Quantum Mechanics
Part I: Wave Properties of Matter Louis De Broglie 1892-1987 Nobel Prize in Physics 1929
So Electromagnetic radiations shown to behave like: Waves: Interferences, diffraction Particles: Photoelectric & Compton effects Particles shown to behave like: Particles (duh!): Rutherford scattering, Waves???: stationary orbits in the Bohr model? By the 1920 s, nobody made the fateful leap to assume that particles may also have a wave-like behavior
1924: De Broglie defends his thesis Suggests that particles could have wave-like properties. Uses Einstein s special theory of relativity together with Planck s quantum theory to establish the wave properties of particles.
De Broglie wavelength Energy in the special theory of relativity: E & = pc & + mc & & Photon: m = 0 E = pc Photoelectric Effect: E = hυ Combining both expression (with υ = # $ ): λ = h p De Broglie suggest this relation for photons extends to ALL particles à Matter Waves
Exercise De Broglie Wavelengths Microscopic / macroscopic objects What is the De Broglie wavelength of a 50 ev electron (non-relativistic approach OK)? l = h/p = 0.17 nm ~ 10-10 m (interatomic distance in crystals) What is the De Broglie wavelength of a Tennis Ball (57g) at 25 m/s (about 56 mph) [assuming the tennis ball to be a point particle!]? l = h/p = 4.7 x 10-34 m (about 10 19 times small than the size of a nucleon!)
A new look at Bohr s model of the atom Bohr Quantization: Bohr s hypothesis of stationary orbits: Angular momentum: L = r p = mvr = nh/2π With momentum p = mv and p = 9 9:, we get L = pr = $ $ If electron has a wave-like behavior (stationary orbit standing wave) Orbit: D = 2πr = nλ r = =$ &> L = 9: $ with r = =$ &> : L = nh 2π = nħ With De Broglie s wavelength, Bohr s quantization appears naturally
Electron scattering 1925: C. Davisson and L.H.Germer observe electron scattering and interference effects --> Experimental evidence of the wave-like properties of the electron
Electron diffraction (transmission) 120-keV electron on quasicrystal Diffraction on a powder
A word on particle energies De Broglie Wavelength: l = h/p In the non-relativistic approximation, p = (2mK) with K, the kinetic energy of the electron à The higher K, the higher p Higher electron energies (smaller wavelength) à Sharper probe of the material.
Neutron diffraction The Spallation Neutron Source (SNS) facility Oak Ridge National Laboratory
Part II: Wave or Particle? Niels Bohr 1885-1962 Nobel Prize in Physics 1922 Werner Heisenberg 1901-1976 Nobel Prize in Physics 1932
Wave or particle? A particle is a localized object, a wave is not. How does it go from waves to particles? Wave-particle duality?
Location of a wave? Consider a single wave: y(x,t) = A cos(kx-wt) POSITION: Where is the wave? Everywhere (within the limit of the definition of the wave) MOMENTUM: The wave number k=2p/l and the De Broglie wavelength l = h/p à k = 2pp/h = p/ħ In this case: k is well defined à the momentum p =ħk is also very well defined x (arbitrary units)
Consider 2 waves: y(x,t) = A cos(k 1 x-w 1 t) + A cos(k 2 x-w 2 t) with k 1 close to k 2 [BEATS!] = 2 A cos(½[(k 1 -k 2 )x-(w 1 -w 2 )t]) cos(½[(k 1 +k 2 )x-(w 1 +w 2 )t]) Posing Dk=k 1 -k 2 and Dw=w 1 -w 2 : y(x,t) = 2 A cos(½[dkx-dwt]) cos(½[(k 1 +k 2 )x-(w 1 +w 2 )t]) POSITION: Where is the particle? Still not really localized, but now some regions of space seem more likely to contain the particle than others. l envelope /2 Location of two superposed waves? The particle is between x 1 and x 2 : ½Dk x 2 - ½Dk x 1 = p Posing Dx = x 2 -x 1 : DxDk=2p x 1 x 2 MOMENTUM: The momentum is now only known within Dp=ħDk We gained a better knowledge of x, but we are losing our perfect knowledge of p
k=10 More waves? 1 Waves Dk max =0 Adding more waves: Dx decreases 2 Waves Dk max =2.5 Dk increases Dx.Dk~1 5 Waves Dk max =5 10 Waves Dk max =10 20 Waves Dk max =20 One can show similarly that: Dw.Dt~1 (Energy: E = hn = ħ(2p)n = ħw) Uncertainty relations between: Position and Momentum Energy and Time Particles can be represented by wave packets
The double-slit experiment with photons From: Quantum (J. Al-Khalili) What about sending one photon at a time? INTERFERENCE EFFECT!
