Applications of Quantum Mechanics, Leszek Frasiński, 23 Jan 2012 Classwork 1
Classwork 1 – Atomic units
Note: To solve the following problems you may use any information in the “Useful Formulae” sheet.
Atomic units make quantum-mechanical calculations simpler and less prone to mistakes. In this classwork you will practise unit conversion.
1. The electronvolt is neither an SI nor an atomic unit of energy but it is widely used in quantum mechanics. The “Useful Formulae” sheet gives the relation between the photon wavelength and its energy as λE = 1240 nm⋅eV. (a) Derive this relation taking care that the physical quantities you write have not only the correct numerical values but also the correct units. (b) The light emitting diode (LED) operates by applying a potential difference across a p-n junction. When crossing the junction electrons gain energy and release it as photons. What is the minimum voltage that needs to be applied to a blue LED? Hint: The blue part of the spectrum is in the range 450–500 nm.
2. Covert hartree, Eh, to electronvolts.
3. Frequency ν and angular frequency ω are often mistaken one for the other because they have the same units, s–1. To avoid the mistakes it is recommended to write the units of ω as rad/s and use s–1 (or Hz) only for ν. Following this idea of keeping a distinction between units of different physical quantities, find out the atomic units of: (a) time, (b) frequency, (c) angular frequency.
4. The “Useful Formulae” sheet gives a relation between the amplitude F (in atomic units) of a monochromatic plane-polarised EM wave and its intensity I in units of International prototype of the kilogram kept W/cm2: at the Bureau International des Poids et Mesures near Paris. This is the only one of the seven basic SI units defined by a physical artifact E W F h ×= 10338.5 −9 I rather than a natural phenomenon. There is a 2 proposal* to redefine kilogram by referring it ea0 cm –9 either to the Planck constant or the Avogadro Derive the numerical constant 5.338×10 . constant. The atomic units are all based on Hint: Put the square root in the above physical constants. 2 equation on the LHS and use I = (cε0/2) F . * I.M. Mills et al, Metrologia 42 71–80 (2005)
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Classwork 1 – Atomic units Classwork 1 – Re: problem 1 (a) Classwork 1 – Re: problem 1 (a) The relation between photon The relation between the joule energy E and its wavelength is: and the electronvolt is: 1. E = h 1. J = (C/e)·eV 2. E = ħ 2. J = (C/(V·s))·eV 3. E = hc / 3. J(/C)VJ = (e/C)·eV 4. E = ħc / 4. J = (V·s/e)·eV
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Classwork 1 – Re: problem 1 (b) Classwork 1 – Re: problem 2 Classwork 1 – Re: problem 2 The minimum voltage in question The formula for the hartree is: The value of the hartree is: corresponds to the wavelength of: 1. E = ħ /(/ (m 2 a ) 1. E = 27. 2 eV 1. 450 nm h e 0 h 2 2 2. (450 + 500)/2 nm = 475 nm 2. Eh = ħ / (me a0) 2. Eh = 13.6 eV 2 3. (500 nm − 450 nm) = 50 nm 3. Eh = ħ /(/ (me a0) 3. Eh = 24. 6 eV 4. 500 nm 2 2 4. Eh = ħ / (me a0 ) 4. Eh = −13.6 eV
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Classwork 1 – Re: problem 3 (a) Classwork 1 – Re: problem 3 (b) Classwork 1 – Re: problem 3 (c) The atomic unit of time is: The atomic unit of frequency is: The atomic unit of angular freqqyuency is: 1. ħ / Eh 1. Eh /(/ (hc) 1. Eh / (rad h) 2. Eh / ħ 2. Eh / (ħc) 2. rad Eh / ħ 3. Eh / a0 3. Eh / h 3. Eh / (rad ħ) 4. a0 / Eh 4. Eh / ħ 4. rad Eh / h
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Classwork 1 – Re: problem 4 The relation between the hartree and the bohr is: 2 1. Eh = e / (40 a0 ) 2 2 2. Eh = e / (40 a0 ) 2 3. Eh = e / (40 a0) 2 4. Eh = e / (0 a0 )
Page 2 Applications of Quantum Mechanics, Leszek Frasiński, 30 Jan 2012 Classwork 2
Classwork 2 – VUV helium lamp
Introduction Air absorbs ultraviolet radiation at wavelengths shorter than 200 nm. Any experiments involving this radiation must be performed in vacuum and this is why the spectral range 10– 200 nm (photon energies 124–6.2 eV) is called vacuum ultraviolet (VUV). The choice of VUV optics is very limited and most methods of VUV generation require sophisticated and expensive instrumentation, such as Diamond synchrotron at the Rutherford Appleton Laboratory. Using an electric discharge in a noble gas offers a simple alternative, albeit limited to a few spectral wavelengths, and helium ion provides the shortest ones. Such a helium lamp is used for experiments at Imperial College. The picture below shows the lamp and the apparatus which uses the VUV radiation in one of the labs.
