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.
in which φ (k) is a function of wavenumber (k) called the wavenumber spectrum.
If you already know the wavenumber spectrum φ (k) and you want to determine the corresponding position functionψ(x), the tool you need is the inverse Fourier transform
ψ(x)= 1
√2π ∞
−∞φ (k)eikxdk. (4.15) Fourier theory (both analysis and synthesis) is rooted in one idea: any well- behaved6function can be expressed as a weighted combination of sinusoidal functions. In the case of a function of position such asψ(x), the constituent sinusoidal functions are of the form coskxand sinkx, withkrepresenting the wavenumber of each component (recall that wavenumber is sometimes called
“spatial frequency” and has dimensions of angle per unit length, with SI units of radians per meter).
To understand the meaning of the Fourier transform, imagine that you have a function of positionψ(x)and you want to know “how much” of each constituent cosine and sine function is present inψ(x)for each wavenumber k. WhatEq. 4.14is telling you is this: to find those amounts, multiplyψ(x)by e−ikx(which is equivalent to coskx−isinkxby Euler’s relation) and integrate the resulting product over all space. The result of that process is the complex functionφ (k). Ifψ(x)is real, then the real part ofφ (k)tells you the amount of coskxinψ(x)and the imaginary part ofφ (k)tells you the amount of sinkx present inψ(x)for each value ofk.
Why does this process of multiplying and integrating tell you the amount of each sinusoid in the functionψ(x)? There are several ways to picture this, but some students find that the simplest visualization comes from using the Euler relation to write the Fourier transform as
φ (k)= 1
√2π ∞
−∞ψ(x)e−ikxdx
= 1
√2π ∞
−∞ψ(x)cos(kx)dx−i 1
√2π ∞
−∞ψ(x)sin(kx)dx. (4.16) Now imagine the case in which the functionψ(x)is a cosine function with single wavenumberk1, soψ(x)=cos(k1x). Inserting this intoEq. 4.16gives
φ (k)= 1
√2π ∞
−∞cos(k1x)cos(kx)dx−i 1
√2π ∞
−∞cos(k1x)sin(kx)dx.
6In this context, “well-behaved” means that the function satisfies theDirichlet conditionsof finite number of extrema and finite number of non-infinite discontinuities.
x
+ +
+ +
- -
-
Multiply
-
Multiply Multiply
Multiply
Multiplication results:
+
+ +
+
Integration over x produces large result
x
x ψ(x) = cos(k x)1
Real part of e–ikx for k=k1
Figure 4.5 Multiplying and integrating cosk1xand the real portion ofe−ikxwhen k=k1.
The next step in many explanations is to invoke the “orthogonality relations,” which say that the first integral is nonzero only when k = k1
(since cosine waves with different spatial frequencies are orthogonal to one another when integrated over all space), while the second integral is zero for all values ofk(since the sine and cosine functions are also orthogonal when integrated over all space). But if you’re not clear on why that’s true, take a look atFig. 4.5, which provides more detail about the orthogonality of sinusoidal functions described inSection 1.5.
The top graph in this figure shows the single-wavenumber wavefunction ψ(x)=cos(k1x)and the center graph shows the real portion of the function e−ikx (which is coskx) for the case in which k = k1. The vertical arrows indicate the point-by-point multiplication of these two functions, and the bottom portion of the figure shows the result of that multiplication. As you can see, since all of the positive and negative portions of ψ(x) align with the portions of the real part of e−ikx with the same sign, the results of the multiplication process are all positive (albeit with varying amplitude due to the oscillations of both functions, as shown in the bottom graph). Integrating the multiplication product overxis equivalent to finding the area under this
curve, and that area will have a large value when the product always has the same sign. In fact, the area under the curve is infinite if the integration extends from−∞to+∞and if the products are all positive, which means that k is precisely equal to k1. But even the slightest difference between k andk1 will cause the two functions to go from in-phase to out-of-phase and back to in-phase at a rate determined by the difference between kand k1, which will cause the product of coskx and cosk1xto oscillate between positive and negative values. That means that the result of integration over all space approaches infinity for k = k1 and zero for all other values ofk, resulting in a function that’s infinitely tall but infinitely narrow. That function is the Dirac delta function, about which you can read more later in this section.