The double-slit experiment with electrons 1961: C.Jonsson manages to produce very narrow slits to observe the interference effects due to the wavelike behavior of the electrons.
Which slit? (I) Pictures from: Quantum (J. Al-Khalili) One slit blocked No interference
A apparatus is placed to detect when one photon goes through the top slit. Detector On Which slit? (II) Pictures from: Quantum (J. Al-Khalili) NO INTERFERENCE! Detector Off INTERFERENCE!
Measurements and quantum systems To determine which slit the electron went through: We set up a light shining on the double slit and use a powerful microscope to look at the region. After the electron passes through one of the slits, light bounces off the electron; we observe the reflected light, so we know which slit the electron came through. Use a subscript ph to denote variables for light (photon). Therefore the momentum of the photon is: p @9 = h > h λ @9 d The momentum of the electrons will be on the order of p CD = h ~ h λ CD d The difficulty is that the momentum of the photons used to determine which slit the electron went through is sufficiently great to strongly modify the momentum of the electron itself, thus changing the direction of the electron! The attempt to identify which slit the electron is passing through will in itself change the interference pattern.
Mach-Zender interferometer
Principle of complementarity Bohr s principle of complementarity: it is not possible to describe physical observables simultaneously in terms of both particles and waves Consequence: once you measure the wave- (particle-) like behavior of a phenomenon, you cannot measure a property linked to its particle- (wave-) like behavior. e.g. once you determine that the photon/electron has made it through a given slit, you reveal its particle-like behavior, therefore you cannot observe the interference phenomenon anymore (linked to the wave behavior)! Niels Bohr s Coat of Arms (1947) Opposites are Complementary
Heisenberg uncertainty principle (I) It is impossible to measure simultaneously, with no uncertainty, the precise values of k and x for the same particle. The wave number k may be rewritten as k = 2π λ = 2π h p = p 2π h = p ħ For the case of a Gaussian wave packet we have: k x = p K ħ x = 1 2 Thus for a single particle we have Heisenberg s uncertainty principle: p K x ħ 2
Heisenberg uncertainty principle (II) p K x ħ 2 with x~ l & [we know the particle is located between 0 and l] If we are uncertain as to the exact position of a particle, for example an electron somewhere inside an atom, the particle can t have zero kinetic energy. K UV= = p & UV= 2m p & 2m ħ& 2ml & The energy uncertainty of a Gaussian wave packet is: E = h ν = h Δω 2π = ħδω combined with the angular frequency relation: ω t = E ħ t = 1 2 Energy-Time Uncertainty Principle: E t ħ 2
Exercise A proton is confined in a Uranium nucleus (diameter: d=16fm). Determine the minimum kinetic energy [non-relativistic] of the proton confined within the diameter of the uranium nucleus.
Exercise It is possible for some fundamental particles to violate conservation of energy by creating and quickly reabsorbing another particle. For example, a proton can emit a p + according to p à n + p + when the n represents the neutron. The p + has a mass of 140 MeV/c 2. The re-absorption must occur within a time Dt consistent with the uncertainty principle. Considering that example, by how much DE is energy conservation violated? [Ignore kinetic energy] For how long Dt can the p + exist? Assuming that the p + is moving at nearly the speed of light, how far from the nucleus could it get in the time Dt?
