SouthernMiss Math Forum » Tag: Mathematical Physics - Recent Posts
http://www.math.usm.edu/mathforum/
SouthernMiss Math Forum » Tag: Mathematical Physics - Recent PostsenFri, 18 Apr 2014 08:47:19 +0000lee on "Strings, Particles, and Surfaces"
http://www.math.usm.edu/mathforum/topic.php?id=163#post-328
Sun, 01 Jan 2012 13:27:43 +0000lee328@http://www.math.usm.edu/mathforum/<p><strong>Conformal Surfaces in $\mathbb{R}^3$</strong></p>
<p>Let $D\subset \mathbb{R}^3$ be a domain (connected open set) and $\varphi: D(x,y)\longrightarrow\mathbb{R}^3$ a parametric surface with metric $ds^2=e^u(dx^2+dy^2)$. It is often convenient to work with complex variables $z,\bar z$, so let $z:=x+iy$. Then the metric can be written as $ds^2=e^udzd\bar z$.</p>
<p>Let us introduce the following quantities that are typical in differential geometry of surfaces:<br />
\begin{align*}<br />
E&=\langle\varphi_x,\varphi_x\rangle,\ F=\langle\varphi_x,\varphi_y\rangle,\ G=\langle\varphi_y,\varphi\rangle\\<br />
l&=\langle\varphi_{xx},N\rangle,\ m=\langle\varphi_{xy},N\rangle,\ n=\langle\varphi_{yy},N\rangle<br />
\end{align*}<br />
Then as well-known in differential geometry the Gaussian curvature $K$ and the mean curvature $H$ are computed by the formulas:<br />
$$K=\frac{ln-m^2}{EG-F^2},\ H=\frac{Gl+En-2Fm}{2(EG-F^2)}\ \ \ \ \ (1)$$<br />
Definition. $\varphi: D\longrightarrow\mathbb{R}^3$ is called conformal if<br />
$$\langle\varphi_x,\varphi_x\rangle=\langle\varphi_y,\varphi_y\rangle=e^u,\ \langle\varphi_x,\varphi_y\rangle=0$$</p>
<p>Let us introduce the differential operators (Wirtinger operators)<br />
$$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right),\ \frac{\partial}{\partial\bar z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$$<br />
Then $\varphi_z$, $\varphi_{\bar z}$, $N$ (unit normal) satisfy<br />
\begin{align*}<br />
\langle\varphi_z,\varphi_z\rangle&=\langle\varphi_{\bar z},\varphi_{\bar z}\rangle=0,\ \langle\varphi_z,\varphi_{\bar z}\rangle=\frac{1}{2}e^u,\\<br />
\langle\varphi_z,N\rangle&=\langle\varphi_{\bar z},N\rangle=0,\ \langle N,N\rangle=1<br />
\end{align*}</p>
<p>The moving frame $(\varphi_z,\varphi_{\bar z},N)$ satisfies the Gauss-Weingarten equations<br />
\begin{align*}<br />
\sigma_z&=\mathcal{U}\sigma,\ \sigma_{\bar z}=\mathcal{V},\ \sigma=(\varphi_z,\varphi_{\bar z},N)^T\\<br />
\mathcal{U}&=\begin{pmatrix}<br />
u_z & 0 & Q\\<br />
0 & 0 & \frac{1}{2}He^u\\<br />
-H & -2e^{-u}Q & 0<br />
\end{pmatrix},\ \mathcal{V}=\begin{pmatrix}<br />
0 & 0 & \frac{1}{2}He^u\\<br />
0 & u_{\bar z} & \bar Q\\<br />
-2e^{-u}\bar Q & -H & 0<br />
\end{pmatrix}<br />
\end{align*}<br />
where $H$ and $Q$ are , respectively, the mean curvature and the coefficient of the Hopf differential $Qdz^2$. $Q$is defined to be<br />
$$Q=\langle F_{zz},N\rangle$$<br />
In terms of complex variables, the mean curvature $H$ can be written as<br />
$$\langle F_{z\bar z},N\rangle=\frac{1}{2}He^u$$<br />
by the formula in (1). The Gaussian curvature $K$ is calculate to be<br />
