Quantum Mechanics PDF

Safia Ahmad Quantum Mechanics Physics of the 19th century and earlier is called Classical Physics. Examples are Newtonian mechanics, Electricity and Magnetism, etc. To describe the motion of objects like a football, a cricket ball, the Newton’s equations of motion ar...

Safia Ahmad Quantum Mechanics Physics of the 19th century and earlier is called Classical Physics. Examples are Newtonian mechanics, Electricity and Magnetism, etc. To describe the motion of objects like a football, a cricket ball, the Newton’s equations of motion are used. It can also be used to describe the motion of tiny bullets fired from an air-gun. But it is found that the Newton’s laws of motion are not sufficient to describe everything. In the early 20th century, when discoveries were made about the composition of atoms, it became clear that the classical physics do not work on a very small scale of the atoms. Size of the atom is only 1 Angstrom (= 10−10 m). So, if 10 million atoms are lined up, then its only 1 mm. Quantum mechanics is the description of the behavior of matter and light in all its details, in particular, their behavior on an atomic scale. Things on a very small scale behave like nothing that one has any direct experience about. They do not behave like waves, they do not behave like particles, they do not behave like cricket ball, or like anything that one has ever seen. For instance, Newton thought that light was made up of particles, but then it was discovered that it behaves like a wave. However, in the early 20th century, it was observed that light does indeed sometimes behave like a particle. Similarly, electron, for example, was thought to behave like a particle, and then it was found that in many respects, it behaved like a wave. The confusion caused by the gradual accumulation of information about atomic and small scale behavior during the first quarter of 20th century, which gave some indications about how microscopic particles do behave, was finally resolved in 1926 and 1927 by Schrodinger, Heisenberg and Born. They finally obtained a consistent description of the behavior of matter on a small scale. An experiment with bullets In order to understand the quantum behavior of electrons, we shall compare their behavior with the more familiar behavior of particles like bullets. Suppose that a machine gun shoots a stream of bullets onto a screen kept at a large distance. The machine gun is not very 1  Safia Ahmad Figure 1: Double-slit experiment with bullets. accurate so it sprays the bullets randomly over a fairly large angular spread. If hundred bullets are fired on the screen - all of them will not hit the screen at the same spot. Now suppose a screen with two holes is kept in between the gun and the screen; the holes are just about big enough to let a bullet through. If one of the holes is blocked, the bullets are only able to pass through the open hole, and a lump on the screen is observed at a point in line with the gun and the open hole. If the second hole is opened and the first one is closed then a lump on the screen at a spot in line with the second hole and the gun is observed. If now one opens both the holes and again fires lots of bullets, one would see that the effect with both holes open is the sum of the effects with each hole open alone obtained in the previous cases i.e. the probabilities will just be the sum of two cases: P12 = P1 + P2 The probabilities just add together. There is no interference. An experiment with electrons Later, the similar experiment was repeated with electrons. It was found that if either of the two holes is closed, one gets a lump as in the case of bullets, but when both the holes 2  Safia Ahmad Figure 2: Double-slit experiment with electrons. are open one gets a pattern which looks like a long array of lumps, an interference pattern. This pattern is not just the sum of the patterns obtained with only one hole open. If an electron cannot be split, one imagines that an electron fired from the electron gun passes through one of the two holes, and should behave as if only that hole was open. So, the net pattern should be a sum of the two patterns with only one hole open. But this does not happen which means the electron somehow passes through both the holes simultaneously. That is, electrons seem to interfere with themselves. The fact that the interference pattern is obtained from the double slit experiment proves that electrons exhibit both particle-like properties and wave-like properties. Now, in order to detect the slit from which the electron passes through, a light source is placed behind the screen consisting the slits. As we know that electric charges scatter light. So, when an electron passes through a slit, it will scatter off some light near the slit from which it passes through i.e. whenever an electron strikes the screen, light is flashed near