Soφ (k), the Fourier transform of the constant-amplitude (and thus infinitely wide) wavefunction ψ(x) = cosk1x has an infinitely large real value at wavenumber k1. What about the imaginary portion of φ (k)? Sinceψ(x)is a pure (real) cosine wave, you can probably guess that no sine function is included inψ(x), even at the wavenumberk1. That’s exactly what the Fourier transform produces, as shown inFig. 4.6.
+ +
+ +
- -
Multiply Multiply Multiply
Multiplication results:
+ +
+
Integration over x produces small result
x
x ψ(x) = cos(k x)1
Imaginary part of e–ikx for k=k1
- - - -
x-
+ +
+
-
+ +
- - - -
Figure 4.6 Multiplying and integrating cosk1xand the imaginary portion ofe−ikx whenk=k1.
Notice that even though the oscillations ofψ(x)and the imaginary portion ofe−ikx (which is−sinkx) have the same spatial frequency in this case, the phase offset between these two functions makes their product equally positive and negative. Hence integrating over xproduces a small result (zero if the integration is done over an integer number of cycles, as explained later in this section). So φ (k) will have small or zero imaginary portion, even at wavenumberk=k1.
Sinceψ(x)is a pure cosine wave in this example, will the Fourier transform produce precisely zero for the imaginary portion of φ (k)? It will if you integrate over an integer number of cycles ofψ(x), because in that case the result of the multiplication will have exactly as much positive as negative contribution toφ (k)(that is, the area under the curve will be exactly zero).
But if you integrate over, say, 1.25 cycles ofψ(x), there will be some residual negative area under the curve, so the result of the integration will not be exactly zero. Note, however, that you can make the ratio of the imaginary part to the real part arbitrarily small by integrating over the entire x-axis, since that will cause the real portion ofφ (k)(with its all-positive multiplication results) to greatly exceed any unbalanced positive or negative area in the imaginary portion. That’s why the limits of integration on Fourier orthogonality relations are−∞to∞in the general case, or−T/2 toT/2 for periodic functions (where Tis the period of the function being analyzed).
So the multiply-and-integrate process of the Fourier transform produces the expected result when the wavenumberk of thee−ikx factor matches the wavenumber of one of the components of the function being transformed (k1
in this case). But what happens at other values of the wavenumberk? Why does this process lead to small values ofφ (k)for wavenumbers that are not present inψ(x)?
To understand the answer to that question, consider the multiplication results shown in Fig. 4.7. In this case, the wavenumberkin the multiplying factore−ikxis taken to be half the value of the single wavenumber (k1) inψ(x).
As you can see in the figure, in this case each spatial oscillation of the real portion of thee−i(12)k1xfactor occurs over twice the distance of each oscillation ofψ(x). That means that the product of these two functions alternates between positive and negative, making the area under the resulting curve tend toward zero. The changing amplitudes ofψ(x)ande−i(12)k1x cause the amplitude of their product to vary overx, but the symmetry of the waveforms ensures the equality of the positive and negative areas over any integer number of cycles.
A similar analysis shows that the Fourier transform also produces a small result when the wavenumber in the multiplying factor e−ikx is taken to be
- - x
+
- +
-
-
Multiply
Multiply
- +
+ +
x+ +
- -
- -
Multiply
Multiply Multiply
Multiplication results:
+
+ +
+
Integration over x produces small result
x
x ψ(x) = cos(k x)1
Real part of e–ikx for k= (½)k1
Figure 4.7 Multiplying and integrating cosk1xand the imaginary portion ofe−ikx whenk=12k1.
larger than the value of the single wavenumber (k1) inψ(x).Fig. 4.8shows what happens for the case in whichk = 2k1, so each spatial oscillation of thee−i(2)k1x factor occurs over half the distance of each oscillation of ψ(x).
As in the previous case, the product of these two functions alternates between positive and negative, and once again the area under the resulting curve tends toward zero (and is precisely zero over any integer number of cycles ofψ(x)).
You should make sure you understand that in this example, the imaginary portion ofφ (k)is zero becauseψ(x)is a pure cosine wave, not becauseψ(x) is real. Ifψ(x)had been a pure (real) sine wave, the resultφ (k)of the Fourier transform process would have been purely imaginary, because in that case only one sine and no cosine components are needed to make upψ(x). In general, the result of the Fourier transform is complex, whether the function you’re transforming is purely real, purely imaginary, or complex.
Multiplying the function being transformed by the real and imaginary parts ofe−ikx is one way to understand the process of Fourier transformation, but there’s another way that has some benefits due to the complex nature ofψ(x), φ (k), and the transformation process. That alternative approach is to represent the components ofψ(x)and the multiplying factore−ikxas phasors.