The Copenhagen interpretation From: Quantum (J. Al-Khalili) No one knows what happens behind the quantum curtain but we should not care! Physics only depends on the outcomes of measurements (in other words: only the results count )
Other interpretations De Broglie - Bohm Many-Worlds
Part III: The Schrödinger Equation Erwin Schrödinger 1887-1961 Nobel Prize in Physics 1933
The Schrödinger Wave Equation The Schrödinger wave equation in its time-dependent form for a particle of energy E moving in a potential V in one dimension is: iħ Ψ(x, t) t = ħ& & Ψ x, t 2m x & + VΨ(x, t) The extension into three dimensions is iħ Ψ t = ħ& 2m & Ψ x & + & Ψ y & + & Ψ z & + VΨ(x, y, z, t) where i = 1 is an imaginary number.
General solution The general form of the wave function is: Ψ x, t = Ae V dkefg = A cos kx ωt + i sin kx ωt which also describes a wave moving in the x direction. In general the amplitude may also be complex. The wave function is also not restricted to being real. Notice that the sine term has an imaginary number. Only the physically measurable quantities must be real. These include the probability, momentum and energy.
Y: The Wave Function People gave some long thoughts to what was Y, the wave-function (even Schrödinger himself): Erwin [Schrödinger] with his psi can do Calculations quite a few But one thing has not been seen Just what does psi really mean From: Walter Hückel, translated by Felix Bloch The Schrödinger equation allows to calculate analytically [exactly] quantum problems, but it does not reveal the nature of Y. Schrödinger called it field scalar. Shut up and Calculate! from R.Feynman (and many others!)
Solvay Congress 1927
Properties of the wave function When squared, the wave function is a probability density (Max Born 1926). The probability P(x) dx of a particle being between x and x+dx is given in the equation: P x dx = Ψ x, t Ψ x, t dx The probability of the particle being between x 1 and x 2 is given by K p P = o Ψ Ψdx The wave function must also be normalized so that the probability of the particle being somewhere on the x axis is 1. K q o Ψ x, t Ψ x, t dx = 1 rs es
Boundary conditions In order to avoid infinite probabilities, the wave function must be finite everywhere. In order to avoid multiple values of the probability, the wave function must be single valued. For finite potentials, the wave function and its derivative must be continuous. This is required because the second-order derivative term in the wave equation must be single valued. (There are exceptions to this rule when V is infinite.) In order to normalize the wave functions, they must approach zero as x approaches infinity. Solutions that do not satisfy these properties do not generally correspond to physically realizable circumstances
Time Independent Schrödinger Wave Equation (I) Simplification possible if the potential is not time-dependent [happens in many cases] The dependence on time and position can then be separated in the Schrödinger wave equation. Let: Ψ x, t = ψ x f(t) which yields: iħψ x f(t) t Now divide by the wave function: = ħ& f t 2m iħ 1 f(t) & ψ x x & df(t) dt + V x ψ x f(t) = ħ& 2mψ x d & ψ x dx & + V x The left side depends only on time, and the right side depends only on spatial coordinates. Hence each side must be equal to a constant. The time dependent side is: iħ 1 df(t) = B f(t) dt
We integrate both sides and find: where C is an integration constant that we may choose to be 0. Therefore ln f = Bt iħ This determines f(t) to be Renaming the constant B E: Time Independent Schrödinger Wave Equation (II) iħ o df f f t = e {g Vħ = e ev{g ħ iħ 1 df(t) f(t) dt = o Bdt iħ ln f = Bt + C = E ħ& d & ψ x 2m dx & + V x ψ x = Eψ x This is known as the time-independent Schrödinger wave equation, and it is a fundamental equation in quantum mechanics. (Energies E are the eigenvalues of the equation)
Solution stationary state The wave function can be written as: Ψ x, t = ψ(x)e evfg The probability density becomes: Ψ x, t Ψ x, t = ψ & x e rvfg e evfg = ψ & x The probability distributions are constant in time. This is a standing wave phenomena that is called the stationary state.