S1 or S2 but never near both at the same time. After many observation, it was found that the intensity distribution with both slits open is the same as that in the case of bullets. This is to say that if the light source is placed to detect the trajectory of electrons, interference pattern disappears; if the light source is removed, the interference pattern appears again. It is, therefore, concluded that if the electrons are being watched, their intensity distribution 3  Safia Ahmad on the screen gets affected i.e. the interference pattern disappears. So, for the electrons that display interference, it is impossible to identify the slit that each electron had gone through. That is, it is impossible to trace the path of an electron without disturbing the interference pattern. These results inspired Heisenberg to propose the uncertainty principle which states that ”It is impossible to design an apparatus to determine which slit the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern. Photoelectric Effect The photoelectric effect provides a direct confirmation for the energy quantization of light. In 1887 Hertz discovered that electrons are ejected from a metal surface when light is incident on it. Moreover, the following experimental laws were discovered prior to 1905: → If the frequency of the incident radiation is smaller than the metal’s threshold frequency, no electron can be emitted regardless of the radiation’s intensity. → No matter how low the intensity of the incident radiation, electrons will be ejected instantly the moment the frequency of the radiation exceeds the threshold frequency, ν0. → At any frequency above ν0 , the number of electrons ejected increases with the intensity of the light but does not depend on the light’s frequency. 4  Safia Ahmad → The kinetic energy of the ejected electrons depends on the frequency but not on the intensity of the beam; the kinetic energy of the ejected electron increases linearly with the incident frequency. Classical physics fails to provide the explanation for these experimental findings. According to classical physics, any amount of energy can be exchanged with matter. If a weak light source is incident on a metal then according to classical physics, an electron would keep on absorbing energy at a continuous rate until it gained a sufficient amount then it would leave the metal. If this argument is valid, then when a very weak radiation is used, the photoelectric effect would not take place for a long time, possibly hours, until an electron gradually accumulated the necessary amount of energy. This disagrees with experimental observation. Experiments were conducted with a light source that was so weak it would have taken several hours for an electron to accumulate the energy needed for its ejection, and yet some electrons were observed to leave the metal instantly. Further experiments showed that an increase in intensity alone can in no way dislodge electrons from the metal. But by increasing the frequency of the incident radiation beyond a certain threshold, even at very weak intensity, the emission of electrons starts immediately. These experimental facts indicate that the concept of continuous absorption of energy by the electron, as predicated by classical physics, is inaccurate. In 1905, Einstein succeeded in giving a the- oretical explanation for the dependence of photoelectric emission on the frequency of the incident radiation. Inspired by Planck’s quantization of electromagnetic radiation, he assumed that light is made of corpuscles each carrying an energy hν, called photons. When a beam of light of frequency ν is in- cident on a metal, each photon transmits all its energy hν to an electron near the surface. So, the photon is entirely absorbed by the electron i.e. the electron will absorb energy only in quanta of energy hν, irrespective of the 5  Safia Ahmad intensity of the incident radiation. If hν is larger than the metal’s work function, W (the minimum energy required to free the electron from the metal is called the work function of that metal), the electron will be knocked out of the metal. Hence no electron can be emitted from the metal’s surface unless hν > W : hν = W + K where K represents the kinetic energy of the electron leaving the material. Thus, photoelec- tric effect provides the confirmation for the particle-like behavior of light. Wave-Particle Duality As discussed above in the photoelectric effect, radiation exhibit particle-like characteristics in addition to its wave nature. In 1923, de Broglie suggested that this wave–particle duality is not restricted to radiation, but must be universal: “All material particles should also display a dual wave–particle behavior.” That is, the wave–particle duality present in light must also occur in matter. The fact that electromagnetic radiation exhibit particle - like behavior, the