Attosecond pulse generation
Spectrometer VUV lamp
VUV helium lamp in the attosecond laboratory (Huxley room 005) of the Laser Consortium. The lamp is used to calibrate a VUV spectrometer, which in turn will be used to characterise attosecond pulses synthesised from high harmonics generated by an intense laser in noble gases.
In the problems that follow you will need to consider the process of emission of the VUV radiation from a helium ion. You may find the “Useful Formulae” sheet helpful.
Page 1 of 2 Applications of Quantum Mechanics, Leszek Frasiński, 30 Jan 2012 Classwork 2
Problems In the helium lamp an electron from the discharge impacts on a He atom, ionises it and at the same time excites it to a superposition of several states. For simplicity, let us consider only the states that are of practical importance and assume that the He+ ion immediately after the impact is left in the following superposition of states:
ψ(r, t = 0) = ψ0(r) = A [10 u100(r) + 2 u210(r) + u310(r)], where unlm(r) are energy eigenfunctions of the hydrogen atom.
Use the Dirac notation, e.g. ψ0(r) = 〈 r | ψ0 〉 and unlm(r) = 〈 r | n l m 〉, in solving the following problems:
1. Find the normalisation constant A.
2. Is ψ0 an eigenfunction of the parity operator?
3. What are the probabilities of finding the He+ ion in each of the following states at t = 0: (a) | 1 0 0 〉 , (b) | 2 0 0 〉 , (c) | 2 1 0 〉 , (d) | 3 1 0 〉 ?
4. Assuming an infinitely heavy nucleus (i.e. the reduced mass m = me) calculate the energies (in atomic units) of each of the following states: (a) | 1 0 0 〉 , (b) | 2 1 0 〉 , (c) | 3 1 0 〉 .
5. What are the expectation values in atomic units for the state | ψ0 〉 of: (a) the energy? (b) the L2 operator?
(c) the Lz operator?
6. After t = 0 the excited states can decay to the ground state emitting photons. (a) Calculate the photon energies in both the atomic units and the electronvolts, and the corresponding wavelengths in nanometres.
(b) Considering that only a fraction of ψ0 is in the excited states and the energy expectation (see 5a) is not much above the ground state (see 4a), explain how can the photon energy (see 6a) be so large? How is the energy conserved? Clarify this issue and explain the meaning of ψ0.
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Classwork 2 – VUV helium lamp Classwork 2 – Re: problem 1 Classwork 2 – Re: problem 1 In the Dirac notation: The orthonormality relation of (r) = A [10 u (r) + 2 u (r) + u (r)] 0 100 210 310 the eigenstates is: 1. |0(r) = A(10 r |100| 1 0 0 + 2 r | 2 1 0 + r | 3 1 0 ) .1 mlnmln ll 2. 0(r) = A(10 | 1 0 0 + 2 | 2 1 0 + | 3 1 0 ) .2 llmlnmln )( 3. | = A(10|100(10 | 1 0 0 0 + 2 | 2 1 0 + | 3 1 0 ) .3 mlnmln mmllnn 4. |0(r) = A(10 | 1 0 0 .4 mlnmln 1ˆ + 2 | 2 1 0 + | 3 1 0 )
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Classwork 2 – Re: problem 1 Classwork 2 – Re: problem 2 Classwork 2 – Re: problem 2 – Parity in Dirac notation The normalisation constant can The parity operator transforms an be calculated from: eigggenvector in the following way: 2 ˆ l 1. ( |0) = 1 .1 )1( mlnmlnP .2 ˆ mlnP )1( ll )1( mln 2. 0 |0 = 1 n1 2 .3 ˆ mlnP )1( mln 3. 0 |0 = A ˆ ll )1( ˆ 2 .4 mlnP )1( mln P 4. |0 |0| = 1
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Classwork 2 – Re: problem 3 Classwork 2 – Re: problem 4 Classwork 2 – Re: problem 5 The probability of finding the He+ The energies of the He+ ion are: The square of the angular ion in state |100 is ggyiven by: E 1 momentum operator has the 1. E h n 2 eigenvalues given by: 2 2 n .1 P r 001 100 1 .1 ˆ2 )1( 2 mlnllmlnL .2 n EE h n2 P100 001.2 0 ˆ2 l 2 .2 nL l (1) nm l m 2 .3 EE n h 2 ˆ2 P100 001.3 0 n .3 mlnmmlnL 4 .4 EE ˆ2 l .4 P r 001 n h 2 .4 mlnL )1( mln 100 n nm
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Classwork 2 – Re: problem 6 The wavelength of the photon emitted in the transition |3 1 0 → |100 is:
1. λ31 = 102 nm
2. λ31 = 30.4 nm
3. λ31 = 122 nm
4. λ31 = 25.6 nm
Page 2 Applications of Quantum Mechanics, Leszek Frasiński, 17 Feb 2012 Classwork 3
Classwork 3 – Comet
Note: To solve the following problems you may use any information in the “Useful Formulae” sheet.
