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6 The quantum wave equation
6.1 Waves and particle characteristics _{}^{}
Before studying modern physics, we think of particles and waves as beloning to fundamentally different categories of objects. Here's a summary of some of the ditterentiating characteristics between particles and waves.
Occupying space. Particles exist in a welldefined amound of space: This property means particles are "localized". But waves exist over an extended region of space. For example a harmonic functions exist over all valuse of 𝑥 from ¡Ä to +¡Ä. So a singlefrequency wave is inherently nonlocalized. These difference are illustrated in Fig. 6.1.
Traveling through openings. As shown in Fig. 6.2, a partcle passes through an opening larger than the partilce and is not affected, but if the aperture is smaller than particle size it can't pass through. Waves, however, behave quite differently. If the aperture is much larger than wavelength, it passes through and is unaffected. But if we reduce the opening so that the aperture is similar to the wavelength of the wave, the portion of the wave that passes through it is no longer plane wave, The surfaces of constant phase are somewhat curved and the wave spreads out after passing through the opening. And if we make the aperture smaller than the wavelength, the wavefront curvature abd spreading out of the wave becomes even greater. This effect is called "diffraction".
Interacting with other particles and waves. Particles interact with other particles vis collision and when particles collide they can exchange momentum and energy. When particles collide, they quickly exchange a discrete amount of energy, and momentum and according to conversation laws the total momntum and energy of the system is the same before and after. Wave, however, interact with other waves vis superposition. When two or more waves occupy the small region of space, the resuting wave is the sum of all contributing waves. The waves are said to "interfere" with another. Wave can also interact with objects and in such interactions some of the wave's energy can be transferred and cocurs over an extended period of time.
A summary of the classicalmechnics model of some distiction between particles and waves is given in the diagram in Fig. 6.3.
6.2 Waveparticle duality _{}^{}
To understand waveduality, it's helpful to consider the type of experiment we could perform on a particle to test for wave behavior. However, in order to detect a particle's wave behavoir, the aperture must be comparable to the wavelength of the wave. That means that it's necessary to predict the wavelength from the properties of the particle.
Due to the work of Louis de Broglie who hypeothesized the existence of matter waves in 1924 the prediction is possible. Almost two decades before Albert Einstein had shown that light exhibits characteristics of both waves and particles, behveing like waves when passing through slits, but acting like particle (called photon) during interactions such as the photoelectric effect. De Broglie made the inspired guess that the waveparticle duality extends to electrons and other particles with mass. As he wrote in his PhD thesis: "My essential idea was to extend to all particles the coexistence of waves and particles discovered by Einstein in 1905 in the case of light and photons."
To understand de Broglie's expression for the wavelength of a particle, consider Einstein's relationship for the energy of a photon with frequency 𝜈 and wavelength 𝜆:
(6.1) 𝛦 = 𝘩𝜈 = 𝘩𝑐/𝜆,
where 𝘩 is Planck's constant (6.626 ¡¿ 10^{24} m^{2} kg/s) and 𝑐 is the speed of light. Although photons have no rest mass, they do carry momentum, and the magnitude of the momentum is related to the photon's energy by
(6.2) 𝑝 = 𝛦/𝑐. This gives
𝛦 = 𝘩𝑐/𝜆 = 𝑐𝑝, 𝑝 = 𝘩/𝜆, 𝜆 = 𝘩/𝑝.
If this relationship holds also for particles with mass, then since 𝑝 = 𝑚𝑣, we have
(6.3) 𝜆 = 𝘩/𝑝 = 𝘩/𝑚𝑣.
Consider the implication of Eq (6.1). This is why only moving particles exhibit wave behavior. Also because Planck's constant is such a small number, the wave characteristic of everyday objects are not appeared since their high mass makes de Boglie wavelength too small to measure.
Example 6.1 What is the de Broglie wavelength of a 75kg human walking at speed of 1.5 m/s?
The human has momenum 𝑝 = 𝑚𝑣 = 75 kg ¡¿ 1.5 m/s = 113 kg m/s, so the de Broglie wavelength is
(6.4) 𝜆 = 𝘩/𝑝 = (6.626 ¡¿ 10^{34} J/s)/(113 kg m/s) = 5.9 ¡¿ 10^{36}.
That is billions times smaller than the protons and neutrons. Hence a human body does not demonstrate the wave behavior of matter.
Example 6.2 What is the de Broglie wavelength of an electron that has passes through a potential difference of 50 volts?
After passing through a potential difference of 50 volts, an electron has 50 eV of energy. 1 eV = 1.6 ¡¿10^{19} J, so
(6.5) (50 eV)(1.6 ¡¿10^{19} J/1 eV) = 8 ¡¿10^{18} J.
