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Symmetry and Quantum Physics

(3 posts)

  1. lee
    Key Master

    When we study differential geometry, we are particularly interested in geometric properties that are independent of change of coordinates. The same is true for physics. When we develop a physical theory, we require certain physical properties to be invariant under change of coordinates. Mathematically, change of coordinates (change of basis of a vector space) is given by linear isomorphisms of a vector space (a physical space) over a field $F$. Physicists usually call such invariance symmetry. The set of all linear isomorphisms of a vector space $V$ of dimension $n$ with a positive definite inner product is denoted by $\mathrm{GL}(n,F)$. $\mathrm{GL}(n,F)$ is a Lie group with function composition $\circ$. The Lie group $\mathrm{GL}(n,F)$ is called the general linear group of $V$. The dimension of $\mathrm{GL}(n,F)$ is $n^2$. The set of orientation preserving isomorphisms of $V$ forms a Lie subgroup of $\mathrm{GL}(n,F)$ and is denoted by $\mathrm{SL}(n,F)$. $\mathrm{SL}(n,F)$ is called the special linear group of $V$. In fact, a linear isomorphism $A\in\mathrm{SL}(n,F)$ if and only if $\det(A)=1$. The set of isometries (distance or length preserving linear isomorphisms) of $V$ forms a Lie subgroup of $\mathrm{GL}(n,F)$ and is denoted by $\mathrm {O}(n,F)$. $\mathrm {O}(n,F)$ is called the orthogonal group and it coincides with the set of all $n\times n$ orthogonal matrices, i.e. $$\mathrm {O}(n,F)=\{A\in\mathrm{GL}(n,F): AA^t=I\}.$$ In particuluar, orthogonal matrices whose determinants are equal to 1 preserve the orientation, and they form a Lie subgroup of $\mathrm{O}(n,F)$ called the special orthogonal group and it is denoted by $\mathrm{SO}(n,F)$.

    Symmetry plays an important role in quantum physics. For instance, we expect that the angular momentum is conserved in quantum mechanics. The angular momentum is conserved if the hamiltonian $H$ is invariant under rotations in 3-space $\mathbb R^3$. Here $H$ being rotationally invariant means that $[R,H]=0$ or $RH=HR$, where $R$ is a rotation. Suppose that a wave function $\psi$ satisfies the Schrodinger equation $H\psi=E\psi$. If $H$ is rotationally invariant, then a rotated wave function $R\psi$ also satisfies the Schrodinger equation. Rotations in $\mathbb R^3$ form the orthogonal group $\mathrm{SO}(3)$. Here we have a problem. Rotations in $\mathbb R^3$ are $3\times 3$ orthogonal matrices while $\psi$ is a complex-valued map. So mathematically a rotation $R\in\mathrm{SO}(3)$ would not act on $\psi$. (The Lie group action would take place by a matrix multiplication.) For example, let us consider a two-state quantum mechanical system such as spin-up and spin-down states of an electron. In that case, the hamiltonian would be a $2\times 2$ complex matrix and states (wave functions) are $\mathbb C^2$-valued maps. In order for a matrix $R$ to act on $\psi$, it needs to be a $2\times 2$ complex matrix! It turns out that there are $2\times 2$ complex matrices that may represent rotations. The rotation group $\mathrm{SO}(3)$ is topologically not simply connected. This means that not all the loops (closed curves) in $\mathrm{SO}(3)$ can be continuously shrink to a point. The universal cover of $\mathrm{SO}(3)$ is $\mathrm{SU}(2)$ and that the covering map $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a 2:1 map. This means that $\mathrm{SU}(2)$ is a simply connected space that consists of two pieces each of which is identified with $\mathrm{SO}(3)$. $\mathrm{SU}(2)$ is called the special unitary group and is defined by $$\mathrm{SU}(2)=\{U\in\mathrm{SL}(2,\mathbb C): UU^\ast=I\}.$$ The covering map $\rho$ being a 2:1 map plays an important role in Dirac's theory of electrons. That means rotating an electron by $360^\circ$ does not bring back the original state of the electron (spin-up or spin-down). One needs to rotate the electron by $720^\circ$ in order to bring back the original state of the electron. If $F:\mathbb R^3\longrightarrow\mathrm{SO}(3)$ is a frame in $\mathbb R^3$, then we can lift this frame to $\mathrm{SU}(2)$. The map $\tilde{F}:\mathbb R^3\longrightarrow\mathrm{SU}(2)$ is called a lifted frame of $F:\mathbb R^3\longrightarrow\mathrm{SO}(3)$ if $F=\rho\circ \tilde{F}$. This relatioship can be simply represented by the following diagram:
    $$\begin{array}{ccc}
    & & \mathrm{SU}(2)\\
    & \nearrow &\downarrow\\
    \mathbb R^3 & \longrightarrow & \mathrm {SO}(3)
    \end{array}$$
    If we assume that the hamiltonian $H$ is time-independent, the Schrodinger equation $H\psi=E\psi$ or $i\frac{\partial\psi}{\partial t}=H\psi$ has solution $$\psi(\mathbf r,t)=U(t)\psi(\mathbf r,0),$$ where $U(t)=\exp\left(-\frac{i}{\hbar}Ht\right)$. As mentioned earlier, we want $U(t)$ to be a special unitary matrix, i.e. $U(t)\in\mathrm{SU}(2)$. As well-known in differential geometry, in order for $U(t)$ to be a special unitary matrix, the matrix $-iH$ needs to be a trace-free skew-hermitian matrix i.e. $-iH$ satisfies $$-iH+(-iH)^\ast=0$$ or equivalently $H=H^\ast$ and that $\mathrm{tr}H=0$. That is, the hamiltonian $H$ is hermitian! The space of all $2\times 2$ skew hermitian matrices is denoted by $\mathfrak{su}(2)$. It is the Lie algebra of the Lie group $\mathrm{SU}(2)$ and the exponential map $$\exp: \mathfrak{su}(2)\longrightarrow\mathrm{SU}(2);\ A\longmapsto\exp(A)$$ relates a Lie algebra element to a Lie group element. More generally, if we assume that the hamiltonian $H$ is invariant under isometries of $\mathbb R^3$ i.e. $[H,R]=0$ for $R\in\mathrm{U}(2)$, then the trace-free condition can be omitted and $H$ is required to be just hermitian since the Lie algebra $\mathfrak{u}(2)$ is simply the set of all $2\times 2$ skew-symmetric matrices. We learn in quantum mechanics that we require the hamiltonian $H$ to be hermitian in order to ensure that its eigenvalues (energies) are real. I have given another reason why $H$ is required to be hermitian from a geometric point of view. In order for the angular momentum to be conserved in quantum mechanics, $H$ must be required to be hermitian. In general, $U(t)=\exp\left(-\frac{i}{\hbar}Ht\right)$ is a unitary transformation on the Hilbert space $\mathcal H$ of states with hermitian hamiltonian $H$. The matrix corresponding to the unitary transformation $U(t)$ is called S (Scattering) matrix in quantum mechanics.