$$K=H^2-4Q\bar Qe^{-2u}\ \ \ \ \ (2)$$</p>
<p><strong>Quaternionic Description</strong></p>
<p>Let $\mathbb{H}$ be the algebra of quaternions with standard basis $\{\mathbf{1},\mathbf{i},\mathbf{j},\mathbf{k}\}$ where<br />
\begin{align*}<br />
\mathbf{i}^2&=\mathbf{j}^2=\mathbf{k}^2=-\mathbf{1},\\<br />
\mathbf{i}\mathbf{j}&=-\mathbf{j}\mathbf{i}=\mathbf{k},\ \mathbf{j}\mathbf{k}=-\mathbf{k}\mathbf{j}=\mathbf{i},\ \mathbf{k}\mathbf{i}=-\mathbf{i}\mathbf{k}=\mathbf{j}<br />
\end{align*}<br />
This basis can be represented by the pauli matrices $\sigma_\alpha$ as<br />
\begin{align*}<br />
\sigma_1&=\begin{pmatrix}<br />
0 & 1\\<br />
1 & 0<br />
\end{pmatrix}=i\mathbf{i},\ \sigma_2=\begin{pmatrix}<br />
0 & -i\\<br />
i & 0<br />
\end{pmatrix}=i\mathbf{j}\\<br />
\sigma_3&=\begin{pmatrix}<br />
1 & 0\\<br />
0 & -1<br />
\end{pmatrix}=i\mathbf{k},\ \mathbf{1}=\begin{pmatrix}<br />
1 & 0\\<br />
0 & 1<br />
\end{pmatrix}<br />
\end{align*}<br />
$\mathbb{H}$ is then identified with $\mathbb{R}^4$:<br />
$$q=q_0\mathbf{1}+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}\longleftrightarrow(q_0,q_1,q_2,q_3)\in\mathbb{R}^4$$<br />
The sphere $S^3\subset\mathbb{R}^4$ is naturally identified with the group of unitary quaternions $\mathbb{H}_1=\mathrm{SU}(2)$. $\mathbb{R}^3$ is identified with the space of imaginary quaternions $\mathrm{Im}\mathbb{H}$:<br />
$$X=-i\sum_{\alpha=1}^3 X_\alpha\sigma_\alpha\in\mathrm{Im}\mathbb{H}\longleftrightarrow X=(X_1,X_2,X_3)\in\mathbb{R}^3$$<br />
The scalar product of vectors in terms of quaternions and matrices:<br />
$$\langle X,Y\rangle=-\frac{1}{2}(XY+YX)=-\frac{1}{2}\mathrm{tr}XY$$</p>
<p><strong>Conformal Framing</strong></p>
<p>The moving frame $(e^{-u/2}\varphi_x,e^{-u/2}\varphi_y,N): D\longrightarrow\mathrm{SO}(3)$ can be lifted to a map $F: D\longrightarrow\mathrm{SU}(2)$ using the fact that $\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$<br />
$$\begin{array}{ccc}<br />
& & \mathrm{SU}(2)\\<br />
& \nearrow &\downarrow\\<br />
F:D & \longrightarrow & \mathrm{SO}(3)<br />
\end{array}$$<br />
Take the Weyl gauge fixing $\Phi=e^{u/4}F: D\longrightarrow\mathbb{H}_\ast$, where $\mathbb{H}_\ast=\mathbb{H}\setminus\{0\}$. Then<br />
\begin{align*}<br />
\varphi_x&=e^{u/2}\Phi^{-1}\mathbf{i}\Phi,\ \varphi_y=e^{u/2}\Phi^{-1}\mathbf{j}\Phi,\ N=\Phi^{-1}\mathbf{k}\Phi,\\<br />
\varphi_z&=-ie^{u/2}\Phi^{-1}\begin{pmatrix}<br />
0 & 0\\<br />
1 & 0<br />
\end{pmatrix}\Phi,\ \varphi_{\bar z}=-ie^{u/2}\Phi^{-1}\begin{pmatrix}<br />
0 & 1\\<br />
0 & 0\end{pmatrix}\Phi<br />
\end{align*}<br />
Thus we have the moving frame $(\varphi_z,\varphi_{\bar z},N)$ in terms of the quaternion $\Phi$.</p>
<p>The quaternion $\Phi$ satisfy linear differential equations called Lax equations:<br />
$$U=\Phi_z\Phi^{-1},\ V=\Phi_{\bar z}\Phi^{-1}$$<br />
where<br />
$$\begin{pmatrix}<br />
u_z/2 & -Qe^{-u/2}\\<br />
\frac{1}{2}He^{u/2} & 0<br />
\end{pmatrix},\ V=\begin{pmatrix}<br />
0 & -\frac{1}{2}He^{u/2}\\<br />