electromagnetic field must be associated with a particle, the photon, whose energy E and the magnitude of momentum p are related to the frequency ν and wavelength λ of the electromagnetic radiation by h E = hν, p= (1) λ Similarly, the fact that particles exhibit wave-like properties imply that we associate with each particle a wave or matter field. The de Broglie relations which link the frequency ν and wavelength λ of the wave with the particle energy E and magnitude of momentum p are given by the above expressions. Thus, we can assume that the relations (1) hold for all types of particles and field quanta. If the angular frequency, ω = 2πν, wave number, k = 2π/λ and the reduced Planck constant, ℏ = h/(2π), the relations (1) may be rewritten as E = ℏω, p = ℏk (2) 6  Safia Ahmad If a particle of mass m is constrained to move along an x-axis subject to some force F (x, t) then in classical mechanics, in order to determine the position of the particle at any given time, x(t), the Newton’s second law, F = ma is applied. This together with the initial conditions, determines x(t). Once x(t) is known, the velocity v(t) = dx/dt, the momentum, p = mv, the kinetic energy, T = (1/2)mv 2 or any other dynamical variable of interest can be determined. In quantum mechanics, the state (or one of the states) of a particle is described by a wave function ψ (x, t) corresponding to the de Broglie wave of a particle which can be obtained by solving the Schrodinger’s equation: ∂ψ ℏ2 ∂ 2 ψ iℏ =− +V ψ ∂t 2m ∂x2 Here, ℏ = h/(2π) = 1.054572 × 10−34 J s. Thus, Schrodinger equation plays a role logically analogous to the Newton’s second law. Wave Function and its Probabilistic Interpretation As a particle, by its nature, is localized at a point whereas the wave function is spread out in space; it’s a function of x at any given time t. Generally, wave function is a complex function and therefore and therefore cannot represent a measurable quantity which should be a real number. In 1927, Max Born gave the Born’s probabilistic interpretation of the wave function which says that “the wave function ψ(x, t) itself has no physical significance but the quantity |ψ(x, t)|2 gives the probability of finding the particle at point x at time t, or more precisely,   Z b  probability of finding the particle between  |ψ(x, t)|2 dx = a  a and b at time t  The quantity |ψ(x, t)|2 dx represents the probability of finding a particle, at time t, between x and x + dx. The probability of finding a particle somewhere in space must then be equal to 1, i.e. Z +∞ |ψ(x, t)|2 dx = 1 (3) −∞ 7  Safia Ahmad This is known as the normalization condition. If ψ(x, t) is a solution to Schrodinger equation so is Aψ(x, t) where A can be determined from the above normalization condition. This process is called normalizing the wave function. The wave function that represents particles have to be square-integrable and normalizable. Any wave function which is not square- integrable has no physical meaning in quantum mechanics. And if ψ(x, t) is normalized at time t = 0, it stays normalized for all future times. Thus, an acceptable wave function must be normalized and satisfy the following require- ments: 1) Wave function ψ(x, t) must be finite everywhere. If ψ(x, t) is infinite, it would imply an infinitely large probability of finding the particle at that point and this would violate the uncertainty principle. 2) The wave function ψ(x, t) must be single-valued. Any physical quantity can have only one value at a point. So, the wave function related to a physical quantity cannot have more than one value at that point. 3) The wave function ψ(x, t) and its spatial derivative, ∂ψ/∂x must be continuous across any boundary. Operators An operator  is the mathematical rule which when applied to a wave function gives you another wave function,  ψ = ϕ (4) A unit operator may be represented by Iˆ which acts on any function, and yields the same function. For every operator, there can be a function such that the operator acts on the function and gives back the same function times a number,  ψ = a ψ (5) In such a situation, this function is called the eigenfunction of that operator Â, and the number, a, is its corresponding eigenvalue. In quantum mechanics, all observable quantities 8  Safia Ahmad are represented by operators. However, when a quantity is measured, say momentum, we always get a number. So, there should be a number associated with a measured value of the observable. Thus, eigenvalues correspond to the measured values of the operator. Operators are denoted by using a hat, so x̂ is the operator that corresponds to position, x, the operator p̂ corresponds to momentum p and operator Ê corresponds to total energy. They are given as ∂ ∂ p̂ = −iℏ , Ê = iℏ ∂x ∂t The operator for energy for a particle of mass m, experiencing a potential can be obtained by simply taking the classical expression for