We can use the classical expression for kinetic energy 𝐾𝐸:
(6.6) 𝐾𝐸 = (1/2)𝑚𝑣^{2}
(6.7) 𝐾𝐸 = (1/2)𝑚𝑣^{2} = (1/2𝑚)𝑚^{2}𝑣^{2} = 𝑝^{2}/2𝑚.
In this case, the electron's energy 𝐸 is all kinetic, so 𝐸 = 𝐾𝐸:
(6.8) 𝑝 = ¡î(2𝑚𝐸).
𝑝 = ¡î(2𝑚𝐸) = ¡î(2 ¡¿ 9.1 ¡¿10^{31} kg ¡¿ 8 ¡¿10^{18} J) = 3.8 ¡¿10^{24} kg m/s. So the de Broglie wavelength is
(6.9) 𝜆 = 𝘩/𝑝 = (6.626 ¡¿ 10^{34} J/s)/(3.8 ¡¿10^{24} kg m/s) = 1.7 ¡¿ 10^{10} m.
This is similar to the spacing between atoms in a crystal array, such an array can be used to experimentally determine a moving electron's wavelenghth.
That experiment was done by Clinton Favisson and Lester Germer in 1927. Knowing that waves scattered from crystals produce a dffraction pattern, they bombarded a nickel crystal with electrons and looked for evidence of diffraction. They found it, and then they used the scattering angle and knowing spacing of the atoms in the crystal to calculate the wavelength of the electrons. Their results were in good agreement with de Broglie's expression for the wavelength of particles with the mass and velocity of the electrons used in the experiment.
A similar, but conceptually more straightfoward experiment in double slit, which is often used to ddemonstrate the wave properties of light. as in Fig. 6.4 with a source of waves or particles facing a barrier with two small slits a short distance apart: far behind the barrier is a detector.
First consider when the source emits a continous stream of wave. According to Huygens's principle, every point on a wavefront can be considered to be the source of another wave that spreads out spherically from that point. When the wave encounters the barrier, most of the secondary waves are blocked; only two small portions of wavefront pass throgh the slits. The spherically expanding secondary waves from thos two portions of the wavefront then travel to the detector, where they superimpose to produce the resultant waveform. But the waves from the two slits arrive at the detector with different phase. Depending on the the amount of that phase difference, the two waves may add or substract and the interference may be constructive or destructive. At points of constructive interference, a bright fringe appears; and at points of destructive interference, a dark fringe appears.
Now imagine that the source fires a stream of particles such as electrons at the barrier. Since streaming electrons also have characteristic of waves, the detector registers a continous accumulation of energy in an interference pattern, as shown in the right portion in Fig. 6.5.
Finally, consider when individual electrons are sent through the slits one at a time. The accumulation of the points creates the dark and light fringes of an interference pattern. Evidently each elctron interfere with itself as it pass through both slits.
The modern understanding of waveparticle duality is shown in Fig. 6.6. When a quantummechanical object is traveling, it acts like a wave, and it bends around corners, diffracts through small openings, and interferes. Wnen a quantummechanical object is depositing or accepting energy, it acts like a particle. And a quantummechanical object behaves like a traveling wave packet that occupies a small, but finite, amount of space.
6.3 The Schrödinger equation _{}^{}
Understanding that electrons can behave like waves comes from the interpretation of wavefunction, 𝜓(𝑥, 𝑡), the solution to the Schrödinger equation which governs these matter waves. Although Richard Feynmann once said that it's not possible to derive the Schrödinger equation from anything we know, we can get a sence of Schrödinger's reasoning by starting with an expression for the energy of a particle. If that particles is moving nonrelativistically (slowly relative to the speed of the light 𝑐), and the forces acting on the particle are conservative, the total mechanical energy 𝛦 of that particle may be written as
(6.10) 𝛦 = 𝐾𝛦 + 𝑉 = (1/2)𝑚𝑣^{2} + 𝑉 = 𝑝^{2}/2𝑚 + 𝑉,
where 𝐾𝛦 is the kinetic energy and 𝑉 is the potential energy.
Schrödinger refered to Einstein's relationship of energy of a photon 𝛦 = 𝘩𝜈 Eq. (6.1), and wrote a version this equation for energy associated with matter waves as follows:
(6.11) 𝛦 = 𝘩𝜈 = (𝘩/2¥ð)(2¥ð 𝜈) = ℏ𝜔,
where "reduced Planck constant" ℏ = 𝘩/2¥ð, frequency 𝜆 = 𝜈, and the angular frequency 𝜔 = 2¥ð 𝜈. So we have
(6.12) ℏ𝜔 = (1/2)𝑚𝑣^{2} + 𝑉.