\bar Qe^{-u/2} & u_{\bar z}/2<br />
\end{pmatrix}$$<br />
The matrices $U$ and $V$ are called Lax pair. The Lax pair $U$ and $V$ satisfy<br />
$$U_{\bar z}-V_z+[U,V]=0$$<br />
which is equivalent to the Gauss-Codazzi equations<br />
\begin{align*}<br />
&\mbox{Gauss equation}\ u_{z\bar z}+\frac{1}{2}H^2e^u-2|Q|^2e^{-u}=0,\ \ \ \ \ (3)\\<br />
&\mbox{Codazzi equation}\ Q_{\bar z}=\frac{1}{2}H_ze^u\ \ \ \ \ (4)<br />
\end{align*}<br />
The Gauss-Codazzi equations are necessary and sufficient conditions for the existence of the corresponding surface with metric $e^udzd\bar z$, Hopf differential $Qdz^2$, and mean curvature $H$.</p>
<p>From Codazzi equation (4) it is clear that the mean curvature $H$ is constant if and only if $Q_{\bar z}=0$ i.e. $Q$ is holomorphic. Constant mean curvature surfacess will play an important role in our discussion.</p>
<p>Using the Gauss equation (3) and (2), we obtain the Liouville's equation<br />
$$\triangle u=-Ke^{2u}\ \ \ \ \ (5)$$</p>
<p><strong>2-D Dirac Equation</strong></p>
<p><em>Definition</em>. A first-order differential operator $\mathcal{D}$ on a Riemannian manifold is called a Dirac operator if $\mathcal{D}^2=-\triangle$ where $\triangle$ denotes the Laplacian operator.</p>
<p><em>Example</em>. [2-D Dirac Operator]<br />
\begin{align*}<br />
\mathcal{D}&=-i\sigma_1\partial_y+i\sigma_2\partial_x\\<br />
&=2\begin{pmatrix}<br />
0 & \partial_z\\<br />
-\partial_{\bar z} & 0<br />
\end{pmatrix}\ \ \ \ \ (6)\end{align*} is a 2-D Dirac operator. The 2-D Dirac equation<br />
$$\hbar c\mathcal{D}\psi=mc^2\psi$$ describes a fermion with spin-$\frac{1}{2}$ and mass $m$ confined in a plane. Here $\hbar$ is the reduced Planck constant and $c$ is the speed of light in vacuum. If $m\to 0$, we obtain 2-D Weyl equation<br />
$$\mathcal{D}\psi=0$$ which describes a neutrino with spin-$\frac{1}{2}$ confined in a plane. Hereafter (6) is the Dirac operator we will use.</p>
<p><em>Theorem</em>. The conformal frame $\Phi: D\longrightarrow\mathbb{H}_\ast$ satisfies the Dirac equation<br />
$$\mathcal{D}\Phi=He^{u/2}\Phi$$ where $\mathcal{D}=2\begin{pmatrix}<br />
0 & \partial_z\\<br />
-\partial_{\bar z} & 0<br />
\end{pmatrix}$.</p>
<p>Note that the conformal framing $\Phi$ can be written as<br />
$$\Phi=\begin{pmatrix}<br />
s_1 & -s_2\\<br />
\bar s_2 & \bar s_1<br />
\end{pmatrix}$$ for some $s_1,s_2: D\longrightarrow\mathbb{C}$. Let $\psi:=\begin{pmatrix}<br />
s_1\\<br />
\bar s_2<br />
\end{pmatrix}: D\longrightarrow\mathbb{C}^2$. Then one can readily see that $\psi$ satisfies the Dirac equation<br />
$$\mathcal{D}\psi=He^{u/2}\psi$$<br />
We may regard $s_1$, $\bar s_2$ as spinors and the wave function $\psi$ as a Dirac spinor field (2-spinor) with $s_1$ and $\bar s_2$ indicating spin-up and spin-down states. The Dirac spinor $\psi$ describes a fermion confined in a plane with spin-$\frac{1}{2}$ and mass<br />
$$m=\frac{\hbar}{c}He^{u/2}$$<br />
or the energy<br />
$$E=\hbar cHe^{u/2}$$<br />