Probabilities and expectation values Because we are dealing with probabilities, we are likely to get different results for many measurements of the physical observables [position, momentum, energy etc ] The average measurement [called expectation value] of a given quantity can be calculated using wave functions. The expectation value of quantity x is denoted x in quantum mechanics.
Continuous expectation values We can change from discrete to continuous variables by using the probability P(x, t) of observing the particle at a particular x. Discrete: Continuous: x = N V Vx V V N V x = rs es rs es xp x dx P x dx Using the wave function, the expectation value is: The expectation value of any function g x for a normalized wave function: x = rs es rs es xψ x, t Ψ x, t dx Ψ x, t Ψ x, t dx rs g(x) = o Ψ x, t g(x)ψ x, t dx es Abbreviated notation: ψ g ψ
Operators Operator Expectation value Momentum Energy p = iħ x E = iħ t p = iħ o E = iħ o rs es rs es Ψ x, t Ψ x, t Ψ x, t x Ψ x, t t dx dx Exercise Show that the operators p and E indeed lead to a measurement of the momentum and energy respectively.
Part IV: Solving the Schrödinger equation Erwin Schrödinger 1887-1961 Nobel Prize in Physics 1933
1D-infinite square well potential The simplest such system is that of a particle trapped in a box with infinitely hard walls that the particle cannot penetrate. This potential is called an infinite square well and is given by: V x = ƒ + x 0, x L 0 0 < x < L
Solving the Schrödinger equation V x = ƒ + x 0, x L 0 0 < x < L The wave function must be zero where the potential is infinite. Where the potential is zero inside the box, the Schrödinger wave equation becomes: d & ψ d & x = 2mE ħ & ψ = k& ψ with k = 2mE ħ & The general solution is: ψ x, t = A sin kx + B cos kx
Quantization Boundary conditions of the potential dictate that the wave function must be zero at x = 0 and x = L. This yields valid solutions for integer values of n such that kl = nπ. The wave function is now We normalize the wave function rs o ψ = x ψ = x dx = 1 es ψ = x = A sin nπx L A & o sin & ˆ nπx L dx = 1 The normalized wave function becomes ψ = x = (with n=1,2,3 ) 2 L sin nπx L These functions are identical to those obtained for a vibrating string with fixed ends.
Quantized energy The quantized wave number now becomes Solving for the energy yields E = = n & π& ħ & 2mL & k = = nπ L = 2mE = ħ & with n=1,2,3 Note that the energy depends on the integer values of n. Hence the energy is quantized and nonzero. The special case of n = 1 is called the ground state energy: wave functions probability density position energy E Š = π& ħ & 2mL &
Finite square-well potential ħ& d & ψ x 2m dx & + V x ψ x = Eψ x Study of a particle trapped between x=0 and x=l in a realistic potential (no infinite walls). V x = V x 0 0 0 < x < L V x L Region I Region II Region III
Finite square-well potential ħ& d & ψ x 2m dx & + V x ψ x = Eψ x In regions I and III, the Schrödinger equation becomes: ħ& 1 d & ψ 2m ψ dx & = E V By posing α & = 2m V E ħ &, the equation can be written: d & ψ dx & = α& ψ Solutions has to tend to 0 when x tends ± : ψ x = Ae ŽK Region I, x<0 ψ x = Be ežk Region III, x>l In region II, the Schrödinger equation is: d & ψ d & x = k& ψ with k = 2mE ħ & Solution: ψ x = Ce VdK + De evdk Region II, 0 x L