When a comet passes by the Earth, sunlight heats the comet and the gases evaporate from its frozen nucleus forming the head and the tail. A water molecule dissociates and one of the hydrogen atoms is excited by the sunlight. Describe the following aspects of this excitation:
1. Show that the transition 1s → 2s is not allowed, by explicitly calculating the transition dipole moment between the two states for the polarization: (a) parallel to the z axis, (b) parallel to the x or y axis.
2. Estimate the order of magnitude of the excitation rate (per second) of the atom by considering its strongest transition, 1s → 2p, in the following steps: (a) Substitute the typical transition dipole moment, −ea0, into the relevant Einstein coefficient. (b) Combine the result from part (a) with the spectral density of black body radiation (Planck’s law) to obtain the transition rate for an atom exposed to isotropic sunlight illumination. (c) Take into account that the sunlight comes to the comet only from a small solid angle: the (linear) angular diameter of the Sun is 0.5º.
3. Calculate the de-excitation rate of the reverse transition, 2p → 1s. How does it compare with the excitation rate? What fraction of the hydrogen atoms in the comet head will be in the excited state at any time? Comet Hale–Bopp in the evening sky above 4. Check the units of the results in points Woodley near Reading. Photo by Leszek 2 and 3. Frasiński, April 1997.
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Classwork 3 – How to calculate selection rules? Classwork 3 – Re: problem 1(a) Classwork 3 – Re: problem 1(a)
Comet For ε||z, μ = −e2s| ε·r |1s = u2s(r) = 2 3 2 * i 1. e dr r d sin d u2s r cos u1s 1. R21(r) sin e 1. Show that the transition 8 1s 2s in not allowed by 00 0 explicitly calculating the 2 * 5 electric dipole moment rre cosdcosdd.2 uru 2 2s 1s 20 rR )(.2 )1cos3( between the two states for 00 0 16 the radiation polarization: 2 1 (a) parallel to the z axis, rre * cosdsindd.3 uru 2s 1s 20 rR )(.3 (b) parallel to the x or y axis. 00 0 4 2 2 * 3 rre 2s cosdsindd.4 uru 1s 21 rR )(.4 cos 00 0 4
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Classwork 3 – Re: problem 1(b) Classwork 3 – Hydrogenic wavefunctions Classwork 3 – Re: problem 2(a)
tiE / ut rr e)(),( n ZE 2 2 Einstein For ε||x, ε·r = nlm nlm E h , E eV2.27 n 2 n2 h am 2 2 r YrRu ),()()( 0e coefficient for nlm lmnl 1. B absorption 12 r sinsin.1 4 a00 depends on the r cossin.2 transition dipole .2 B12 moment μ in the 0 r sincos.3 2 following way: .3 B12 2 r coscos.4 0 2 .4 B12 2 40
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Classwork 3 – Re: problem 2(b) Classwork 3 – Re: problem 2(b) Classwork 3 – Interaction with electromagnetic radiation The spectral density of black body The temperature of Sun’s F ε 0 tF )cos( radiation is given by: photosphere is: e 2 ε r 1 3 3 A B 1 K4800.1 21 32 21 )(.1 c /32 kT F0 c 1e Ω 2 2 3 B / kT 12 2 K6800.2 c 2 32 e)(.2 0 0 c I F0 2 I() c() 2 2 / kT e)(.3 Iindoor W/m1 3 22 K5800.3 1 c 2 )( I kW/m1 /32 kT 2 sunlight ec 1 1 )(.4 T surfaceSun 5800 K c /22 kT 1e K7800.4
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Classwork 3 – Re: problem 2(c) Classwork 3 – Re: problem 2(c) Classwork 3 – Re: problem 3 The solid angle In this problem, the processes res- subtended by the ponsible for the de-excitation are: Sun is the following Ω 2.1 1. equally spontaneous and stimulated emissions function of the 2 angular diameter .2 Ω 2. mainly stimulated emission of the Sun, : 4 2 .3 Ω 3. mainly spontaneous emission 2 .4 Ω 4. mainly absorption 4
Page 2 Applications of Quantum Mechanics, Leszek Frasiński, 24 Feb 2012 Classwork 4
Classwork 4 – Negative ion
Note: To solve the following problems you may use any information in the “Useful Formulae” sheet.