Mathematically spinors $s_1$ and $s_2$ are the smooth sections of a spin bundle $K^\frac{1}{2}$ over $D$, where $K$ is the canonical bundle. If $m=0$ or equivalently $H=0$ (i.e. $\varphi$ is a minimal surface), then $\psi$ is a Weyl spinor field in which case $s_1$ and $s_2$ are holomorphic sections. Thus for every surface with mean curvature of the form $H=\frac{c}{\hbar}me^{-u/2}$ for some constant $m>0$, there is a corresponding fermion with spin-$\frac{1}{2}$ and mass $m$. For every minimal surface, there is a corresponding neutrino with spin-$\frac{1}{2}$.</p>
<p>The whole construction can be reversed, namely</p>
<p><em>Theorem</em>. Let $D\subset\mathbb{R}^2$ be a simply connected domain and the wave function (field) $\psi=(s_1,\bar s_2)^T: D\longrightarrow\mathbb{C}^2$ a solution of the Dirac equation with the mass $m\leq 0$<br />
$$\hbar c\mathcal{D}\psi=mc^2\psi$$<br />
Then<br />
$$\Phi:=\begin{pmatrix}<br />
s_1 & -s_2\\<br />
\bar s_2 & \bar s_1<br />
\end{pmatrix}: D\longrightarrow\mathbb{H}_\ast$$<br />
is a conformal frame of the surfaces<br />
\begin{align*}<br />
\varphi_1+i\varphi_2&=\int s_1^2dz-\int\bar s_2^2dz,\\<br />
\varphi_3&=\int s_1s_2dz+\bar s_1\bar s_2d\bar z<br />
\end{align*}<br />
The metric and the mean curvature are given by<br />
$$e^udzd\bar z=(|s_1|^2+|s_2|^2)^2dzd\bar z,\ H=\frac{c}{\hbar}me^{-u/2}$$</p>
<p>What the theorem says is that given a fermion with mass $m>0$ (or a neutrino with $m=0$) and spin-$\frac{1}{2}$ there is a corresponding surface represented by the integral formula (Weierstrass representation) as stated in the Theorem.</p>
<p>Now, we have seen a peculiar correspondence between certain particles (neutrinos or fermions with spin-$\frac{1}{2}$) and surfaces, and the question is why? The reason is that those surfaces are (Euclidean) string worldsheets. (I'll discuss more details in a separate thread.) Hence what we have seen so far is a correspondence between strings and particles. There can be infintely many surfaces with prescribed mean curvature, so does that mean there are infinitely many neutrinos and fermions with spin-$\frac{1}{2}$? The answer is no and here is why. The conformal factor $e^u$ is written in terms of spinors $s_1$, $s_2$ as<br />
$$e^{u/2}=|s_1|^2+|s_2|^2=\psi^\dagger\psi$$<br />
On the other hand, the wave function $\psi=(s_1,\bar s_2)^T$ describes a quantum mechanical system of two states (spin-up or spin-down). Since the particle with spin-$\frac{1}{2}$ must exist, $e^{u/2}=\psi^\dagger\psi=1$ i.e. $u=0$. This implies that $K=0$, i.e. the surface is flat and $H$ is constant which gives rise to the mass-mean curvature relation<br />
$$m=\frac{\hbar}{c}H\ \ \ \ \ (7)$$<br />
and the energy-mean curvature relation<br />
$$E=\hbar cH\ \ \ \ \ (8)$$<br />