- -
+ +
- -
Multiply
Multiply
+
x- + -
+
-
+
+ +
-
+
- - - + +
- - - - -
+ +
+ + -
- -
Multiply
Multiply Multiply
Multiplication results:
+
+ +
+
Integration over x produces small result
x
x ψ(x) = cos(k x)1
Real part of e–ikx for k= 2k1
-
Figure 4.8 Multiplying and integrating cosk1xand the imaginary portion ofe−ikx whenk=2k1.
If it’s been a while since you’ve seen phasors, and even if you never really understood them, don’t worry. Phasors are a very convenient way to represent the sinusoidal functions at the heart of the Fourier transform, and the next few paragraphs will provide a quick review of phasor basics if you need it.
Phasors dwell in the complex plane described inSection 1.4. Recall that the complex plane is a two-dimensional space defined by a “real” axis (usually horizontal) and a perpendicular “imaginary” axis (usually vertical). A phasor is a type of vector in that plane, typically drawn with its base at the origin and its tip on the unit circle, which is the locus of points at a distance of one unit from the origin.
The reason that phasors are helpful in representing sinusoidal functions is this: by making the angle of the phasor from the positive real axis equal to kx(proceeding counterclockwise as kxincreases), the functions cos(kx)and sin(kx)are traced out by projecting the phasor onto the real and imaginary axes. You can see this inFig. 4.9, in which the rotating phasor is shown at eight randomly selected values of the anglekx. The phasor rotates continuously as xincreases, with the rate of rotation determined by the wavenumberk. Since
Real axis Imaginary axis
kx
kx
kx Phasor
cos(kx)
sin(kx) Imaginary part of rotating phasor (projection onto imaginary axis) produces sine wave
Real part of rotating phasor (projection onto real axis) produces cosine wave Complex
Plane
Unit circle
Angle with real axis is kx, so phasor rotates as x increases, representing eikx= cos(kx)+ i sin (kx)
Figure 4.9 Rotating phase relation to sine and cosine functions.
k= 2πλ, an increase in the value ofxof one wavelength causeskxto increase by 2πradians, which means the phasor will make one complete revolution.
To use phasors to understand the Fourier transform, imagine that the functionψ(x)under analysis has a single, complex wavenumber component represented byeik1x. The phasor representingψ(x)is shown inFig. 4.10aat 10 evenly spaced values ofk1x, which means that the value ofxis increasing by10λ (angle ofπ5 radians or 36◦) between each position of the phasor.
Now look atFig. 4.10b, which shows the phasor representing the Fourier- transform multiplying factore−ikx for the case in whichk = k1. The phasor representing this function rotates at the same rate as the phasor representing ψ(x), but the negative sign in the exponent of this function means that its phasor rotates clockwise. To appreciate what this means for the result of the Fourier-transform process, it’s important to understand the effect of multiplying two phasors (such asψ(x)timese−ikx).
Multiplying two phasors produces another phasor, and the amplitude of that new phasor is equal to the product of the amplitudes of the two phasors (both equal to one if their tips lie on the unit circle). More importantly for this application, the direction of the new phasor is equal to the sum of the angles of the two phasors being multiplied. And since the two phasors in this case
ψ(x)=eik x e–ikx for k=k1
Product ψ(x)e–ikx
Real Imag
Angle =k x1
Imag
Real
Imag
Real
Rotation
Rotation kx
(a) (b) (c)
Figure 4.10 Phasor representation of (a) the functioneik1x, the multiplying factor e−ikxfork=k1, and (c) the product.
are rotating in opposite directions at the same rate, the sum of their angles is constant. To see why that’s true, begin by defining the direction of the real axis to represent 0◦. The positions shown for theψ(x)phasor are then 36◦, 72◦, 108◦, and so on, while the positions shown for the clockwise-rotating phasor representinge−ikx are −36◦,−72◦,−108◦. That makes the sum of the two phasors’ angles equal to zero for all values ofx.7 Hence the multiplications performed at the ten phasor angles shown inFig. 4.10a all result in a unit- length phasor pointing along the real axis, as shown inFig. 4.10c.
