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Chapter 4 Quantum Mechanics in 3D
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Examples of representations
Let us consider a set of functions: The set is orthonormal: Kets can be expanded:
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Examples of representations
Closure relation
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Examples of representations
For two kets a scalar product is:
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Examples of representations
Useful relationship:
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Examples of representations
Let us consider a set of functions: The set is orthonormal: Kets can be expanded:
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Examples of representations
Closure relation
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Examples of representations
For two kets a scalar product is:
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Change of representations
Recall: Choosing we obtain:
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r and p operators For we obtain: where Similarly “Vector” operator r:
4.1 r and p operators For we obtain: where Similarly “Vector” operator r:
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4.1 r and p operators “Vector” operator p: Then:
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4.1 r and p operators Analogously: Then:
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4.1 r and p operators Calculating a commutator: Similarly:
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r and p operators Calculating a matrix element: Thus: Similarly:
4.1 r and p operators Calculating a matrix element: Thus: Similarly: Since |r > and |p > constitute complete bases operators r and p are observables
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Orbital angular momentum
4.3 Orbital angular momentum In classical mechanics the angular momentum of a point mass relative to some axis is defined as: In quantum mechanics the orbital angular momentum is defined by applying a quantization operation to the classical expression – replacing classical physical quantities with the corresponding observables (operators): This definition does not require symmetrization with respect to non-commuting operators, e.g.:
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Orbital angular momentum
4.3 Orbital angular momentum For a system of (spinless) particles the total orbital angular momentum is defined as:
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Commutation relations
4.3 Commutation relations Let us consider: Similarly, one can obtain: Thereby:
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Tullio Levi-Civita (1873 – 1941) Levi-Civita symbol
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Angular momentum As we will see later, the so called spin can be treated as an intrinsic (non-orbital) angular momentum, satisfying similar commutation relations: Thus both the orbital and intrinsic angular momenta are manifestation of the observable called angular momentum J, defined via the commutation relations:
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Angular momentum Let us introduce an operator:
4.3 Angular momentum Let us introduce an operator: This operator is Hermitian, and we will assume it is an observable
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Angular momentum Let us consider: Similarly, one can obtain: Thereby:
4.3 Angular momentum Let us consider: Similarly, one can obtain: Thereby:
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4.3 Angular momentum What is the physical meaning of the commutation relations? It is impossible to measure simultaneously the three components of the angular momentum, however, J2 and any component of J are compatible and could be measured simultaneously Therefore, there is a possibility to find simultaneous eigenstates of J2 and any component of J (e.g., Jz)
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Angular momentum We introduce (non-Hermitian) operators:
4.3 Angular momentum We introduce (non-Hermitian) operators: Let us consider: Also
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Angular momentum We introduce (non-Hermitian) operators:
4.3 Angular momentum We introduce (non-Hermitian) operators: Let us consider: Synopsizing:
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4.3 Angular momentum Let us also calculate: Similarly: Therefore:
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Eigenvalues Let us consider an eigenvalue problem:
4.3 Eigenvalues Let us consider an eigenvalue problem: Recalling the expression Moreover
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Eigenvalues Let us consider an eigenvalue problem:
4.3 Eigenvalues Let us consider an eigenvalue problem: This also can be written as: And this also can be written as:
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Eigenvalues Let us consider an eigenvalue problem:
4.3 Eigenvalues Let us consider an eigenvalue problem: This also can be written as: And this also can be written as:
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4.3 Eigenvalues Since: On the other hand, the square of the eigenvalue of the z-component of the angular momentum cannot exceed the eigenvalue of its magnitude squared: Therefore, there should be top and bottom “rungs” for integers n and p
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Eigenvalues Let us assume that for the top “rung” the eigenstate is:
4.3 Eigenvalues Let us assume that for the top “rung” the eigenstate is: Then: And: Recall:
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4.3 Eigenvalues Let us assume that for the bottom “rung” the eigenstate is: Then: And: Recall:
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Eigenvalues Combining: There are two solutions: Thereby:
4.3 Eigenvalues Combining: There are two solutions: Thereby: j must be integer or half-integer
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Eigenvalues Using: So:
4.3 Eigenvalues Using: So: We will therefore use indices j and m to label the eigenstates common to J2 and Jz
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Eigenvalues Synopsizing:
4.3 Eigenvalues Synopsizing: We thus have found the eigenvalues of the angular momentum What are the eigenstates?
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Eigenproblem for orbital momentum
4.3 Eigenproblem for orbital momentum We return to the orbital angular momentum of a spinless particle Let us find relevant eigenstates in the r-representation The Cartesian components of the orbital angular momentum operator:
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Eigenproblem for orbital momentum
4.3 Eigenproblem for orbital momentum It is convenient to work in spherical coordinates
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Eigenproblem for orbital momentum
4.3 Eigenproblem for orbital momentum Then
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Eigenproblem for orbital momentum
4.3 Eigenproblem for orbital momentum Then And
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Eigenproblem for orbital momentum
4.3 Eigenproblem for orbital momentum Recall: For the orbital angular momentum: In the r-representation (and spherical coordinates):
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum In these equations r does not appear in the differential operators, so we will consider it as a parameter Thus, the wavefunction can be written as: We try separating the variables:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum In these equations r does not appear in the differential operators, so we will consider it as a parameter Thus, the wavefunction can be written as: We try separating the variables:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum In these equations r does not appear in the differential operators, so we will consider it as a parameter Thus, the wavefunction can be written as: We try separating the variables:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Then:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Then:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Then:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Then: Since m is integer, l shoud be also an integer (not a half-integer)
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum We successfully separated variables but still need to solve this equation: Solutions: Here Pl are the Legendre polynomials: Adrien-Marie Legendre (1752 –1833)
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Legendre polynomials: Adrien-Marie Legendre (1752 –1833)
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum The resulting solutions: They have to be normalized This yields: Functions Y are called spherical harmonics
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Spherical harmonics:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Spherical harmonics:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum These harmonics constitute an orthonormal basis: Any function of θ and ϕ can be expanded in terms of the spherical harmonics:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum Spherical harmonics are functions with a definite parity:
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Eigenproblem for orbital momentum
4.1 Eigenproblem for orbital momentum The original eigenproblem: In the r-representation it looks like: Where
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Particle in a central potential
4.1 Particle in a central potential This equation describes the behavior of a single particle in a central potential: Here:
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Particle in a central potential
4.1 Particle in a central potential The Laplacian in spherical coordinates: Therefore:
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Particle in a central potential
4.1 Particle in a central potential Since the orbital angular momentum operator depends only on the angular coordinates: And: Since H, L2 and Lz commute there is a common basis for all three of them
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Particle in a central potential
4.1 Particle in a central potential Using the theory of the orbital angular momentum operator: And:
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Particle in a central potential
4.1 Particle in a central potential Using the theory of the orbital angular momentum operator: And: This equation and its solutions depend on the quantum number l as well as index k (that represents different eigenvalues for the same l) and does not depend on the quantum number m; thus:
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Particle in a central potential
4.1 Particle in a central potential The behavior of the R functions at the origin should be sufficiently regular in order to represent a physical solution The equation can be simplified via this substitution:
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Particle in a central potential
4.1 Particle in a central potential The behavior of the R functions at the origin should be sufficiently regular in order to represent a physical solution The equation can be simplified via this substitution:
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Particle in a central potential
4.1 Particle in a central potential Therefore: This function must be square-integrable: Since the spherical harmonics are normalized Quantum number l is called azimuthal, whereas quantum number m is called magnetic
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4.2 The hydrogen atom A system of a proton and an electron can form a hydrogen atom In this case the potential energy is: Therefore, the radial eigenproblem becomes:
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The hydrogen atom Let us make substitutions:
4.2 The hydrogen atom Let us make substitutions: This yields a dimensionless equation: Therefore, the radial eigenproblem becomes:
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The hydrogen atom Let us make substitutions:
4.2 The hydrogen atom Let us make substitutions: This yields a dimensionless equation: What are the asymptotes of the solutions?
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The hydrogen atom Let us make substitutions:
4.2 The hydrogen atom Let us make substitutions: This yields a dimensionless equation: What are the asymptotes of the solutions?
