1. Lecture from Quantum Mechanics

3. Marek Zrałek Field Theory and Particle Physics Department. Silesian University Lecture 5

6. How to build a mathematical picture of quantum systems ??? How to build a mathematical picture of quantum systems ???  Build a classic image of the microscopic systems ( in te case of classical analogies) and all observables expressed as a function of positions (x) and momentum (p).  Replace the numbers on operators, x Q, p P and imposed on those operators Heisenberg [Q, P] = i  For systems that do not have a classical analogy entire algebra of observables we are building from the beginning.

7. Experimental use of the Quantum Mechanics postulates (Relations between the experimental data and Quantum Mechanics observables) Source of electrons stream with specified energy E e Gas of carbon monoxide Energy Analyzer Detector On the example of absorption spectrum of carbon monoxide CO, based on the original Franck - Hertz experiment from 1914. The spectrum depends on the energy E e and measurement accuracy e e’ Small temperature Measurement accuracy E ≈ 0.005-0.05eV It measures the intensity of electrons with energy E

8. The intensity of electrons in detector Energy E = (E e -E e’ ) ΔE = 0.26 eV For diatomic molecules CO molecules cannot be excited to any energy. Only discrete energy values are possible, the CO molecule has a discrete energy levels. They are almost equally distant. For higher line the distances are reduced C O Vibrating dipole 0 ΔE 2ΔE 3ΔE For E e ≈ 2.05eV and ΔE ≥ 0.06 eV G.J. Schulz, Phys. Rev. 135, A988 (1964) E0E0 E1E1 E2E2 E3E3 This system we can describe approximately by the quantum-mechanical harmonic oscillator.

10. CO  the space of states can be set by the algebra of three operators H, P, Q satisfying the relationship: By measuring the energies ΔE we always get the value: 1/2hν, 3/2hν, 5/2hν,...... (0.26 eV,.52 eV,.78 eV,....), so the eigenvalues of energy operator. The prepared state of the system: P n =, dim(h n )=1,. Probabilities that measurement of energy gives an eigenvalue E n is equal to:: p n = Tr(ρP n ) = w n,

11. Annihilation and creation operators: The operator of the number of "particles" (excitations) and the commutation relation: There are ground state (lowest energy state):

12. Let us assume that at least one eigenvector of the N operator exist: Let us define a new vector: So, now we can mark: And similarly: Now we see that:

13. We can restrict the value of λ and m: It follows that λ must be a positive integer so that there was always the smallest number m = λ such that λ - m = 0: As usually we need orthonormalized states:

14. And for the subsequent vectors: We can now find the normalizing coefficients c n : We assumed that the state is normalized, i.e. c 0 = 1, in addition we assume that c n are real, we get:

15. So, we have: Where normalized energy eigenvectors have the form: In the following we will need: From which we can get: And:

16. Space Φ for harmonic oscillator states : Now we can find our oscillator states in the position base : So we have to find a form of base vectors in the positional representations: As we would like to have not only: but also: We assume that our functional is reflexive

17. At first we look for : From which we obtain: So, we have: Defining the new variable, m = n+1, we obtain: for: (*)

18. So, we have a recursive relationship to find eigenstates of the operator of energy in the positional representation. We need to know: For m = 0 there is: From which we get: In the following we use denotation: Dimensionless: and: (*) Dividing the equation by, we obtain:

19. Using now the new denotations: We get: This is the recurrence relation for Hermite polynomials.

20. From the definition on the previous page: We can check that the Hermite polynomials can be written as: The full basis vectors in the positional representation have the form: This factor we choose in such a way in order the total wave function be normalized to 1.

21. As the eigenvectors of the Hermitian operator they are orthogonal, but we have to normalize them. We would like to have: We use the property: And our normalization condition looks like: As we know(from literarure) that:

22. Formally, we find the value of the functional for the vector: Similarly, you can find the energy state in the momentum representation Now also we have:

23. We can find the eigenstates of energy in momentum representation: We know

25. A quantum system will remain in stationary state as long some external do not act forces on it. Under the action of these forces the system is changing. In practice, any quantum system is subject to external forces (weak electromagnetic field, some inside fields arising from the movement of local charge). Such weak interaction does not change significantly our system but makes the transition between their states (perturbation theory, time-dependent perturbation theory can be used). Transitions between energy states the original system can be observed. Knowing the interaction we can find the probability of transitions and selection rules. The system can absorb or radiate electromagnetic wave with frequency: Transition probability is equal: Our diatomic molecule behaves like an electric moment: -- + -q +q Oscillating charges radiate electromagnetic wave