De Broglie's equation Eq. (6.3) relating momentum to wavelength can be expressed
(6.13) 𝑝 = 𝘩/𝜆 = 𝘩/(2¥ð/𝑘) = (𝘩/2¥ð)𝑘 = ℏ𝑘
So we can write Eq. (6.12) as
(6.14) ℏ𝜔 = ℏ^{2}𝑘^{2}/2𝑚 + 𝑉.
To get from this equation to the Schrödinger equation, assumme the matter wave associated eith the particle can be written as the harmonistic wavefunction 𝜓(𝑥, 𝑡):
(6.15) 𝜓(𝑥, 𝑡) = 𝛢𝑒^{𝑖(𝑘𝑥𝜔𝑡)} = 𝑒^{𝑖𝜔𝑡}(𝛢𝑒^{𝑖𝑘𝑥}).
Notice that we've divided the expontial term the timedependent term from the spacedependent term. Then we can get the first and and second derivatives with respect to tme 𝑡 or space 𝑥:
(6.16) ¡Ó𝜓/¡Ó𝑡 = [¡Ó(𝑒^{𝑖𝜔𝑡})/¡Ó𝑡](𝛢𝑒^{𝑖𝑘𝑥}) = 𝑖𝜔(𝑒^{𝑖𝜔𝑡})(𝛢𝑒^{𝑖𝑘𝑥}) = 𝑖𝜔𝜓.
(6.17) ¡Ó𝜓/¡Ó𝑥 = 𝑒^{𝑖𝜔𝑡}¡Ó(𝛢𝑒^{𝑖𝑘𝑥})/¡Ó𝑥 = 𝑖𝑘𝑒^{𝑖𝜔𝑡}(𝛢𝑒^{𝑖𝑘𝑥}) = 𝑖𝑘𝜓.
(6.18) ¡Ó^{2}𝜓/¡Ó𝑥^{2} = 𝑒^{𝑖𝜔𝑡}¡Ó^{2}(𝛢𝑒^{𝑖𝑘𝑥})/¡Ó𝑥^{2} = (𝑖𝑘)^{2}(𝛢𝑒^{𝑖𝑘𝑥}) = 𝑘^{2}𝜓.
Before fitting these derivatives into Eq. (6.14) we are to multiply Eq. (6.16) by 𝑖ℏ at Eq. (6.16)
𝑖ℏ(¡Ó𝜓/¡Ó𝑡) = 𝑖ℏ(𝑖𝜔𝜓) = ℏ𝜔𝜓 or ℏ𝜔 = (𝑖ℏ/𝜓)(¡Ó𝜓/¡Ó𝑡).
Now we can plug this into Eq. (6.14):
(𝑖ℏ/𝜓)(¡Ó𝜓/¡Ó𝑡) = ℏ^{2}𝑘^{2}/2𝑚 + 𝑉 or 𝑖ℏ(¡Ó𝜓/¡Ó𝑡) = (ℏ^{2}𝑘^{2}/2𝑚)𝜓 + 𝑉𝜓.
But we know that 𝑘^{2}𝜓 = ¡Ó^{2}𝜓/¡Ó𝑥^{2} from Eq. (6.18), so
(6.19) 𝑖ℏ(¡Ó𝜓/¡Ó𝑡) = (ℏ^{2}/2𝑚)(¡Ó^{2}𝜓/¡Ó𝑥^{2}) + 𝑉𝜓.
This is the onedimensional timedependent Schrödinger equation. Note that the presence of "𝑖" as a multilicative factor means the the solutions will generally complex. We can derive a "timeindepenent" version by using Eq. (6.16) to relace ¡Ó𝜓/¡Ó𝑡 by 𝑖𝜔𝜓:
(6.20) 𝑖ℏ(𝑖𝜔𝜓) = (ℏ^{2}/2𝑚)(¡Ó^{2}𝜓/¡Ó𝑥^{2}) + 𝑉𝜓 or ℏ𝜔𝜓 = (ℏ^{2}/2𝑚)(¡Ó^{2}𝜓/¡Ó𝑥^{2}) + 𝑉𝜓.
Since 𝛦 = ℏ𝜔 Eq. (6.11) we may also see this equation written as
(6.21) 𝛦𝜓 = (ℏ^{2}/2𝑚)(¡Ó^{2}𝜓/¡Ó𝑥^{2}) + 𝑉𝜓 or
(6.22) (𝛦 𝑉)𝜓 = (ℏ^{2}/2𝑚)(¡Ó^{2}𝜓/¡Ó𝑥^{2}).