the energy, and replacing the momentum and position by their respective operators. The Hamiltonian, the operator for energy, is then given by p̂2 Ĥ = + V (x̂) 2m The energy of the particle can be found out by solving the eigenvalue equation ∂ψ(x) Ĥψ(x) = Eψ(x) =⇒ Ĥψ(x) = iℏ ∂t Heisenberg Uncertainty Principle In quantum mechanics, since a particle is represented by a wave function corresponding to the particle’s wave, and since wave functions are not localized in space,the microscopic particles cannot be described with accuracy, for waves give at best a probabilistic account. In addition, double-slit experiment shows that it is impossible to determine the slit that the electron went through without disturbing it. Therefore, classical concepts of exact position, exact momentum, and unique path of a particle make no sense at the microscopic scale. These experimental findings inspired Heisenberg to postulate the indeterministic nature of the microphysical world. In its original form, Heisenberg’s uncertainty principle states that: “If the x-component of the momentum of a particle is measured with an uncertainty ∆px , then its x-position cannot, at the same time, be measured more accurately than ∆x = ℏ/(2∆px )”. 9  Safia Ahmad Thus, ℏ ∆x ∆px ≥ (6) 2 This principle indicates that, although it is possible to measure the momentum or position of a particle accurately, it is not possible to measure these two observables simultaneously to an arbitrary accuracy. That is, a microscopic particle cannot be localized without giving to it a rather large momentum. The Uncertainty principle indicates the the limit to the accuracy with which the measurements can be made. To measure the position of an electron in an atom, one must use radiation of very short wavelength (the size of the atom). The energy of this radiation (E = hc/λ) is high enough to change tremendously the momentum of the electron; the mere observation of the electron affects its motion so much that it can knock it entirely out of its orbit. It is therefore impossible to determine the position and the momentum simultaneously to arbitrary accuracy. According to de Broglie’s relation p = h/λ, the momentum of this particle will be rather high. This means that if a particle is accurately localized (i.e., ∆x → 0), there will be total uncertainty about its momentum i.e., ∆px → ∞). That is, the location and momentum of a particle cannot be simultaneously determined with certainty. The similar relations (6) for other components of position and momentum are ℏ ℏ ∆y ∆py ≥ , ∆z ∆pz ≥ (7) 2 2 Wave Packets In classical physics, a particle is well localized in space as its position and velocity can be calculated simultaneously to arbitrary precision. While in quantum mechanics, material particle is described by a wave function corresponding to the matter wave associated with the particle. However, wave functions spread over a whole space and hence cannot be localized and therefore cannot represent a highly localized particle. Schrodinger postulated that rather than a single wave, a particle is represented by a wave packet. That is, the particle is represented not by a single de Broglie wave of well-defined frequency and wavelength, but by a wave packet that is obtained by adding a large number of waves of different frequencies. 10  Safia Ahmad A localized wave function is called a wave packet. It can be thought of as a ”packet” or a bundle of waves that travel together. A wave packet consists of a group of waves of slightly different wavelengths, with phases and amplitudes so chosen that they interfere constructively over a small region of space and destructively elsewhere. The construc- tive interference enhances the amplitude of the combined wave in that region while reducing its amplitude elsewhere. As a result, the wave packet is concentrated and localized around that specific region. The velocity with which the wave packet propagates is called the group velocity. The individual waves forming the wave packet propagate at a velocity known as phase velocity. Localized wave packet can be constructed by superposing waves of slightly different wave- lengths but with phases and amplitudes so chosen that the superposition is constructive in the desired region and destructive outside it. Consider a one dimensional wave-packet i.e. a packet that describes a particle confined to a one-dimensional region, say, particle moving along x-axis. The packet ψ(x, t) can be constructed by superposing plane waves (propagating along the x-axis) of different frequencies or wavelengths, Z +∞ 1 ψ(x, t) = √ ϕ(k)eikx−ωt)/ dk 2π −∞ where ϕ(k) is the amplitude of the wave packet. Using the fact that the energy of a particle propagating in x -direction is E = ℏω and the x -component of momentum is px = ℏk which √ implies that k = px /ℏ, ω = E/ℏ and dk = dpx /ℏ. Redefining ϕ(px ) = ϕ(k)/ ℏ, the above