The relation (8) is a reminiscence of the Planck relation or the Planck-Einstein equation $E=h\nu=h\frac{c}{\lambda}$ and it hints us that $H=\frac{2\pi}{\lambda}$ for non-zero constant mean curvature $H$. For minimal case ($H=0$), the only flat minimal surface is a plane, so that means the string that corresponds to a neutrino with spin-$\frac{1}{2}$ is a straight line. For non-zero constant mean curvature case, since $K=k_1k_2=0$ either one of principal curvatures $k_1$, $k_2$ is zero and the other is not. So, the constant mean curvature surface is a right circular cylinder, i.e. a tube like worldsheet swept by a closed string which corresponds to a fermion with spin-$\frac{1}{2}$. A right circular cylinder can be given by the parametric equation<br />
$$\varphi(\sigma,\tau)=(R\cos\sigma,R\sin\sigma,R^2\tau)$$<br />
where $0\leq\sigma\leq 2\pi$ and $-\infty<\tau<\infty$. The mean curvature is computed to be $H=\frac{1}{R}$. Thus (7) and (8) can be written as<br />
\begin{align*}<br />
m&=\frac{\hbar}{Rc}=\frac{h}{2\pi Rc},\ \ \ \ \ (9)\\<br />
E&=\frac{\hbar c}{R}=\frac{hc}{2\pi R}\ \ \ \ \ (10)<br />
\end{align*}<br />
Richard Feynman did not like string theory and he said to string theorists <code></code>if you can calculate the mass of electron, I will believe you.'' Using (10) we sort of attempt to calculate the mass of electron. We actually do not know the radius of the closed circular string. Instead we can try the classical electron radius $r_e=2.8179\times 10^{-15}\mbox{m}$ for $R$.<br />
Now,<br />
\begin{align*}<br />
m&=\frac{\hbar}{r_ec}\\<br />
&=\frac{1.0546\times 10^{-34}\mbox{J}\cdot\mbox{s}}{(2.8179\times 10^{-15}\mbox{m})(3\times 10^8\mbox{m/s})}\\<br />
&=1.248\times 10^{-28}\mbox{kg}<br />
\end{align*}<br />
The known electron mass is $m_e=0.9109\times 10^{-30}\mbox{kg}$, so the mass I obtained is about 100 times heavier.
</p>lrmead on "PT-Symmetric Quantum Mechanics 3"
http://www.math.usm.edu/mathforum/topic.php?id=46#post-131
Mon, 23 May 2011 10:12:05 +0000lrmead131@http://www.math.usm.edu/mathforum/<p>The three dimensional case is especially interesting = it is the real-word case.<br />
LRM
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-130
Thu, 19 May 2011 23:27:54 +0000lee130@http://www.math.usm.edu/mathforum/<p>In a private conversation, Larry pointed out that I was confused with thermal expansion and thermal diffusivity, and he did so rightly. The confusion arose due to my lack of understanding on thermodynamics.</p>
<p>I thought I finally found a physical reality of PT-symmetric quantum mechanics. It is a bit disappointing that it was not the case.
</p>lee on "PT-Symmetric Quantum Mechanics 3"
http://www.math.usm.edu/mathforum/topic.php?id=46#post-129
Mon, 16 May 2011 14:52:02 +0000lee129@http://www.math.usm.edu/mathforum/<p>Larry,</p>
<p>Yes, you can consider higher dimensional case in principle, though I am not sure if such case would be physically or mathematically interesting.</p>
<p>For higher dimensional case, the Hamiltonian $H$ would be $H=-\frac{\hbar^2}{2m}\Box+V(r^2)$, where $\Box$ is the d'Almebertian operator. For example, $\Box=-\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ and $r^2=-x^2+y^2+z^2$ for 3-dimensional case.