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4.2 The hydrogen atom Taking into account the asymptotes the solution could besought in the following form:
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4.2 The hydrogen atom Taking into account the asymptotes the solution could besought in the following form:
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4.2 The hydrogen atom
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4.2 The hydrogen atom Let us look for the solution in the following form of a polynomial:
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4.2 The hydrogen atom Equating the coefficients of like powers of ρ yields: The polynomial terminates at:
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The hydrogen atom Therefore: Let us recall that:
4.2 The hydrogen atom Therefore: Let us recall that: Combining the two equations: Defining:
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The hydrogen atom Therefore: Let us recall that:
4.2 The hydrogen atom Therefore: Let us recall that: Combining the two equations: Defining: We obtain:
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4.2 Spectrum Since Conventionally and conveniently n is used to label the energy spectrum n is called a principal quantum number A given value of n characterizes an electron shell Defining: We obtain:
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4.2 Spectrum Since Conventionally and conveniently n is used to label the energy spectrum n is called a principal quantum number A given value of n characterizes an electron shell Defining: We obtain:
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4.2 Spectrum Since There is a finite number of values of l associated with the same value of n: Each shell contains n sub-shells each corresponding to a given value of l Defining: We obtain:
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4.2 Spectrum Since There is a finite number of values of l associated with the same value of n: Each shell contains n sub-shells each corresponding to a given value of l Each sub-shell contains (2l + 1) distinct states associated with the different possible values of m for a fixed value of l
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4.2 Spectrum The total degeneracy of the energy level with a value of En is: Conventionally, different values of l are (spectroscopically) labelled as follows: Subshell notations:
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4.2 Spectrum
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4.2 Spectrum
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4.2 Eigenfunctions Let us synopsize all the transformations and assumptions for the eigenfunctions
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Eigenfunctions Let us also recall the normalization conditions:
4.2 Eigenfunctions Let us also recall the normalization conditions: Now, using all this information, let us calculate the eignefucntions for the problem
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4.2 Eigenfunctions We will start with the ground level, a nondegenerate 1s subshell Normalizing:
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Eigenfunctions We obtained the eigenfunction of the ground state!
4.2 Eigenfunctions We obtained the eigenfunction of the ground state! It is completely spherically symmetric
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4.2 Eigenfunctions What is the probability density of finding an electron in an elementary volume? The probability of finding an electron between r and r + dr is proportional to For the ground state this probability is thus proportional to
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Eigenfunctions The maximum of this probability occurs at
4.2 Eigenfunctions The maximum of this probability occurs at Parameter a0 is known as the Bohr radius Niels Henrik David Bohr (1885 – 1962)
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4.2 Eigenfunctions The ground state function can be used to generate the rest of the eigenfunctions E.g.,
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Eigenfunctions The most general expression: Where
4.2 Eigenfunctions The most general expression: Where Is the associated Laguerre polynomial Edmond Nicolas Laguerre (1834 – 1886)
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Eigenfunctions Is the associated Laguerre polynomial 4.2
Edmond Nicolas Laguerre (1834 – 1886)
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4.2 Eigenfunctions The radial parts of the eigenfunctions:
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4.2 Eigenfunctions Probability density plots for the wave functions:
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Angular momentum Let us recall key results for the angular momentum
4.3 Angular momentum Let us recall key results for the angular momentum
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Angular momentum Let us recall key results for the angular momentum
4.3 Angular momentum Let us recall key results for the angular momentum
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4.4 Angular momentum Let us now introduce 2×2 Hermitian matrices representing some operators in some basis: They satisfy the following commutation relations: Therefore, these matrices represent three components of an angular momentum operator in a basis for which the Jz operator is diagonalized with eigenvalues ±ћ/2
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4.4 Angular momentum Let us now introduce 2×2 Hermitian matrices representing some operators in some basis: Using definitions: One obtains: Therefore, in the same basis the J2 operator is also diagonalized with eigenvalues 3ћ2/2
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4.4 Spin angular momentum The matrices that we just introduced represent one of the examples of a spin vector operator It turns out that such observable indeed exists in nature Moreover, it is one of the most fundamental properties of an electron and other elementary particles
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4.4 Spin angular momentum Spin can be measured experimentally, and it gives rise to many macroscopic phenomena (such as, e.g., magnetism) If one imagines that a particle with a spin has a certain spatial extension, then a rotation around its axis would give rise to an intrinsic angular momentum However, if it were the case, the value of j would necessarily be integral, not half-integral
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4.4 Spin angular momentum Therefore, the spin angular momentum has nothing to do with motion in space and cannot be described by any function of the position variables Spin has no classical analogue! Here we will introduce spin variables satisfying the following postulates: 1) The spin operator S is an angular momentum: 2) It acts in a spin state space Es
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4.4 Spin angular momentum The space Es is spanned by the set of eigenstates common to S2 and Sz: Spin quantum number s can take both integer and half-integer values Every elementary particle has a specific and immutable value of s