Finite square-well potential Boundary Conditions Continuity: ψ x = 0 = ψ x = 0 & ψ x = L = ψ x = L Smoothness : d dx ψ x = 0 = d dx ψ x = 0 d & dx ψ x = L = d ψ x = L dx The wave function is non-zero outside of the box
A more realistic potential: the harmonic oscillator Simple harmonic oscillators describe many physical situations: springs, diatomic molecules and atomic lattices. Substituting V(x)=½k(x-x 0 ) 2 [assuming x 0 =0 for simplification] into the wave equation: d & ψ dx & = 2m ħ & E κx& 2 ψ = 2mE ħ & + mκx& ħ & ψ Introducing: α & = U &U ħp and β = ħp, we get: d & ψ dx & = α& x & β ψ
Wave function solutions: ψ = x = H = x e ežkp & Solutions H n (x): Hermite polynomial functions Energy Solutions: E = = n + 1 2 ħ κ m = n + 1 2 ħω E = 1 2 ħω
Barriers and tunneling Barrier: region II (V=V 0 >0) Assuming the energy of the particle E larger than V 0 d & ψ dx & = k& ψ k = k = 2mE ħ k = 2m E V ħ Region I (V=0): d & ψ dx & + 2m ħ & Eψ = 0 Region II (V=V 0 ): Region III (V=0): d & ψ dx & d & ψ dx & + 2m ħ & E V ψ = 0 + 2m ħ & Eψ = 0
Solutions From the 3 preceding equations: Region I (V=0): ψ x = Ae Vd K + Be evd K Region II (V=V 0 ): Region III (V=0): ψ ψ x = Ce Vd K + De evd K x = Fe Vd K + Ge evd K Then boundary conditions, etc
Transmitted / reflected waves ψ x, incident = Ae Vd K ψ x, reflected = Be evd K ψ x, transmitted = Fe Vd K = Fe Vd K The probability of the particles being reflected R or transmitted T is: R = T = ψ ψ reflected & ψ incident & transmitted & ψ incident & = B B A A = F F A A And after much calculations: T = 1 + V & sin & k L 4E E V eš
What if What if the energy of the particle considered is smaller than the potential energy (E<V 0 )? Quantum Mechanics?
Tunneling Region II solution: ψ x = Ce K + De e K κ = 2m V E with ħ Transmission: T = 1 + V & sinh & κl 4E V E eš If κl 1, T can be simplified: T = 16 E V 1 E V e e& ˆ Exponential
Example: a-decay a-particle
Example: scanning tunneling microscope Heinrich Rohrer & Gerd Binnig Nobel Prize in Physics 1986 Electrons tunnel through the gap to be collected by the tip. Current collected is very sensitive to the distance between the tip and the surface.
Part V: The Hydrogen atom
Schrödinger equation and the Hydrogen atom Hydrogen = proton + electron system Potential: e& V r = 4πε r The 3D time-independent Schrödinger Equation: ħ& 1 2m ψ x, y, z & ψ x, y, z x & + & ψ x, y, z y & + & ψ x, y, z z & = E V r
Radial symmetry of the potential The Coulomb potential has a radial symmetry V(r): switch to the spherical polar coordinate system. Wave function: ψ r, θ, φ e& V r = 4πε r Schrödinger Equation (in spherical polar coordinate system): 1 r & r r& ψ r + 1 r & sin θ θ ψ sin θ θ + 1 & ψ r & sin & θ φ & + 2μ E V ψ = 0 ħ &
Separation of variables (I) In the spherical polar coordinate, the wavefunction: ψ r, θ, φ Solution may be the product of three functions: ψ r, θ, φ = R r f θ g φ Derivatives of y (the ones useful to solve the Schrödinger equation): ψ r = fg R r ψ θ = Rg f θ & ψ φ & = Rf & g φ &