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-127
Fri, 13 May 2011 23:22:53 +0000lee127@http://www.math.usm.edu/mathforum/<p>Dirac was actually the first person who invented path integral formulation. His work was published in "The Lagrangian in Quantum Mechanics," Physikalische Zeitschrift der Sowjetunion 3, 64–72 (1933). Dirac was the one who formulated the amplitude of a particle to propagate from a point $q_I$ to anothe point $q_F$ in time $T$ by $$\langle q_F|e^{-iHT}|q_I\rangle=\int Dq(t)e^{i\int_0^TdtL(\dot{q},q)}.$$ On the other hand, Feynman was the one who provided a precise prescription to calculate the sum over paths. He is also the one who showed that Schrodinger's equation and the canonical commutation relation can be recovered from path integral formulation.
</p>lrmead on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-126
Fri, 13 May 2011 12:44:10 +0000lrmead126@http://www.math.usm.edu/mathforum/<p>I am unaware of any work done by Dirac. The earliest I am aware of is Feynman's thesis<br />
of 1948.<br />
LRM
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-124
Wed, 11 May 2011 16:43:59 +0000lee124@http://www.math.usm.edu/mathforum/<p>Larry,<br />
I knew about Feynman's Ph.D. thesis but I did not know that his thesis work contributed to his Nobel prize. By the way, wasn't path integral formulation first invented by P.A.M. Dirac long before Feynman did?
</p>lrmead on "PT-Symmetric Quantum Mechanics 3"
http://www.math.usm.edu/mathforum/topic.php?id=46#post-123
Tue, 10 May 2011 07:16:36 +0000lrmead123@http://www.math.usm.edu/mathforum/<p>Nonetheless, there are PT-symmetric systems in 2 or 3 dimensions. Here is an example in<br />
3D, one of the systems I want to study (still need a student laborer, though):</p>
<p>$$H=-1/2 \vec P^2 + iz r^2,$$</p>
<p>where $r=\sqrt{x^2+y^2+z^2}$.
</p>lrmead on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-122
Tue, 10 May 2011 07:10:48 +0000lrmead122@http://www.math.usm.edu/mathforum/<p>Two comments. First, that the Schr\"odinger equation is a heat equation with<br />
imaginary diffusivity is known. It is, however, a great idea to search for<br />
evidence of PT symmetry in thermal systems with negative k. As another elementary example,<br />
consider water. As its temperature is lowered below $4^o$ C, it expands! This is the<br />
reason why ice floats; it becomes less dense than water as it freezes.<br />
Lastly, the third person to get a Nobel for his dissertation was R. Feynman, whose thesis<br />
on path integrals was groundbreaking work. I do not know exactly what the citation was<br />
for Feynman, but physicists always consider path integrals (applied to QED) as his greatest<br />
contribution to physics.
</p>lee on "PT-Symmetric Quantum Mechanics 3"
http://www.math.usm.edu/mathforum/topic.php?id=46#post-121
Sun, 08 May 2011 12:07:28 +0000lee121@http://www.math.usm.edu/mathforum/<p>In the previous <a href="http://www.math.usm.edu/mathforum/topic.php?id=41">thread</a>, I pointed out that PT-symmetric quantum mechanics is a quantum theory in $\mathbb R^{2+2}$, not in this world (the real physical world $\mathbb R^{3+1}$), and that PT-symmetric quantum mechanics is a paraquaternionic quantum mechanics. Bender and his collaborators made a comparison between PT-symmetric quantum mechanics and the standard hermitian quantum mechanics. Such a comparison is indeed pointless because they are quantum theories in two different spaces. One may wonder if we can also consider hermitian quantum mechanics, i.e. quantum mechanics with hamiltonians that are hermitian in $\mathbb R^{2+2}$. The target space $\mathbb R^{2+2}=\mathbb C\oplus\mathbb C{\mathbf j}\cong\mathbb C^2$ of wave functions is equipped with the $\mathbb C$-hermitian symmetric form $$\langle(z_1,z_2),(w_1,w_2)\rangle=z_1w_1-\bar z_2\bar w_2.$$ Note that the complex unitary group of signature $(1,1)$ \begin{eqnarray*}\mathrm S\mathrm U(1,1)&=&\left\{\mathrm U\in\mathrm S\mathrm L(2,\mathbb C): \mathrm U\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\mathrm U^\ast=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\right\}\\&=&\left\{\begin{pmatrix}a & \bar b\\b &\bar a\end{pmatrix}:a,b\in\mathbb C, |a|^2-|b|^2=1\right\}\end{eqnarray*} preserves the $\mathbb C$-hermitian symmetric form.</p>