We must note that the wavefunction 𝜓 is still 𝜓(𝑥, 𝑡) and it's essentially an expression of the conservation of enegy: 𝛦 𝑉 is the kinetic energy. So theequation tells us that the kinetic energy of the particle is proportinal to the second spatial derivative of 𝜓, which is the curvature of the wavefunction. Greater curvature with 𝑥 means that the waveform has a higher spatial frequency, which means it has shorter wavelength, and de Brogle's equation relates shorter wavelength to higher momentum.
Example 6.3 What is the timeindependent Schrödinger equation for a free particle?
In this context "free" means the particle is free from the influence of external force. Because a free particle travels in a region of constant potential energy and reference location of zero potential energy is arbitrary, we can set 𝑉 = 0 in Eq. (6.22). So
(6.23) 𝛦𝜓 = (ℏ^{2}/2𝑚)(¡Ó^{2}𝜓/¡Ó𝑥^{2}) or
(6.24) ¡Ó^{2}𝜓/¡Ó𝑥^{2}=  (2𝑚𝛦/ℏ^{2}) 𝜓.
Since the total energy of a free particle equals the particle's kinetic energy, we can set 𝛦 = 𝑝^{2}/2𝑚:
(6.25) ¡Ó^{2}𝜓/¡Ó𝑥^{2}=  (𝑝^{2}/ℏ^{2}) 𝜓.
This is the Schrödinger equation for a free particle. It's instructive to compare with Eq. (3.22) for standing wave, which is assumed to be the product of a function 𝑋(𝑥) that depends only on 𝑥 and another function 𝑇(𝑡) that depends only on 𝑡. The equation is
(3.22) ¡Ó^{2}𝑋/¡Ó𝑥^{2} = 𝛼𝑋 = 𝑘^{2},
which is identical to Eq. (6.25), provided that
𝑝^{2}/ℏ^{2} = 𝑘^{2}. so 𝑝^{2}/ℏ^{2} = (2¥ð/𝜆)^{2} or
𝑝/ℏ = 𝑝/𝘩/(2¥ð) = (2¥ð/𝜆) or 𝜆 = 𝘩/𝑝,
which is de Broglie's expression Eq. (6.3) for the wavelength of matter wave. For a free particles, oscillation frequency and energy may take on any value, but particles in the regions in which the potential energy varies may be constrained to certain values of frequency and energy that depend on the boundary conditions. Thus the allowed energies of a particle in a potential well can be quantized.
This means that the wave characteristic of a particle is analogous to a standing wave, and the energy of a particle is proportional to the oscillational frequency of that standing wave. We may see solutions to the timeindependent Schrödinger equation reffered to as"stationary states", but we should note that this means that the energy of the system is constant over time, not that the wavefunction is stationary.
6.4 Probability wavefunctions _{}^{}
At first Schrödinger thoght that the wavefunction 𝜓(𝑥, 𝑡) was the charge density of electron. However, MaxBorn was able to offer his own explanation, which is now the modern interpretation of wavefunction. It is a "prpbability amplitude" related to the probability of finding the particle in a given region of space. The quantuty is called an amplitude because just as we must square the amplitudeto get its energy in Eq. (4.22), so we must square the wavefunctin to get a probability density 𝓟. Since the wavefunction is generally complex, we square it by multiplying it by its complex conjugate:
(6.2627) 𝓟(𝑥, 𝑡) = 𝜓^{*}(𝑥, 𝑡)𝜓(𝑥, 𝑡) or 𝓟(𝑥, 𝑡) = ∣𝜓(𝑥, 𝑡)∣^{2}.
The probabilty density is the probability per unit length in one dimension, per unit area in two dimension and per unit volumee in three dimension. 𝓟 tells us how the probability of finding the particle in a particular place is spread out through space at any given time. This answers the question of what's waving for an electron and other quantummechanical objects: A traveling particle is actually traveling lump of probabilty amplitude. When it encounter an obstacle (such as the two slits descrobed in Section 6.2), the waving probabilty amplitude is diffracted according to the wavelength. When interact (which is how it is measured or detected), this wavefunction collapses to the single measured result. This result is discrete, which is consistent with particle behavior.
So we can write the wavefunction for a free particle using harmonic function. Since 𝓟(𝑥, 𝑡) represents the probability density, integrating overall space (𝑥 = ¡Ä to 𝑥 = ¡Ä) should give a probabilty of one. So
(6.2831) 𝜓(𝑥, 𝑡) = 𝐴𝑒^{𝑖(𝑘𝑥𝜔𝑡)}, 1 = ¡ò_{¡Ä}^{ ¡Ä}𝜓^{*}(𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑𝑥, 1 = ¡ò_{¡Ä}^{ ¡Ä}𝐴^{*}𝐴𝑒^{𝑖(𝑘𝑥𝜔𝑡)}𝑒^{𝑖(𝑘𝑥𝜔𝑡)}𝑑𝑥, 1 = 𝐴^{*}𝐴(¡Ä).