equation may be rewitten as Z +∞ 1 ψ(x, t) = √ ϕ(px )ei(px x−Et)/ℏ dpx 2πℏ −∞ The wave packet at a given time, say at t = 0, is given by Z +∞ 1 ψ(x, 0) = √ ϕ(px )eipx x/ℏ dpx 2πℏ −∞ 11  Safia Ahmad where ϕ(px ) is the Fourier transform of ψ(x, 0), Z +∞ 1 ϕ(px ) = √ ψ(x, 0)e−ipx x/ℏ dk 2πℏ −∞ Thus, ϕ(px ) determines ψ(x, 0) and vice versa. Here, |ψ(x, 0)| peaks at x = 0 and vanishes far away from x = 0. Since eipx x/ℏ → 1 as x → 0, the waves of different frequencies interfere constructively i.e. various k- integration add constructively. While far away from x = 0, i.e. |x| ≫ 0, the phase eipx x/ℏ goes through many periods leading to violent oscillations, thereby yielding destructive interference i.e. various k- integration add up to zero. This is to say that particle has a greater probability of being found near x = 0 and a very small chance of being found far away from x = 0. Hence |ψ(x, 0)|2 gives the probability density for finding the particle at x, and P (x) dx = |ψ(x, 0)|2 dx gives the probability of finding the particle between x and x + dx. The Gaussian wave packet gives the lowest limit of Heisenberg’s inequality. Therefore, the Gaussian wave packet is called the minimum uncertainty wave packet. All other wave packets yield higher values for the product of x and p uncertainties, ∆x ∆px > ℏ/2. In conclusion, the value of the uncertainties product ∆x ∆px varies with the choice of ψ(x), but the lowest bound, ℏ/2, is provided by a Gaussian wave function. Schrodinger equation Consider a free particle of mass m, having a well defined momentum, p = px x̂ directed along x direction and a non-relativistic energy, E = p2x /(2m). So, the particle is associated with a wave function, Ψ(x, t) = A exp (i(kx − ωt)) (8) where k is the wave number, ω is the angular frequency. Differentiating the above wave function with respect to x gives ∂Ψ i = ikA exp (i(kx − ωt)) = i k Ψ(x, t) = p Ψ(x, t) ∵ p = ℏk ∂x ℏ ∂Ψ =⇒ −i ℏ = p Ψ(x, t) ∂x ∂ =⇒ p = −i ℏ → momentum operator (9) ∂x 12  Safia Ahmad If we now differentiate the wave function with respect to the time t, we get ∂Ψ i = −iωA exp (i(kx − ωt)) = −i ω Ψ(x, t) = − E Ψ(x, t) ∵ E = ℏω ∂t ℏ ∂Ψ =⇒ i ℏ = E Ψ(x, t) ∂t ∂ =⇒ E = i ℏ → energy operator (10) ∂t As we know that the total energy is a sum of kinetic and potential energy,  2  ∂Ψ p̂ iℏ = + V (x̂, t) Ψ(x, t) ∂t 2m Substituting for the momentum operator, ℏ2 ∂ 2   ∂Ψ iℏ = − + V (x̂, t) Ψ(x, t) (11) ∂t 2m ∂x2 This is the time-dependent Schrodinger equation for a particle moving in a potential V (x̂, t) in one dimension. Now, if we consider a particular case of time-independent potentials: V (x, t) = V (x). In this case, the Hamiltonian operator is time-independent and therefore Schrodinger equation can be solved by the method of separation of variables i.e. the solution consist of a product of two functions, one depending only on x and the other only on time t, Ψ(x, t) = ψ(x) ϕ(t) (12) Substituting this in (11), we get dϕ ℏ2 d2 ψ iℏ ψ(x) =− ϕ(t) 2 + V (x̂) ψ(x) ϕ(t) dt 2m dx Dividing throughout by (ψ ϕ), we have 1 dϕ ℏ2 1 d2 ψ iℏ =− + V (x) ϕ dt 2m ψ dx2 Now, the left side is a function of time only and the right side is a function of x only. This can only be possible if both sides are constant (say textitE), otherwise, by varying t, the left side could be changed without touching the right side and the two sides would no longer be equal. Therefore, 1 dϕ ℏ2 1 d2 ψ iℏ = E, − + V (x) = E ϕ dt 2m ψ dx2 dϕ iE ℏ2 d2 ψ =⇒ = − ϕ, − +V ψ =E ψ (13) dt ℏ 2m dx2 13  Safia Ahmad Thus, method of separation of variables has turned a partial differential equation into two ordinary differential equations. The solution of first differential equation is dϕ iE =− dt =⇒ ϕ(t) = C e−iEt/ℏ ϕ ℏ so that Ψ(x, t) = ψ(x) e−iEt/ℏ (14) where the constant C has been absorbed into ψ(x) which can be obtained by solving the second equation, ℏ2 d2 ψ(x) − + V (x) ψ(x) = E ψ(x) (15) 2m dx2 This is called the time-independent Schrodinger equation. Expectation Value Till now we have associated the eigenvalue of an operator with the value of the corresponding observable quantity. But there are situations when the state of the system is not an eigenstate of the operator of one’s interest. How does one talk of the value of an observable in such a situation. For this general situation one can use the expectation value or mean value of an operator  is defined as ψ ∗ (x)  ψ(x) dx R ⟨Â⟩ = R ∗ ψ (x) ψ(x) dx Problem: A particle is represented at time t = 0 by a wave function,   A (a2 − x2 ) if − a ≤ x ≤ +a ψ(x, 0) =  0 otherwise (a) Determine the normalization constant A. The normalization condition is Z +∞ |ψ(x, 0)|2 dx = 1 −∞ Z a Z a 2 2 2 2 2 |A| (a − x ) dx = 1 =⇒ |A| (a4 + x4 − 2a2 x2 ) dx = 1 Z−aa −a 2|A|2 (a4 + x4 − 2a2 x2 ) dx = 1 0 14  Safia Ahmad a x5 2a2 x3 a5 2a5    2 4 25 2|A| a x+ − = 1 =⇒ 2|A| a + − =1 5 3 0 5 3   2 5 1 2 16 2 5 2|A| a 1 + − = 1 =⇒ |A| a = 1 5 3 15 