Why is it significant that the phasor resulting from the multiplication of the phasors representingψ(x)ande−ikx has constant direction? Because the Fourier-transform process integrates the result of that multiplication over all values ofx:
φ (k)= 1
√2π ∞
−∞ψ(x)e−ikxdx, (4.14) which is equivalent to continuously adding the resulting phasors. Since those phasors all point in the same direction when k = k1 (that is, when ψ(x) contains a wavenumber component that matches the wavenumber in the multiplying functione−ikx), that addition yields a large number (phasors follow the rules of vector addition, so the largest possible sum occurs when they all point in the same direction). That large number becomes infinite when the
7If you prefer to use positive angles, the clockwise-rotating phasor’s angles count down from 360◦to 324◦, 288◦, 252◦, and so on, which sum with the angles of theψ(x)phasor to give a constant value of 360◦, the same direction as 0◦.
ψ(x)=eik x e–ikx for k= 2k1
Product ψ(x)e–ikx
Real Imag
Angle =k x1
Imag
Real
Imag
Real
Rotation
Rotation kx
(a) (b) (c)
Figure 4.11 Phasor representation of (a) the functioneik1x, (b) the multiplying factore−ikxfork=2k1, and (c) their product.
value ofkprecisely equals the value ofk1and the integration extends from
−∞to+∞.
The situation is quite different when a wavenumber contained withinψ(x) (such ask1) does not match the wavenumber of the multiplying functione−ikx).
An example of that is shown inFig. 4.11, for which thekine−ikxis twice the value of the wavenumberk1 present inψ(x). As you can see inFig. 4.11b, the larger wavenumber causes the phasor representing this function to rotate at a higher angular rate. That rate is twice as large in this case, so this phasor advances 72◦as theψ(x)phasor advances 36◦, and it completes two cycles as theψ(x)phasor shown inFig. 4.11acompletes one.
The important consequence of these different angular rates is that the angle of the phasor produced by multiplyingψ(x) bye−ikx is not constant. With both phasors starting at 0◦whenx=0, after one increment theψ(x)phasor’s angle is 36◦, while thee−ikx phasor’s angle is−72◦, so their product phasor’s angle is 36◦+(−72◦)= −36◦. Asxincreases one more of these increments, theψ(x)phasor’s angle becomes 72◦, while thee−ikxphasor’s angle becomes
−144◦, making their product phasor’s angle 72◦+(−144◦)= −72◦. When the increase inxhas caused theψ(x)phasor to complete one revolution (and thee−ikx phasor to complete two revolutions), their product phasor will also have completed one clockwise cycle, as shown inFig. 4.11c.
The changing angles of those phasors means that their sum will tend toward zero, as you can determine by lining them up in the head-to-tail configuration of vector addition (since they will form a loop and end up back at the starting point). And the result of integrating the product of ψ(x) ande−ikx will be
ψ(x) =eik x e–ikx for k= (¹ k1
Product ψ(x)e–ikx
Real Imag
Angle =k x1
Imag
Real
Imag
Real
Rotation
Rotation kx
(a) (b) (c)
Figure 4.12 Phasor representation of (a) the functioneik1x, (b) the multiplying factore−ikxfork= 12k1, and (c) their product.
exactly zero if the integration is performed over a sufficiently large range of xto cause the product phasor to complete an integer number of cycles.
As you may have guessed, a similar analysis applies when the wavenumber component in the functione−ikx is smaller than the wavenumberk1inψ(x).
The case for whichk = 12k1 is shown inFig. 4.12; you can see the phasor representinge−ikx taking smaller steps (18◦in this case) in Fig. 4.12b. And sincekdoes not matchk1, the direction of the product phasor is not constant;
in this case it rotates counter-clockwise. But as long as the product phasor completes an integer number of cycles in either direction, the integration of the products will give zero.
You can use this same type of phasor analysis for functions that cannot be represented by a single rotating phasor with constant amplitude. For example, consider the wavefunctionψ(x) = cos(k1x)discussed earlier in the section.
Since this is a purely real function, a single rotating phasor won’t do. But recall the “inverse Euler” relation for cosine:
cos(k1x)= eik1x+e−ik1x
2 . (4.17)
This means that the function ψ(x) = cos(k1x)may be represented by two counter-rotating phasors with amplitude of12, as shown inFig. 4.13a. As these two phasors rotate in opposite directions, their sum (not their product) lies entirely along the real axis (since their imaginary components have opposite signs, and cancel). Over each complete rotation of the two component phasors, the amplitude of the resultant phasor varies from +1, through 0 (when they