But since nothing can multiply to give one, this waveform is "nonnormalizable". To modify the wavefunctin of Eq. (6.28) and to form quantum wave packet, the Fourier concepts are very useful as we'll see in the next section.
6.5 Quantum wave packets _{}^{}
Since a particleis localized in space, so it seems reasonable to expect that its wave should also be spatially limited: that is, it should be a wave packet. To form a wave pocket some waves with different but similar wavelength must be included. A range of wvelength means arange of wavenumber, and a range of wavenumber meansa range of momenta (since 𝑝 = ℏ𝑘). So the challenge is to find a wave packet that is localized in space overa region 𝛥𝑥 but that travels with a welldefined momentum 𝑝 = ℏ𝑘_{0}, where 𝑘_{0} represents the dominant wavenumber.
It's a bit easier to write 𝜓(𝑥, 𝑡) = 𝑓(𝑡)𝜓((𝑥) allow us to concentrate on localizing the spatial terrm 𝜓((𝑥); the effect of the time term 𝑓(𝑡) will be considered later. And one approach to limiting the spatial extent of the wavefunction is to write 𝜓((𝑥) as the product of two functions: the exterior" envelope 𝑔(𝑥), and the "interior" oscillations 𝑓(𝑥),
(6.32) 𝜓((𝑥) = 𝑔(𝑥)𝑓(𝑥).
If the evelope function goes to zero everywhere except at a certain range of 𝑥 values, then the oscillations of the wave packet will be localized to that range.
For example, consider the singlewavelength oscillating function 𝑓(𝑥) = 𝑒^{𝑖𝑘𝑥} (= cos𝑘𝑥 + 𝑖sin𝑘𝑥) as the plot of the real part of 𝑓(𝑥) in Fig. 6.7 (a)(𝑘 = 10), this function extends over all space. And consider the envelope function 𝑔((𝑥) given by
(6.33) 𝑔(𝑥) = 𝑒^{𝑎𝑥2}.
As we can see in Fig. 6.7 (b)(𝑎 = 1), this function reaches a peak of 𝑔(𝑥) at 𝑥 = 0 and decreases both directions at a rate determined by 𝑎.
Multiplying this envelope function 𝑔(𝑥) by the oscillating function 𝑓(𝑥) causes the product function 𝑓(𝑥)𝑔(𝑥) as (the real part) shown in Fig. 6.8. So the plrobability density is
𝜓 = 𝑒^{𝑖𝑘𝑥}𝑒^{𝑎𝑥2}, 𝜓^{*} = 𝑒^{𝑖𝑘𝑥}𝑒^{𝑎𝑥2},
𝓟 = ∣𝜓^{*}𝜓∣ = (𝑒^{𝑖𝑘𝑥}𝑒^{𝑎𝑥2})(𝑒^{𝑖𝑘𝑥}𝑒^{𝑎𝑥2}) = 𝑒^{2𝑎𝑥2}.
Integrating this over all space gives
𝓟_{all space} = ¡ò_{¡Ä}^{ ¡Ä}𝑒^{2𝑎𝑥2}𝑑𝑥 = ¡î(¥ð/2𝑎). [Re Wolframalpha]
So to set 𝓟_{all space} = 1, 𝜓 must be scaled by the square root of the inverse of ¡î¥ð/(2𝑎)}, so
(6.34) 𝜓((𝑥) = ¡î[1/{¡î¥ð/(2𝑎)}]𝑒^{2𝑎𝑥2}𝑒^{𝑖𝑘𝑥} = (2𝑎/¥ð)^{1/4}𝑒^{2𝑎𝑥2}𝑒^{𝑖𝑘𝑥}.
This function is normalized to give an allspace probability of one.
Example 6.4 Determine the probability of finding a particle at a given location if the particle's wavefunction is defined as
𝜓((𝑥) = (0.2/¥ð)^{1/4} 𝑒^{0.1𝑥2} 𝑒^{𝑖𝑘𝑥}.
𝜓^{*}((𝑥)𝜓((𝑥) = [(0.2/¥ð)^{1/4} 𝑒^{0.1𝑥2} 𝑒^{𝑖𝑘𝑥}][(0.2/¥ð)^{1/4} 𝑒^{0.1𝑥2} 𝑒^{𝑖𝑘𝑥}] = (0.2/¥ð)^{1/2} 𝑒^{0.2𝑥2},
which is a Gaussian distribution, as shown in Fig. 6.9.