r 2 15 15 |A| = 5 =⇒ |A| = 16 a 16 a5 (b) What are the expectation values of x̂, x̂2 , p̂, p̂2 at time t = 0? The expectation value of x̂ is Z +∞ Z a ∗ 2 ⟨x̂⟩ = ψ (x, 0) x̂ ψ(x, 0) dx = |A| x (a2 − x2 )2 dx =⇒ ⟨x̂⟩ = 0 −∞ −a since integrand is an odd function. Expectation value of x̂2 is Z +∞ Z a 2 ∗ 2 2 ⟨x̂ ⟩ = ψ (x, 0) x̂ ψ(x, 0) dx = |A| x2 (a2 − x2 )2 dx −∞ −a Z a Z a 2 2 4 4 2 2 2 = 2|A| x (a + x − 2a x ) dx = 2|A| (a4 x2 + x6 − 2a2 x4 ) dx 0 −a 5 a (2)15 4 x3 x7 3 (a7 − 0) 5     2x 15 4 (a − 0) 2 (a − 0) = a + − 2a = a + − 2a 16 a5 3 7 5 0 8 a5 3 7 5 15a7 1 1 2 15a2 8 a2   2 = + − = =⇒ ⟨x̂ ⟩ = 8 a5 3 7 5 8 105 7 Expectation value of p̂ is Z ∞ Z ∞   ∗ ∗ ∂ ⟨p̂⟩ = ψ (x, 0) p̂ ψ(x, 0) dx = ψ (x, 0) −iℏ ψ(x, 0) dx −∞ −∞ ∂x Z a Z a 2 2 2 ∂ 2 2 2 = |A| (−iℏ) (a − x ) (a − x ) dx = |A| (−iℏ) (a2 − x2 )(−2x) dx ∂x Z −aa −a = |A|2 (2iℏ) x (a2 − x2 ) dx =⇒ ⟨p̂⟩ = 0 −a Expectation value of p̂2 is Z ∞ Z ∞  2 2 ∗ 2 ∗ ∂ ⟨p̂ ⟩ = ψ (x, 0) p̂ ψ(x, 0) dx = ψ (x, 0) −iℏ ψ(x, 0) dx −∞ −∞ ∂x Z a  2  Z a 2 2 2 2 ∂ 2 2 2 2 ∂ = |A| (a − x ) −ℏ 2 (a − x ) dx = |A| (−ℏ ) (a2 − x2 ) (−2x) dx −a ∂x −a ∂x Z a 3 a   x = |A|2 (2ℏ2 ) (a2 − x2 ) dx = 2|A|2 (2ℏ2 ) a2 x − −a 3 0 a3 2a3 5ℏ2        15 15 = 4ℏ2 a 3 − = ℏ 2 =⇒ ⟨p̂ 2 ⟩ = 16 a5 3 4 a5 3 2a2 15  Safia Ahmad So, the uncertainty in position, x̂ can be calculated as r p a2 a ∆x = ⟨x̂2 ⟩ − ⟨x̂⟩2 = − 0 =⇒ ∆x = √ 7 7 and the uncertainty in momentum p̂ is r √ p 5ℏ2 5ℏ ∆p = ⟨p̂2 ⟩ − ⟨p̂⟩2 = 2 − 0 =⇒ ∆p = √ 2a 2a Therefore, the uncertainty product is √ r r a 5ℏ 5 10 ℏ ℏ ∆x∆p = √ √ = ℏ= > 7 2a 14 7 2 2 Hence, this is consistent with the Heisenberg uncertainty principle. Problem: A particle of mass m is represented by a wave function, mx2    ψ(x, t) = Aexp −a + it ℏ where A and a are real constants. (a) Find A. Normalization condition is Z ∞ ψ ∗ (x, 0) ψ(x, 0) dx = 1 −∞ Z ∞ mx2 mx2       2 =⇒ A exp −a − it exp −a + it dx = 1 −∞ ℏ ℏ Z ∞ 2amx2   2 =⇒ A exp − dx = 1 −∞ ℏ r  Z ∞ r  2 π −αx2 π =⇒ A =1 ∵ e dx = (2am/ℏ) −∞ α r  1/4 2 πℏ 2am =⇒ A = 1 =⇒ A = 2am πℏ (b) What are the expectation values of x̂, x̂2 , p̂, p̂2 ? The expectation value of x̂ is Z ∞ Z ∞ ∗ ⟨x̂⟩ = ψ (x, t) x̂ ψ(x, t) dx = x |ψ(x, t)|2 dx −∞ Z ∞  2  −∞ 2amx = |A|2 x exp − dx =⇒ ⟨x̂⟩ = 0 −∞ ℏ 16  Safia Ahmad Expectation value of x̂2 is Z ∞ ∞ 2amx2 Z   2 2 2 2 2 ⟨x̂ ⟩ = x |ψ(x, t)| dx = |A| x exp − dx −∞ −∞ ℏ Using the integral Z ∞ 2 Γ((n + 1)/2) xn e−αx dx = −∞ α(n+1)/2 we have ∞ 1/2 2amx2 Z    2 2 2 2am Γ(3/2) ⟨x̂ ⟩ = |A| x exp − dx = −∞ ℏ πℏ (2am/ℏ)3/2  1/2  1/2  √ 2am ℏ ℏ π ℏ = =⇒ ⟨x̂2 ⟩ = πℏ 2am 2am 2 4am Expectation value of p̂ is Z ∞ Z ∞   ∗ ∗ ∂ ⟨p̂⟩ = ψ (x, t) p̂ ψ(x, t) dx = ψ (x, t) −iℏ ψ(x, t) dx −∞ −∞ ∂x   ∂ ψ(x, t) 2amx 2 Since, = A − e−amx /ℏ e−iat ∂x ℏ     ∂ 2 −amx2 /ℏ iat 2amx −amx2 /ℏ −iat 2amx 2 =⇒ ψ ∗ (x, t) ψ(x, t) = |A| e e − e e 2 = |A| − e−2amx /ℏ ∂x ℏ ℏ  Z ∞ 2am 2 ∴ ⟨p̂⟩ = |A|2 (−iℏ) − x e−2amx /ℏ dx =⇒ ⟨p̂⟩ = 0 ℏ −∞ Expectation value of p̂2 is Z ∞  Z 2∞  2 ∗ 2 2 ∂ ∗ ⟨p̂ ⟩ = ψ (x, t) p̂ ψ(x, t) dx = ψ (x, t) −ℏ ψ(x, t) dx −∞ −∞ ∂x2   ∂ ψ(x, t) 2amx 2 Since, = A − e−amx /ℏ e−iat ∂x ℏ 2 ∂2    2am −amx2 /ℏ −iat 2amx 2 =⇒ 2 ψ(x, t) = A − e e +A − e−amx /ℏ e−iat ∂x ℏ ℏ    2 2am 2am = − ψ(x, t) + x2 ψ(x, t) ℏ ℏ " 2 # Z ∞   2am 2am ∴ ⟨p̂2 ⟩ = −ℏ2 ψ ∗ (x, t) − ψ(x, t) + x2 ψ(x, t) dx −∞ ℏ ℏ  Z ∞  2 Z ∞ 2 2am 2 2 2am = −ℏ − |ψ(x, t)| dx − ℏ x2 |ψ(x, t)|2 dx ℏ −∞ ℏ −∞ ℏ = 2amℏ − 4a2 m2 ⟨x̂2 ⟩ = 2amℏ − 4 a2 m2 4 am 2 = 2amℏ − amℏ =⇒ ⟨p̂ ⟩ = amℏ 17  Safia Ahmad So, the uncertainty in position, x̂ can be calculated as r r p ℏ ℏ ∆x = ⟨x̂2 ⟩ − ⟨x̂⟩2 = − 0 =⇒ ∆x = 4am 4am and the uncertainty in momentum p̂ is p √ √ ∆p = ⟨p̂2 ⟩ − ⟨p̂⟩2 = amℏ − 0 =⇒ ∆p = amℏ Therefore, the uncertainty product is ℏ √ r r ℏ2 ℏ ∆x∆p = amℏ = = 4am 4 2 Hence, this is just barely consistent with the Heisenberg uncertainty principle. The given wave function gives the lowest bound on the uncertainty principle. Particle in a Box One-dimensional potential box is also called infinite potential well. Considering a case where a particle of mass m is not moving freely but is trapped inside a one-dimensional box of length L. Particle is free inside a one-dimensional box. However, as the two walls of the box are rigid, particle can neither go outside nor penetrate through the wall. Thus, the potential is zero inside the box and infinity at the walls, i.e.   0 for 0 < x < L V (x) = (16)  ∞ for x ≤ 0, x ≥ L Thus, we only need to solve the Schrodinger equation inside the potential well. The time- independent Schrodinger equation is ℏ2 d2 ψ − = E ψ(x) 2m dx2 d2 ψ 2mE + 2 ψ=0 dx2 ℏ r 2 dψ 2mE + k 2 ψ = 0, k= dx2 ℏ2 18  Safia Ahmad The general solution is ψ(x) = A cos (kx) + B sin (kx) where, A and B are arbitrary constants which can be fixed by boundary conditions. As potential is infinity at the boundaries, the probability of finding a particle outside the box is zero and therefore the wave function ψ(x) must vanish for x < 0 and x > L. Also, since the particle