It is possible to form the negative ion of hydrogen, H–, by attaching an extra electron to the hydrogen atom.
1. Deduce the spin state of the ground state of the H– ion by considering indistinguishability of the two electrons.
2. What is the energy required to remove both electrons from the H– ion within the independent particle approximation?
3. Use first order perturbation theory to estimate a more accurate value of this energy as follows: 5 2 Ze (a) recall that Handout 5 evaluates the Coulomb integral as J = ; 48 πε a00
(b) express this result in terms of Z and the Hartree energy, Eh; (c) calculate the value of J in eV.
4. Sketch appropriate energy diagrams and comment upon what your results in parts 1–3 say about the stability of this ion (in fact this ion is stable because the experimental value of the energy required to remove both electrons is 14.4 eV).
Tandem accelerator used for many years at Daresbury Laboratory. The diagram shows the principle of a tandem van de Graaff accelerator. Negatively charged ions from an ion source at ground potential are accelerated towards a terminal at high positive potential in the centre, where gas or a thin foil removes two or more electrons from the ions, which then become positively charged and repelled towards the grounded electrode (V = 0). Electric charge is transported on a belt from the ground to the terminal and as a consequence of the charge accumulation, the potential increases. The high voltage (V = +5 MV) is insulated from the ground by high pressure gas, normally SF6. http://nobelprize.org/nobel_prizes/physics/articles/kullander/
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Classwork 4: Negative ion Classwork 4 – Re: problem 1 Classwork 4 – Re: problem 1 In the ground state of He, In terms of single-electron spin 1 functions,,g the singlet state (r ,r21 ) 2 1 uuuu rrrr )2,1())()()()((.1 )2,1( 0011s21s21s11s 00 2
uuuu rrrr )2,1())()()()((.2 ))1()2()2()1((.1 111s21s21s11s M S 2. ( (1) (2) (2) (1)) uuuu rrrr )2,1())()()()((.3 0011s21s21s11s ))1()2()2()1((.3
))1()2()2()1((.4 uuuu rrrr )2,1())()()()((.4 111s21s21s11s M S
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Classwork 4 – Re: problem 2 Classwork 4 – Re: problem 2 Classwork 4 – Re: problem 4 Atomic number for the H− ion: Within IPA, to remove both To have a stable H− ion it is electrons from H − we need : sufficient that the energies in 1. Z = 3 the ground state satisfy: 1. 27.2 eV 2. Z = 4 1. E(H−) < E(H+) 2. 13.6 eV − + 3. Z = 2 2. E(H )>) > E(H ) 3. 108.8 eV 3. E(H−) < E(H) 4. Z = 1 − 4. 54.4 eV 4. E(H ) > E(H)
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Classwork 1 – Solutions
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Classwork 2 – Solutions
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(b) The meaning of ψ0 is that the numerical coefficients in the assumed superposition give us probability amplitudes of finding the He+ ion in a given energy eigenstate. Nearly 95% of the ions are found in the ground state (see answer 3a) and the energy expectation is heavily weighted towards the energy of this state. Individual photons, however, are emitted only from the ~5% of the excited states (see answers 3c and 3d). The photon energy is so large because only a small fraction of the whole ion population emits these photons. The photos are emitted as whole quanta of energy and in a long run, when we detected many photons, the photon energy averaged over the whole population tends to the energy difference between the energy expectation and the ground state energy. On a short run, however, when we detect only a few photons, their average energy may deviate from the expectation-to-ground energy difference. This does not mean the energy is not conserved, it only means that the energy of state | ψ0 〉 is uncertain because it is in a superposition of states of different energy eigenvalues.
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Classwork 3 – Solutions
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Classwork 4 – Solutions
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