To find probability of the particle with this wavefunction being located in a particular spot, we have to intergrate the density around that place. In this example the likelihood of finding the particle at 𝑥 = 1, give or take 0.1 m is
𝓟(1 ∓ 0.1) = (0.2/¥ð)^{1/2}¡ò_{0.9}^{1.1}𝑒^{0.2𝑥2} 𝑑𝑥 = 0.041, [Re Wolframalpha]
or 4.1%. We can check the normalization of this function by intergrating over all space:
𝓟_{all space} = (0.2/¥ð)^{1/2}¡ò_{¡Ä}^{ ¡Ä}𝑒^{0.2𝑥2}𝑑𝑥 = 1. [Re Wolframalpha]
If we encounter other wave functions that aren't normalized then we may write the waveform with a multiplicative factor 𝛢 out front, we can set the integral over all space of propability density equal to one and solve for 𝛢.
Example 6.5 Normalize the triangluarpurlse wavefunction in Fig. 6.10.
The equation for this triangular pulse can be written as
⌈ 𝛢𝑥 0 ¡Â 𝑥 ¡Â 0.5,
𝜓((𝑥) =¦𝛢(1𝑥) 0.5 ¡Â 𝑥 ¡Â 1,
⌊ 0 else,
which can be plugged into the probability density integral:
𝓟_{all space} = 1 = ¡ò_{¡Ä}^{¡Ä}𝜓^{*}(𝑥)𝜓(𝑥)𝑑𝑥. Thus
1 = ¡ò_{0}^{0.5}(𝛢^{*}𝑥)(𝛢𝑥)𝑑𝑥 + ¡ò_{0.5}^{1}(𝛢^{*}(1𝑥))(𝛢(1𝑥))𝑑𝑥, and pulling out 𝛢^{*}𝛢 from each integral leaves
1 = 𝛢^{*}𝛢 {¡ò_{0}^{0.5}𝑥^{2}𝑑𝑥 + ¡ò_{0}^{0.5}(1  𝑥)^{2}𝑑𝑥} = 𝛢^{*}𝛢 (1/24 + 1/24). [Re Wolframalpha]
Here all factors are real, so 𝛢^{*}𝛢 = 𝛢^{2}. Solving for the normalization constant gives
1 = 𝛢^{2}(1/24 + 1/24), 𝛢 = ¡î12.
Fig. 6.11 shows the probability density before and after normaliztion. Note that the shape of the normalized probabilty density haven't changed from nonnormalized one: only its scale has changed.
Additionaly it's important that we also understand the wavenumber (and momentum) range of those waveform. Using the Fourier systhesis approach, we can construct a spatially limited wavefunction from a single wavelength finction 𝑒^{𝑖𝑘0𝑥} by adding in other singlewavelength function in just the right proportion to cause the amplitude of the combined function to roll off over distance at the desired rate.
If we atempt to do this using a discrete set of wavefunctions 𝜓_{n} and represent the amplitude coefficient of each component waveform as 𝜙_{n}, the combined waveform will be
(6.35) 𝜓((𝑥) = ¢²_{n} 𝜓_{n} = [1/¡î(2¥ð)]¢²_{n} 𝜙_{n}𝑒^{𝑖𝑘𝑥}.
The reason for including the scaling factor of 1/¡î(2¥ð) will be made clear below. If we hope to construct a wavefunction with a single region of large amplitude, the wavenumber difference between the component wavefunctions must be infinitesimally small, and the discrete sum must become an integral.
(6.36) 𝜓(𝑥) = [1/¡î(2¥ð)] ¡ò_{¡Ä}^{¡Ä} 𝜙(𝑘)𝑒^{𝑖𝑘𝑥} 𝑑𝑘,
in whichthe discrete coefficient 𝜙_{n} have been replaced by 𝜙(𝑘), a continuous function that determines the amount of each wavenumber component that gets added to the mix.
We may recall Eq. (3.34) which is the equation for the inverse Fourier transform. Thus the spatial wavefunction 𝜓(𝑥) and the wavenumber function 𝜙(𝑘) are a Fourier transform pair. That means we can find the wavenumber function through the foward Fourier transform of 𝜓(𝑥):^{(1)}
(6.37) 𝜙(𝑘) = [1/¡î(2¥ð)] ¡ò_{¡Ä}^{¡Ä} 𝜓(𝑥)𝑒^{𝑖𝑘𝑥} 𝑑𝑥,
The Fouriertransform relationship between 𝜓(𝑥) and 𝜙(𝑘) has powerful implications like all conjugatevariable pairs, these functions obey the uncertainty principle, and that help us determine the wavenumber content of a given spatial wavefunction.