cannot penetrate the boundary, the probability of finding it inside the walls should be zero. As the probability density of finding the particle at a position x is |ψ(x)|2 and since particle cannot penetrate the walls, we have, |ψ(0)|2 = 0 and |ψ(L)|2 = 0 =⇒ ψ(0) = 0 and ψ(L) = 0, i.e. ψ(x) = 0 at x = 0, L i.e. ψ(0) = 0 = ψ(L) Applying the boundary conditions, we get ψ(0) = 0 =⇒ A = 0 therefore, ψ(x) = B sin (kx) ψ(L) = 0 =⇒ B sin (kL) = 0 =⇒ kL = nπ  nπx  =⇒ ψ(x) = B sin L The energy eigenvalues can be obtained from k L = nπ =⇒ k 2 L2 = n2 π 2 2mE n2 π 2 n2 π 2 ℏ2 =⇒ = =⇒ E n = (17) ℏ2 L2 2mL2 Unlike classical particle, a quantum particle in the infinite square well cannot have just any energy; it has to be one of the allowed energy levels. 19  Safia Ahmad The above wave function can be normalized using the normalization condition, Z L ψ ∗ (x) ψ(x)dx = 1 0 Z L 2 nπx   2 =⇒ |B| sin dx = 1 0 L |B|2 L Z    2nπx =⇒ 1 − cos dx = 1 2 0 L L |B|2   L 2nπx =⇒ x− sin =1 2 2nπ L 0 |B|2 =⇒ (L) = 1 2 r 2 =⇒ B = (18) L Therefore, the wave function inside the potential well is r 2  nπx  ψn (x) = sin (19) L L These are the eigenfunctions corresponding to energy eigenvalues, En. The time-independent Schrodinger equation has an infinite set of solutions; one for each positive integer n. The wave function corresponding to n = 1 is the ground state while the wave functions that correspond to n > 1 are the excited states. Thus, a particle confined in a certain region can have only certain values of energies; other energy values are not allowed. That is, energy quantization is a consequence of restricting a microparticle to a certain region. 20  Safia Ahmad This can also be shown that Z L ψ1∗ (x)ψ2 (x)dx = 0 0 2 L Z  πx    2πx i.e. sin sin dx = 0 L 0 L L Thus, these wave functions are orthonormal, Z L ψn∗ (x)ψn′ (x)dx = δnn′ (20) 0 where δnn′ is the kronecker delta function,   1 when n = n′ δnn′ = (21) ̸ n′  0 when n = Thus, a trapped particle cannot have an arbitrary energy, as a free particle can. The fact of its confine- ment leads to restrictions on its wave function that allow the particle to have only certain specific energies and no others. The energies depend on the mass of the particle and on how the particle is trapped. Also, be- cause Planck’s constant is so small (h = 6.626 × 10−34 J.s), quantization of energy is noticeable only when m and L are small. The reason why the quantization of energy is not observed for large scales. Potential Barrier Figure shows the potential barrier of height V0 and thickness L. The region around the barrier can be divided into three regions. The potential energy of a beam of particles of mass m that are sent from the left on a potential barrier is,     0 xL 21  Safia Ahmad Tunneling: E < V0 According to classical mechanics, the particles coming from left, with energy E < V0 , will stop at x = 0 and will bounce back with the same mag- nitude of momentum without penetrating. The classical particles cannot penetrate through the barrier at x = 0 and there will be total reflection. While according to quantum mechanics, particle bounces back at x = 0 but now it has a finite probability of penetrating into the region although E < V0. The particles of mass m coming from left with energy E > V0 , are free for x < 0 and feel a repulsive potential V0 that starts at x = 0. For x < 0, V = 0, therefore the time-independent Schrodinger equation is, ℏ2 d2 ψ1 − = E ψ1 (x) 2m dx2 d2 ψ1 2mE + 2 ψ1 = 0 dx2 ℏ r 2 d ψ1 2mE + k12 ψ1 = 0, k1 = (23) dx2 ℏ2 The general solution of the above differential equation is ψ1 (x) = A eik1 x + B e−ik1 x (24) where, A and B are arbitrary constants. The first term represents the wave moving in positive x -direction and the second term represents the wave moving in negative x -direction. For 0 ≤ x ≤ L, the Schrodinger equation becomes ℏ2 d2 ψ2 − + V0 ψ2 (x) = E ψ2 (x) 2m dx2 d2 ψ2 2mE 2mV0 + ψ2 − ψ2 = 0 dx2 ℏ2 ℏ2 d2 ψ2 2m(V0 − E) − ψ2 = 0 dx2 ℏ2 r d2 ψ2 2m(V0 − E) 2 − k22 ψ2 = 0, k2 = dx ℏ2 22  Safia Ahmad The general solution of the above differential equation is ψ2 (x) = C ek2 x + D e−k2 x (0 ≤ x ≤ L) For x > L, V = 0, therefore the time-independent Schrodinger equation is, ℏ2 d2 ψ3 − = E ψ3 (x) 2m dx2 d2 ψ3 2mE + 2 ψ3 = 0 dx2 ℏ r 2 d ψ3 2mE 2 + k12 ψ3 = 0, k1 = (25) dx ℏ2 The general solution of the above differential equation is ψ3 (x) = E eik1 x + F e−ik1 x (26) where, E and F are arbitrary constants. The first term represents the wave moving in positive x -direction and the second term represents the wave moving in negative x -direction. Since no wave is reflected from the region x > 0 to the left, the constant F must vanish, therefore ψ3 (x) = E eik2 x , x>L As particles are initially incident on the potential barrier from the left, they can be reflected or transmitted at x = 0. Reflection coefficient, which is the fraction of incident particles that are reflected back