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Spin angular momentum Pi-meson: s = 0
4.4 Spin angular momentum Pi-meson: s = 0 Electron, proton, neutron: s = 1/2 Photon: s = 1 Delta-particle: s = 3/2 Graviton: s = 2 Etc. Every elementary particle has a specific and immutable value of s
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4.4 Spin angular momentum 3) All spin observables commute with all orbital observables Therefore the state space E of a given system is: Let us restrict ourselves to the case of the particles with spin 1/2 In this case the space Es is 2D In this space we will consider an orthonormal basis of eigenstates common to S2 and Sz:
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Spin 1/2 The eigenproblem: The basis: Recall: Thus:
4.4 Spin 1/2 The eigenproblem: The basis: Recall: Thus: In this space we will consider an orthonormal basis of eigenstates common to S2 and Sz:
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4.4 Spin 1/2 Any spin state in Es can be represented by an arbitrary vector:
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4.4 Spin 1/2 Any operator acting in Es can be represented by a 2×2 matrix in basis E.g.: σ’s are called Pauli matrices: Their properties: Wolfgang Ernst Pauli (1900 – 1958)
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4.4 Spin 1/2 Any operator acting in Es can be represented by a 2×2 matrix in basis E.g.: Therefore: Their properties: Wolfgang Ernst Pauli (1900 – 1958)
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Observables and state vectors
Since The basis used will be: Then:
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Observables and state vectors
This basis is orthonormal and complete: Any state in E can be expanded as: Where: I.e.: This can be written in a spinor form:
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Observables and state vectors
This basis is orthonormal and complete: An associated bra can be expanded as: Where: I.e.: This can be written in a spinor form:
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Observables and state vectors
A scalar product can be written as: Normalization:
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Observables and state vectors
It may happen that some state vector can be factored as: Then:
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Magnetic Moment The vector is called the magnetic dipole
moment of the coil with current Its magnitude is given by μ = IAN The vector always points perpendicular to the plane of the coil The potential energy of the system of a magnetic dipole in a magnetic field depends on the orientation of the dipole in the magnetic field and is equal to:
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Spin in a magnetic field
4.4 Spin in a magnetic field Analogously, in quantum mechanics a Hamiltonian describing interaction of the spin with the magnetic field is written as: γ is called the gyromagnetic ratio Then the Hamiltonian matrix can be rewritten as:
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Spin in a magnetic field
4.4 Spin in a magnetic field Let’s consider a uniform field B0 and chose the direction of the z axis along the direction of B0 Within the notation So, the eigenvalue problem for H is: The two states correspond to the spin vector parallel and antiparallel to the field
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More properties of spin 1/2
4.4 More properties of spin 1/2 What are the eigenvectors of Sz? One can calculate:
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More properties of spin 1/2
What are the eigenvalues and eigenvectors of Sx? The probabilities are:
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Total angular momentum
4.4 Total angular momentum Let us recall the definition of the classical angular momentum of a system of particles: What is the quantum equivalent of the total angular momentum of the system in which there are parts with their own angular momenta? The answer: What are the eigenvalues and the eigenstates associated with this observable?
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Addition of two spin-½’s
4.4 Addition of two spin-½’s Let us start with the simplest nontrivial case of two angular momenta being added, each of spin ½: The state space is 4D: The orthonormal basis: These are the eigenstates of this set of operators:
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Addition of two spin-½’s
4.4 Addition of two spin-½’s The eigneproblem: S is also an angular momentum operator: Similarly: So:
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Addition of two spin-½’s
4.4 Addition of two spin-½’s Another important operator is:
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Addition of two spin-½’s
4.4 Addition of two spin-½’s Another important operator is:
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Addition of two spin-½’s
4.4 Addition of two spin-½’s Important commutation relations: What about this commutator?
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Addition of two spin-½’s
4.4 Addition of two spin-½’s Let us calculate: Since
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Basis change We just showed that these observables commute: Since
4.4 Basis change We just showed that these observables commute: Since This set is not the same as: What can we say about this set?: Is there a new basis that will satisfy these equations?
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4.4 Basis change Since One can write: Thus: I.e.,
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4.4 Basis change So:
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4.4 Basis change Let us recall: Then:
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4.4 Basis change Let us recall: Then:
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4.4 Basis change Let us recall: Then:
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4.4 Basis change Let us recall: Then:
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4.4 Basis change Synopsizing: Thus:
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+ - Basis change Let us diagonalize this matrix: Secular equation:
4.4 Basis change Let us diagonalize this matrix: Secular equation: Roots: + -
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4.4 Basis change Normalizing:
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Basis change Thus, we found all four eigenstates:
4.4 Basis change Thus, we found all four eigenstates: The corresponding values of M: Thus the corresponding values of S:
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Basis change So, we successfully obtained the new orthonormal basis:
4.4 Basis change So, we successfully obtained the new orthonormal basis: The first three states with S = 1 are called a triplet, and the fourth state with S = 0 is called a singlet
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Basis change So, we successfully obtained the new orthonormal basis:
4.4 Basis change So, we successfully obtained the new orthonormal basis: As a result of the basis change the subspace allocation became different
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