To see how that works, consider a general Gaussian envelope function of width 𝜊_{x}. That function can be written as
(6.38) 𝑔(𝑥) = 𝑒^{𝑥2/(2𝜊x2)}.
this is essentially the same envelope function in Eq. (6.33) and 𝑎 = 1/(2𝜊_{𝑥}^{2}), where 𝜊_{𝑥} is the standard deviation of Gaussian wavefunction. Multiplying this envelope function by interior, singlewavenumber oscillating function 𝑓(𝑥) = 𝑒^{𝑖𝑘0𝑥} and normalizing produces the waveform. That is Eq. (6.34) with 𝑎 = 1/(2𝜊_{𝑥}^{2}:
(6.39) 𝜓(𝑥) = (1/¥ð𝜊_{𝑥}^{2})^{1/4} 𝑒^{𝑥2/(2𝜊𝑥2)}𝑒^{𝑖𝑘0𝑥}.
That Fourier transform is
𝜙(𝑘) = [1/¡î(2¥ð)] ¡ò_{¡Ä}^{¡Ä} 𝜓(𝑥)𝑒^{𝑖𝑘𝑥} 𝑑𝑥 = [1/¡î(2¥ð)](1/¥ð𝜊_{x}^{2})^{1/4} ¡ò_{¡Ä}^{¡Ä} 𝑒^{𝑥2/(2𝜊𝑥2)}𝑒^{𝑖𝑘0𝑥}𝑒^{𝑖𝑘0𝑥} 𝑑𝑥
So the wavenumber (and momentum) distribution is^{(2)}
(6.40) 𝜙(𝑘) = (𝜊_{x}^{2}/¥ð)^{1/4} 𝑒^{(𝜊𝑥2/2)(𝑘0𝑘)2}.
This is a Gaussian distribution around 𝑘_{0} width 𝜊_{𝑘} = 1/𝜊_{𝑥}. And, the spread in those values depends on how much the packet spreads in space, as expexted.
Detailed analysis of uncertainty in position shows that for Gaussian wavevpacket with standard deviation 𝜊_{𝑥}, the position uncertainty is 𝛥𝑥 = 𝜊_{𝑥}/¡î(2¥ð) and the wavenumber uncertainty is 𝛥𝑘 = 𝜊_{𝑘}/¡î(2¥ð). And since 𝜊_{𝑥} =1/𝜊_{𝑘}, the product of the uncertainties in position and wavenumber is
(6.41) 𝛥𝑥𝛥𝑘 = [𝜊_{𝑥}/¡î(2¥ð)][𝜊_{𝑘}/¡î(2¥ð)] = [𝜊_{𝑥}/¡î(2¥ð)][1/𝜊_{𝑥}¡î(2¥ð)] = 1/2.
Likewise, the product of uncertainties in 𝑥 and 𝑝 is (since 𝑝 = ℏ𝑘)
(6.42) 𝛥𝑥𝛥𝑝 = [𝜊_{𝑥}/¡î(2¥ð)][𝜊_{𝑝}/¡î(2¥ð)] = [𝜊_{𝑥}/¡î(2¥ð)][ℏ𝜊_{𝑘}/¡î(2¥ð)] = [𝜊_{𝑥}/¡î(2¥ð)][ℏ/𝜊_{𝑥}¡î(2¥ð)] = ℏ/2.
This is known as"Heisenberg's uncertainty principle" it's a version of the general uncertainty between conjugate varables discussed in Chapter 3.
If we measure the position of a given particle and then measure its monentum at a later time, we can certainly measure a precise value for each. So there's a better way think about Heisenberg's uncertainty principle.
Imagine that we have a large number of identical particles, all in the same state (so they have the same wavefunction). If the spread positions of these "ensemble" of particles is small, measuring the positions of all of the particles will return very similar values. However, if we measure the momentum of each of these particles, the measured momenta will be very different from one another. That's because particles with position have many contributing momentum states {waves with different wavenumbers), and the measurement process cause the wavefunction to "collapse" randomly to one of those states. Conversely, if the spread in position among the particles is large, then there are few contributing momentum states, and the measured values of momentum will bevery similar.