at x = 0, is given as reflected current density Jreflected R= = incident current density Jincident while transmission coefficient, which is the fraction of incident particles that are transmitted into the region x ≥ 0, is given as transmitted current density Jtransmitted T = = incident current density Jincident where J is the probability current density which tells us the rate at which probability is “flowing” past the point x. That is, it tells us about the flux of particle that are moving from one position to other. The probability current density is given as   iℏ d ∗ ∗ d J= ψ(x) ψ (x) − ψ (x) ψ(x) (27) 2m dx dx 23  Safia Ahmad Since incident wave is ψi (x) = A eik1 x =⇒ ψi∗ (x) = A∗ e−ik1 x , therefore d d ∗ ψ(x) = ik1 A eik1 x , ψ (x) = −ik1 A∗ e−ik1 x dx dx Therefore, incident current density is given by iℏ Jincident = (A eik1 x (−ik1 A∗ e−ik1 x ) − A∗ e−ik1 x (ik1 A eik1 x )) 2m iℏ = (−ik1 |A|2 − ik1 |A|2 ) 2m iℏ = (−2ik1 |A|2 ) 2m ℏk1 2 =⇒ Jincident = |A| (28) m And since transmitted wave is ψt (x) = E eik1 x =⇒ ψt∗ (x) = E ∗ e−ik1 x , therefore d d ∗ ψt (x) = ik1 E eik1 x , ψ (x) = −ik1 E ∗ e−ik1 x dx dx t Therefore, transmitted current density is given by iℏ Jtransmitted = (E eik1 x (−ik2 E ∗ e−ik1 x ) − E ∗ e−ik1 x (ik1 E eik1 x )) 2m iℏ = (−ik1 |E|2 − ik1 |E|2 ) 2m iℏ = (−2ik1 |E|2 ) 2m ℏk1 2 =⇒ Jtransmitted = |E| (29) m Thus, transmission coefficient is, ℏk1 2 m |E|2 T = |E| = m ℏk1 |A|2 |A|2 The constants, A, B, C, D and E can be determined from the boundary conditions at x = 0 and x = L, dψ1 (x) dψ2 (x) ψ1 (0) = ψ2 (0), = dx x=0 dx x=0 dψ2 (x) dψ3 (x) ψ2 (L) = ψ3 (L), = dx x=L dx x=L 24  Safia Ahmad These boundary conditions give A+B =C +D (30) −ik2 ik1 (A − B) = k2 (C − D) =⇒ A − B = (C − D) (31) k1 C ek2 L + D e−k2 L = E eik1 L (32) ik1 k2 C ek2 L − D e−k2 L = ik1 E eik1 L =⇒ C ek2 L − D e−k2 L = E eik1 L
(33) k2 Adding and subtraction of conditions (32) and (33) give   E ik1 L ik1 C= e 1+ e−k2 L 2 k2   E ik1 L ik1 D= e 1− ek2 L 2 k2 Substituting the expressions for C and D in boundary condition (30) will give   E ik1 L k2 L −k2 L ik1 −k2 L k2 L A+B = e e +e + (e −e ) 2 k2   E ik1 L 2ik1 A+B = e 2cosh (k2 L) − sinh (k2 L) 2 k2   B E ik1 L ik1 1+ = e cosh (k2 L) − sinh (k2 L) (34) A A k2 Substituting the expressions for C and D in boundary condition (31) will give   ik2 E ik1 L −k2 L k2 L ik1 −k2 L k2 L A−B =− e e −e + (e +e ) k1 2 k2    E ik1 L ik2 ik2 2ik1 A−B = e − (−2sinh (k2 L)) − cosh (k2 L) 2 k1 k1 k2   B E ik1 L ik2 1− = e cosh (k2 L) + sinh (k2 L) (35) A A k1 25  Safia Ahmad Adding (34) and (35) gives     E ik1 L k2 k1 2 = e 2cosh (k2 L) + i − sinh (k2 L) A k1 k2  2 k2 − k12    A 1 ik1 L = e 2 cosh (k2 L) + i sinh (k2 L) E 2 k1 k2 A∗  2 k2 − k12    1 −ik1 L =⇒ ∗ = e 2 cosh (k2 L) − i sinh (k2 L) E 2 k1 k2 |A|2 (k22 − k12 )2   1 2 2 ∴ = 4 cosh (k2 L) + sinh (k2 L) |E|2 4 k12 k22 |A|2 (k22 − k12 )2   2 2 =⇒ = cosh (k2 L) + sinh (k2 L) |E|2 4k12 k22 −1 |E|2 (k22 − k12 )2  2 2 ∴ T = = cosh (k2 L) + sinh (k2 L) |A|2 4k12 k22 Since cosh2 (k2 L) = 1 + sinh2 (k2 L), the expression of transmission coefficient, T can be rewritten as −1 (k22 − k12 )2  2 2 T = 1 + sinh (k2 L) + sinh (k2 L) 4k12 k22  2 −1 (k2 − k12 )2   2 T = 1+ + 1 sinh (k2 L) 4k12 k22 2 !−1 1 k12 + k22  T = 1+ sinh2 (k2 L) 4 k1 k2 Since, 2mE 2 2m(V0 − E) k12 = , k2 = ℏ2 ℏ2 Therefore, 2 2  k12 + k22 ℏ4    2mE 2m(V0 − E) = + k1 k2 ℏ2 ℏ2 (2mE)(2m(V0 − E)) 2 k12 + k22 (2mV0 )2 V02    =⇒ = = k1 k2 4m2 E(V0 − E) E(V0 − E) So, the transmission coefficient, which is the fraction of incident beam that gets through the barrier, can be written as  1  V02  p −1 2 2 2 T = 1+ sinh 2m(V0 − E)L /ℏ (36) 4 E(V0 − E) −1 4E(V0 − E) + V02 sinh2 (k2 L)  =⇒ T = 4E(V0 − E) 4E(V0 − E) =⇒ T = p  (37) 4E(V0 − E) + V02 sinh2 2m(V0 − E)L2 /ℏ2 26  Safia Ahmad Assuming that the potential barrier V0 is high as compared to the energy of the particles E then V02 V02 V02 ≫ 1 =⇒ 1 + ≈ 4E(V0 − E) 4E(V0 − E) 4E(V0 − E) Also assume that the barrier is wide enough for the wave function inside the potential barrier, ψ2 to be severely weakened so that ek2 L ≫ e−k2 L then 1 k2 L 1 sinh (k2 L) = (e + e−k2 L ) ≈ ek2 L 2 2 Therefore, from (36), we have " 2 #−1 −1 V02 V02   1 k2 L 2k2 L T = e = e 4E(V0 − E) 2 16E(V0 − E)   16E(V0 − E) −2k2 L 16E E =⇒ T = e =⇒ T = 1− e−2k2 L (38) V02 V0 V0 The prefactor varies much less with E and V0 than does the exponential. Also, the prefactor is always of the order of magnitude of 1 in value. Therefore, a reasonable approximation for transmission probability is T = e−2k2 L (39) Clearly, T is finite. This means that the probability for the transmission of the particles into the region x ≥ L is not zero. This is a purely quantum mechanical effect known as the tunneling effect which is due to the wave aspect of microscopic objects. That is, quantum mechanical objects can tunnel through classically impenetrable barriers as shown in the figure. 27