brings out a factor ofk, this leads to the conclusion that the ratio of the amplitudes on opposite sides of the boundary is inversely
4Note that this is equivalent toA1cos(kx)+B1sin(kx)and toA2sin(kx+φ); you can see why that’s true as well as the relationship between the coefficients of these equivalent expressions in the chapter-end problems and online solutions.
proportional to the wavenumber ratio.5Thus the wavefunction on the side of the boundary with larger energy difference E−V (which means largerk) must have smaller amplitude than the wavefunction on the opposite side of the boundary between classically allowed regions.
Now consider the case in which the potential energy exceeds the total energy, soE−Vis negative. In that case, the TISE (Eq. 4.7) can be written as
d2[ψ(x)]
dx2 = −2m
¯
h2(E−V)ψ(x)= +κ2ψ(x), (4.11) in which the constantκis given by
κ =
2m
¯
h2(V−E). (4.12)
The general solution to this equation is
ψ(x)=Ceκx+De−κx, (4.13)
in whichCandDare constants to be determined by the boundary conditions.
If the region of interest is a classically forbidden region extending toward +∞(so thatxcan take on large positive values within the region), the first term ofEq. 4.13will become infinitely large unless the coefficientCis set to zero.
In this region,ψ(x)=De−κx, which decreases exponentially with increasing positivex.
Likewise, if the region of interest is a classically forbidden region extending toward−∞, the second term ofEq. 4.13will become infinitely large asxtakes on large negative values, so in that case the coefficientDmust be zero. That makesψ(x)=Ceκxin this region, and the wavefunction amplitude decreases exponentially with increasing negativex.
So once again, even without knowing the precise boundary conditions, you can conclude that quantum wavefunctions decay exponentially in regions in whichE is greater thanV (that is, in classically forbidden regions), again in accordance with the curvature analysis presented earlier.
Additional information can be gleaned fromEq. 4.13: the constantκ is a
“decay constant” that determines the rate at which the wavefunction tends toward zero. And sinceEq. 4.12states that κ is directly proportional to the square root of V −E, you know that the greater the amount by which the potential energyVexceeds the total energyE, the larger the decay constantκ, and the faster the wavefunction decays with increasingx.
5If you need help getting that result, check out the chapter-end problems and online solutions.
a b c
a
c b
V1
V2
V3 E V1
–
E0
V5
E–V3
V5
–
EV4 E–V4
Small V1–E means slow exponential decay here
Small E–V3 means long wavelength and large amplitude here
Large E–V4 means short wavelength and small amplitude here
Large V5–E means fast exponential decay here E=V2 means zero
curvature here V(x)
ψ(x)
Slopes must match at boundary points denoted by circles
x
x
1 2 3 4 5
Region
Figure 4.4 Stepped potential and wavefunction.
You can see all of these characteristics at work inFig. 4.4, which shows five regions over which the potential has different values (butV(x)is constant within each region). Such “piecewise constant” potentials are helpful for understanding the behavior of quantum wavefunctions, and they can also be useful for simulating continuously varying potential.
The potential V1 in region 1 (leftmost) and the potential V5 in region 5 (rightmost) are both greater than the particle’s energyE, albeit by different amounts. In regions 2, 3, and 4, the particle’s energy is greater than the potential, again by a different amount in each region.
In classically forbidden regions 1 and 5, the wavefunction decays exponen- tially, and sinceV−Eis greater in region 5 than in region 1, the decay over distance is faster in that region.
In classically allowed region 2, the total energy and the potential energy are equal, so the curvature is zero in that region. Note also that the slope of the
wavefunction is continuous across the boundary between classically forbidden region 1 and allowed region 2.
In the classically allowed regions 3 and 4, the wavefunction oscillates, and since the difference between the total and potential energy is smaller in region 3, the wavenumberkis smaller in that region, which means that the wavelength is longer and the amplitude is larger. The larger value of E −V in region 4 makes for shorter wavelength and smaller amplitude in that region.
At each boundary between two regions (marked by circles in Fig. 4.4), whether classically allowed or forbidden, both the wavefunctionψ(x)and the slope∂ψ∂x must be continuous (that is, the same on both sides of the boundary).
Another aspect of the potentials and wavefunction shown in Fig. 4.4is worth considering: for a particle with the total energyEshown in the figure, the probability of finding the particle decreases to zero asxapproaches±∞.
That means that the particle is in a bound state – that is, localized to certain region of space. Unlike such bound particles, free particles are able to “escape to infinity” in the sense that their wavefunctions are oscillatory over all space.
As you’ll see inChapter 5, particles in bound states have a discrete spectrum of allowed energies, while free particles have a continuous energy spectrum.