The final aspct of quantum waves we'll consider is the time evolution of the wavefunction of a free particle. To do that, we'll have to put the time term back into the spatially limited wavefunction. At time 𝑡 = 0, the wavefunction 𝛹(𝑥, 0) for Gaussian wave packet can be written as
(6.43) 𝛹(𝑥, 0) = 1/(¥ð𝜊_{𝑥}^{2}) 𝑒^{𝑖𝑘0𝑥}𝑒^{𝑥2/(2𝜊𝑥2)}.
in which 𝜊_{𝑥} is the standard deviation of Gaussian position envelope. At time 𝑡, this wavefunction is
(6.44) 𝛹(𝑥, 𝑡) = [1/¡î(2¥ð)] ¡ò_{¡Ä}^{¡Ä} 𝜙(𝑘) 𝑒^{𝑖[𝑘𝑥𝜔(𝑘)𝑡]} 𝑑𝑘,
in which 𝜙(𝑘) is the wavenumber function that is the Fourier transform of the position function and we've written 𝜔(𝑘) to remind that 𝜔 depends on 𝑘. so the 𝜔(𝑘) = ℏ𝑘^{2}/(2𝑚). Now inserting 𝜙(𝑘) Eq. (6.40) for a Gaussian wavepacket into the expression for 𝛹(𝑥, 𝑡) gives
𝛹(𝑥, 𝑡) = [𝜊_{𝑥}^{2}/(4¥ð^{3})]^{1/4} ¡ò_{¡Ä}^{¡Ä} 𝑒^{(𝜊𝑥2/2)(𝑘0𝑘)2} 𝑒^{𝑖[𝑘𝑥𝜔(𝑘)𝑡]} 𝑑𝑘,
so the wavefunction is consequently.^{(3)}
(6.45) 𝛹(𝑥, 𝑡) = [𝜊_{𝑥}^{2}/(4¥ð^{3})]^{1/4} exp[𝑖(𝑘_{0}𝑥  𝜔_{0}𝑡)] [¥ð/{𝜊_{𝑥}^{2}/2 + 𝑖ℏ𝑡/(2𝑚)}]^{1/2} exp[(𝑥  ℏ𝑘_{0}𝑡/𝑚)^{2}/4(𝜊_{𝑥}^{2}/2 + 𝑖ℏ𝑡/(2𝑚)].
This expression for 𝛹(𝑥, 𝑡) looks quite complicate, programs such as Mathematica, MATLAB and Octave help us explore the behavior of the wavefunction over time. For example, if we choose a particle with the mass of a proton (1.67 ¡¿ 10^{27} kg) at a speed of 4 mm/s, the particle's de Broglie wavelength is just under 100 microns. Forming a Gaussian wave packet with a standard deviation of 250 microns results in the wavefunction in Fig. 6.12.
In this figure, the wavefunction is shown at times of 𝑡 = 0, 1, and 2 seconds over a spatial interval about 10 mm. The particles dominant wavelength is approximately 100 microns, and there are about 2.5 cycles on either side of the central maximum within the standard deviation at time 𝑡 = 0. Each second, the wave packet propagates 4 mm, so the group velocity of the packet equals theparticles's speed, as expected.
The component waves that make up the wave packet all have slightly different velocities, but those velocities are about half the group velocity of the packet. To see the reason, consider that the dispersion relationship for de Broglie waves is (also see eq. (6.14) when 𝑉 = 0)
(6.46) 𝜔(𝑘) = [ℏ/(2𝑚)]𝑘^{2},
so thepacket's group velocity 𝑑𝜔/𝑑𝑘 is
(6.47) 𝑣_{𝑔} = (𝑑𝜔/𝑑𝑘) = ℏ𝑘/𝑚,
which is the classical particle speed. Recall also that the phase velocity is 𝜔/𝑘, which is
(6.48) 𝑣_{𝑝} = 𝜔/𝑘 = ℏ𝑘/(2𝑚).
This is half the wave packet 's group velocity, and half the particles's speed.
Figure 6.13 shows the probability density of the wave packet moving through space as time passes. As we can see both the plot of 𝛹(𝑥, 𝑡) and the plot of the probability density. the wave packet isn't just moving, it's also spreading out as time passes. As explaimed Section 3.4 0f Chapter 3. dispersion occurs whenever the component waves that make up a wave packet travel with different speeds. Given that the quantum dispersion relationship is not linear with respect to 𝑘, quantum objects are dispersive  the different speeds of the components of the wave packet cause the packet to spread out in time.
All of the matter waves we've seen in this chapter have been for free particles in the regions constant energy (which we set to zero): these all have 𝑒^{𝑖𝑘𝑥} as their basic functions. As mentioned in Chapter 4, waves have nonsinusoidal basis functions in an inhomogeneous string. This is also true for matter waves in regions of nonconstant potential energy. And, we may study on them further in the same auther's other textbook.^{(4)}
(1) This is why we included the factor of 1/¡î(2¥ð) to Eq.(6.36)
(2) (3) For this complicate integral, we can calculate with Wolframalpha or refer to textbook's online solution.
(4) D. Fleisch A Student's Guide to Schrödinger Equation (Cambridge University Press 2020)
* Textbook: D. Fleisch & J. Kinnaman A Student's Guide to Waves (Cambridge University Press 2015) 