    Note that states $\psi$ and $e^{i\theta}\psi$ have the same probability density, so they are not physically distinguishable. This means that the possible states are points in the complex projective space $\mathbb CP^1\cong S^2$. This can be seen as follows. Since we do not distinguish $\psi$ and $e^{i\theta}\psi$, distinct states are represented by elements (orbits) in the Riemannian symmetric space $\mathrm{U}(2)/\mathrm{U}(1)=\mathrm{SU}(2)/\mathrm{U}(1)\cong S^2$. Knowing that $\mathrm{SU}(2)=S^3$ and $\mathrm{U}(1)=S^1$, we have an interesting exponential identity (relation) $S^3/S^1=S^2$. In general, the following relations hold for $\mathrm{U}(n)$ and $\mathrm{SU}(n)$: $$\mathrm{U}(n+1)/\mathrm{U}(n)=\mathrm{SU}(n+1)/\mathrm{SU}(n)=S^{2n-1}.$$

    Posted 2 years ago #
  2. lee
    Key Master

    In the above thread, it should be noted that the relation $HU=UH$ or $UHU^{-1}=H$ for $U\in\mathrm{SU}(2)$ is not really an identity as one can easily check. It should rather be understood as an equivalence relation. More speciffically, we say that two hamiltonians $H_1$ and $H_2$ are equivalent if $H_2=UH_1U^{-1}=UH_1U^\ast$ for some $U\in\mathrm{SU}(2)$. That is, we do not distinguish between two Hamiltonians that are related by a similarity relation. This makes sense physically because those hamiltonians would have exactly the same eigenvalues (the same energies), so we would not be able to distinguish two systems derived from those similar hamiltonians.

    Posted 2 years ago #
  3. lee
    Key Master

    The Euclidean 3-space $\mathbb R^3$ can be identified with the set of $2\times 2$ hermitian matrices of the form $\underline{X}=\begin{bmatrix}
    x & y-iz\\
    y+iz & -x
    \end{bmatrix}$. Here the hermitian matrix $\underline{X}$ is identified with the vector $X=(x,y,z)\in\mathbb R^3$. Since $|X|^2=-\det\underline{X}$, the identification is an isometry. The unitary group $\mathrm{SU}(2)$ acts isometrically on $\mathbb R^3$ by the Lie group action $$\mathrm{SU}(2)\times\mathbb R^3\longrightarrow\mathbb R^3;\ (U,X)\longmapsto UXU^{-1}.$$ For fixed $U\in\mathrm{SU}(2)$, the map $$\mathbb R^3\longrightarrow\mathbb R^3;\ X\longmapsto UXU^{-1}$$ is an orientation preserving isometry of $\mathbb R^3$. Thus, the Lie group action induces a Lie group representation $\rho: \mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$. Since both $U$ and $-U$ result the same isometry, the representation $\rho$ is a $2:1$ map. $\mathrm{SU}(2)$ is the universal cover of $\mathrm{SO}(3)$ and $\rho$ is a universal covering map. The kernel of $\rho$ is $\{\pm I\}$ or $\mathbb Z_2$, so we have $\mathrm{SU}(2)/\mathbb Z_2=\mathrm{SO}(3)$. The quotient group $ \mathrm{SU}(2)/\mathbb Z_2$ is denoted by $\mathrm{PSU}(2)$ and is called the projective special unitary group. For the proof that $\mathrm{SU}(2)$ is the universal cover of $\mathrm{SO}(3)$, please see Shlomo Sternberg's wonderful book Group Theory and Physics, for example.

    The double cover of the special orthogonal group $\mathrm{SO}(n)$ is called the spin group and is denoted by $\mathrm{Spin}(n)$. Hence, the double cover $\mathrm{SU}(2)\stackrel{2:1}{\longrightarrow}\mathrm{SO}(3)$ is the spin group $\mathrm{Spin}(3)$.

    Posted 2 years ago #

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