Separation of variables (II) Into the Schrödinger equation: fg r & r r& R r + Rg r & sin θ r & sin & θ Rfg : θ sin θ f θ + Rf r & sin & θ & g φ & + 2μ E V Rfg = 0 ħ & sin& θ R r r& R r 2μ ħ & r& sin & θ E V sin θ f θ sin θ f θ = 1 g & g φ & = C Depends on r and q Depends on f only
Azimuthal equation One can solve the right hand part of the equation Assuming C = m l & (with a little bit of clairvoyance!): d & g dφ & = m l & g Solution: e VU l (with m l = integer (positive/negative) or zero)
Left side of the equation Now looking at the left side of the equation: sin& θ R r r& R r 2μ ħ & r& sin & θ E V sin θ f θ sin θ f θ = m l & And rearranging the equation by separating the variables (again): 1 R r r& R r + 2μr& ħ & E V = m & l sin & θ 1 f sin θ θ sin θ f θ = C Depends on r Depends on q
Radial and angular equations Setting the constant to l l + 1 [with even more clairvoyance!]: 1 d r & dr r& dr dr + 2μ ħ & ħ& l l + 1 E V 2μ r & R = 0 Radial equation 1 sin θ d dθ sin θ df dθ + l l + 1 m l & sin & θ f = 0 Angular equation
Solution of the radial equation (I) Solving for l = 0 m l = 0 : 1 d r & dr r& dr dr + 2μ ħ & ħ& l l + 1 E V 2μ r & R = 0 becomes 1 d r & dr r& dr dr + 2μ E V R = 0 ħ & Solving : r& : and introducing V r : d & R dr & + 2 r dr dr + 2μ e& E + ħ & 4πε r R = 0
Solution of the radial equation (II) d & R dr & + 2 r dr dr + 2μ e& E + ħ & 4πε r R = 0 Solution: R r = Aee: Solving the equation with: 1 & a + 2μ 2μe& E + ħ & 4πε ħ & 2 1 a r = 0 E = ħ& 2μa & = E Ground State Energy Level = 0 = 0 a = 4πε ħ & μe & The Bohr Radius
Boundary conditions lead to: l = 0, 1, 2, 3, m l l Radial Wave Functions R nl (r)
Solution of the angular and azimuthal equations Combining the angular and azimuthal solutions: Y lu θ, φ = f θ g φ Y lm (q,f): Spherical Harmonics
Solution of the Schrödinger equation for the Hydrogen atom ψ =lul r, θ, φ = R =l r Y lul θ, φ The three quantum numbers: n Principal quantum number l Orbital angular momentum quantum number m l Magnetic quantum number The boundary conditions: n = 1, 2, 3, 4,... Integer l = 0, 1, 2, 3,..., n 1 Integer m l = l, l + 1,..., 0, 1,..., l 1, l Integer The restrictions for quantum numbers: n > 0 l < n m l l
3 different electron states Probability distribution functions
Part VI: Quantum numbers
Solution of the Schrödinger equation for the Hydrogen atom ψ =lul r, θ, φ = R =l r Y lul θ, φ The three quantum numbers: n Principal quantum number l Orbital angular momentum quantum number m l Magnetic quantum number The boundary conditions: n = 1, 2, 3, 4,... Integer l = 0, 1, 2, 3,..., n 1 Integer m l = l, l + 1,..., 0, 1,..., l 1, l Integer The restrictions for quantum numbers: n > 0 l < n m l l
Principal Quantum Number n In the hydrogen atom, this is the number of the Bohr orbit (n=1,2,3 no upper limit) Associated with the solution of R(r) Quantized energy: E = = μ 2 e & 4πε ħ & 1 n & = E n & (-) sign: proton-electron system bound
Orbital angular momentum quantum number l Associated with the solutions of R r and f(θ) Boundary Conditions: l = 0, 1,, n 1 Classical Orbital Momentum: L = r p Quantum Orbital Momentum: L = l l + 1 ħ Note: l = 0 state à quantum orbital momentum L=0 This disagrees with Bohr s semi-classical planetary model of electrons orbiting a nucleus L = nħ.