26. In one dimension D = q Q, so So we have a selection rule: | n-m | = 1, and the transition as for harmonic oscillator For molecules CO : ΔE = 0.265 eV and: In spectroscopy the frequency is not given not in seconds, we will us (ν / c = 1 / λ) 1 / cm (the number of waves per cm). The spectrum of vibrating molecules of CO is a single line (near infrared) Similarly for other diatomic molecules, e.g. for HCl λ = 2.46 μ, No spectrum for example. O 2, N 2, H 2 Check it on exercise VIBRATOR

27. Increasing the thickness of the gas absorbent, next absorption lines appear: The intensity of the other lines is significantly smaller The oscillator cannot be good ( ), non-harmonics effects: | n - m | = 2, 3, 4,... If we observe spectral lines through a spectrophotometer with a much better resolution, line ν = 2140/cm can be split into a large number of individual lines: R branch P branch 222021402060 1/cm Absorption REALISTIC VIBRATOR + RIGID ROTATOR

28. After increasing the thickness of the absorbent Vibrating and rotating spectrum for carbon monoxide (Vibrating and rotating structure)

29. If energy << 0.26 eV – The rigid rotator m1m1 m2m2 CM In the CM system: ;

30. Kinetic energy: The kinetic energy expressed by the frequency of rotation and moment of inertia: Angular momentum can be expressed by the moment of inertia and frequency

31. then: In the CM system we have: For oscillation: And we have the new quantum oscillating system :

32. Algebra for rigid rotator: From canonical commutation relation: we obtain:

33. Statistical operator in degeneracy subspace of angular momentum with eigenvalue l (= R l ) Subspaces R l are not invariant under the action of the operators Q j and P k, because we have: Algebra of the rotator contains operators l k, Q i and P m, hence the space of physical states of quantum mechanical rotator is a simple sum of the: Algebra of angular momentum operators has invariant subspaces R l

34. Algebra of the vibrator with rotator is a collection of six operators satisfying commutation relations (algebra E 3 ): ;. The spectrum of the energy operator looks like: E0E0 E1E1 E2E2 E3E3 E4E4

35. The intensity of the radiation emitted by the rotator is proportional to the modulus square of the matrix element of the dipole moment operator (we assume that the electric dipole rotates): And the following selection rule is satisfied (please check it): So the transition between the energy lines with different n and as well as with the same n and will occur. The frequency of such move is specified as usual:

36. Prove, that: This follows from the : Wigner - Eckart theorem, prove this theorem

37. The difference in frequency between successive lines is constant: E.g. in the case of HCl for the first 11 levels the Δν is almost constant and from the numerical matching we obtain Compatibility is not good for higher levels, it would be better if we take: and this corresponds to the energy levels: This can be simple enough to explain. HCl is not a rigid rotator. The distance between H and Cl varies because the moment of inertia is proportional to x and classically we have:

38. This occurs when: So for energy we get: And in the quantum case for energy operator we can propose:

39. And for energy spectrum we obtain where : Full quantum mechanical approach to rotating oscillator system R 1 -- space of quantum oscillator states, R 2 – space of quantum states for oscillator Scalar product: Base of states in R:

40. Product of operators: Any operator working in R can be given by: Let us apply now the postulate VII in our case --space of quantum states for harmonic oscillator: -- space of quantum states for quantum rotator: -- space of quantum spaces for diatomic molecules. Operators in : Base in : Φ

41. From experiment we know that: where: If there is no any correlation between oscillator and rotator movement : with the selection rules: So we obtained two series of equidistant lines:

42. In the middle there is no signal, lack of transition Δl=0. There is reasonable compatibility with the data but not overwhelming, the discrepancy is caused by the impact between vibrational and rotational degrees of freedom. We can take into account this correlation and assume that: And we get: where: ; ;. So then:

43. If we make parameterization: where moment of inertia: n InIn Dla ujemnych g In such case the frequencies of transition in the two series R and P are equal: Better agreement with experiment, but not excellent: ----- anharmonicity is not taken into account, ------ connection between I and distance x is not consider ------ for higher energies atomic excitation are also important.

44. Complete set of commuting observables (csco) (quantum numbers): ----- Important to determine the base states, ----- Important to determine pure states, ----- Question about csco is the experimental question, having experimental spectra we suggest algebra, so also csco, ----- Answer to the question what is the csco, depends on available energy and precision of measurement. When E <10 -2 eV only rotational states, “J, J z “. If the energy increases, additionally oscillator energy operator ”N” must be taken into account. Further with an increase of energy, atomic levels appear.