More on quantum number l Energy is independent of the quantum number l, we say the energy level is degenerate with respect to l. Note: this is only true for the Hydrogen atom. States: l = 0 1 2 3 4 5 Letter s p d f g h (sharp) (diffuse) (principal) (fundamental) Atomic states are referred to by their n and l. A state with n = 2 and l = 1 is called a 2p state. Note: the boundary conditions require l < n.
The magnetic quantum number m l The angle φ is a measure of the rotation about the z axis. The solution for g φ specifies that m l is an integer and related to the z component of L. L = m l ħ Figure: the relationship of L, L, l and m l for l = 2 L = l l + 1 ħ = 6ħ is fixed by the value of l. Only certain orientations of L are possible however. Those are given by the projection of L on the quantization axis L. This is called space quantization. Note: One cannot know L exactly, as this would violate the uncertainty principle.
Intrinsic spin quantum number m Samuel Goudsmit and George Uhlenbeck in Holland proposed that the electron must have an intrinsic angular momentum and therefore a magnetic moment (1925) Paul Ehrenfest showed that the surface of the spinning electron should be moving faster than the speed of light! In order to explain experimental data, Goudsmit and Uhlenbeck proposed that the electron must have an intrinsic spin quantum number s = ½. [ Number of possible values: 2s+1 = 2 à m s =-½ or m s =½]
Intrinsic spin quantum number m Does not appear from the solutions of the Schrödinger equation Appears when solving the problem in a relativistic way For the electron: m s = +½ or m s = -½ The spinning electron reacts similarly to the orbiting electron in a magnetic field.
Exercise If the principal quantum number n for a certain electronic state is equal to 3, what are the possible values of the orbital (angular momentum) quantum number l? If the orbital quantum number l for a certain electronic state is equal to 2, what are the possible values for the magnetic quantum number m l? How many distinct electronic states are there with n=2?
Atomic fine structure Experimentally: By the 1920s, a fine structure in the spectra lines of Hydrogen and other atoms has been observed. Spectra lines appeared to be split in the presence of an external magnetic field. INTERPRETATION: Energy is independent of the quantum number l à the energy level is degenerate with respect to l Example: Considering n=2 and l =1 à m l = -1,0,1 e.g. 3 quantum states are degenerate at the same energy These 3 magnetic states would behave differently under a magnetic field resulting in the degeneracy being lifted!
Magnetic moment Model: electron circulating around the nucleus à Loop of current I = g = ± ² T, time it takes for the electron to make one rotation: T = 2πr v Introducing p = mv à T = 2πmr p Magnetic Moment induced: μ = IA = ±@ πr & &>U: Simplification: μ = ± ± rp, introducing L = rp gives μ = &U Electron magnetic moment: μ = e 2m L &U L
The normal Zeeman effect (I) Potential energy of the dipole created by the electron orbiting around the nucleus (under a magnetic field B): V { = μ ³ B One can only know one component of L: L = m l ħ Along z, the magnetic moment becomes: μ = eħ 2m m l = μ { m l Quantization: V { = μ B = +μ { m l B Bohr Magneton µ B = 9.274 x 10-24 J/T
The normal Zeeman effect (II) When a magnetic field is applied, the 2p level of atomic hydrogen is split into three different energy states with energy difference of E = μ { B m l. Potential energy of the dipole: V { = μ B = +μ { m l B m l Energy 1 E 0 + μ B B 0 E 0 1 E 0 μ B B Fine Structure
Fine structure Transition 2p à 1s
Stern & Gerlach experiment (1922) An atomic beam of particles in the l = 1 state pass through a magnetic field along the z direction. z V { = μ B F = dv { dz = μ db dz The m l = +1 state will be deflected down, the m l = 1 state up, and the m l = 0 state will be undeflected. If the space quantization were due to the magnetic quantum number m l only, and since the number of m l states is always odd (2l + 1), the experiment should produce an odd number of lines à But it doesn t. Why?