<p>It is worth note the following theorem. </p>
<p><em>Theorem</em> Let $V$ be a finite dimensional vector space over $\mathbb C$ with $\mathbb C$-hermitian symmetric form $\langle\ ,\ \rangle$ of signature $(p,q)$. Let $A: V\longrightarrow V$ be a linear operator on $V$. If $A$ is a hermitian operator then $\langle Av,v\rangle$ is real for all $v\in V$. The converse need not be true.</p>
<p><em>Remark</em> If $\langle\ ,\ \rangle$ is positive definite (i.e. $q=0$), the converse is also true. This means that there can be no quantum mechanics other than hermitian quantum mechanics in our world, since the Hilbert space structure on the space $\mathcal H$ of states is induced by the the positive definite hermitian symmetric form on $\mathbb C$, the target space of state functions. That is, the theorem in the positive definite case hints us that PT-symmetric quantum mechanics belongs to another world.</p>
<p>Regardless of the indefinite inner product on the target space, hermitian hamiltonians would still have real eigenvalues, so there is no reason we cannot consider hermitian quantum mechanics in $\mathbb R^{2+2}$. But then the question is how different is it from the standard hermitian quantum mechanics in our world? Well, the answer is "very different." I will give you an example from a finite dimensional case. Let us consider the $2\times 2$ hermitian hamiltonian matrix (also called energy matrix in physics) $$H=\begin{bmatrix}3 & 3+\sqrt{3}i\\3-\sqrt{3}i & 2\end{bmatrix}.$$ It has two real eigenvalues $-1, 6$ and the corresponding eigenvectors are, respectively, $\psi_1=\left(-\frac{3}{2}-\frac{\sqrt{3}}{2}i,2\right)$ and $\psi_2=(\sqrt{3}+i,\sqrt{3})$. Note that $|\psi_1|^2=-1$ and $|\psi_2|^2=1$ due to the indefinite hermitian form on $\mathbb R^{2+2}=\mathbb C\oplus\mathbb C{\mathbf j}$. Hence the hermitian matrix would exhibit the same quantum mechanical structure as that of a PT-symmetric hamiltonian matrix with the same eigenvalues. For instance, $$H_{PT}=\begin{bmatrix}<br />
\frac{5}{2}+\frac{3}{2}i & \sqrt{13}\\\sqrt{13} & \frac{5}{2}-\frac{3}{2}i\end{bmatrix}.$$<br />
The space-time $\mathbb R^{2+2}$ of split-signature $(2,2)$ is a place that can accommodate both hermitian and PT-symmetric quantum mechanics. However Bender at al.'s comparison between them would still be pointless because they exhibit the same quantum physics there.</p>
<p>One last thing I would like to mention is that only 1-dimensional quantum mechanics (hermitian or PT-symmetric) appears to be interesting or appears to make sense in $\mathbb R^{2+2}$.
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-119
Thu, 28 Apr 2011 02:12:24 +0000lee119@http://www.math.usm.edu/mathforum/<p>In the above thread, I said "the wave function describes a particle in spacetime, so it should take its value in spacetime." I just want to make it clear that the above statement is not something commonly agreed or shared by the physics community but is solely my own viewpoint which I believe is a logical assumption. The common physicists' viewpoint on the wave function is that "it is not a real wave but a collection of numbers whose only ability is to predict the probability of a phenomenon that may happen at some point in the future."
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-117
Sun, 24 Apr 2011 21:22:13 +0000lee117@http://www.math.usm.edu/mathforum/<p>Since $\mathcal C^2=\imath_{\mathcal H}$, $\mathcal C$ has two real eigenvalues $\pm 1$. So $\mathcal H$ can be written as the direct sum $\mathcal H=U\oplus W$ where $U$ is the eigenspace corresponding to the eigenvalue 1 and $W$ is the eigenspace corresponding to the eigenvalue $-1$. Notice that $U=\mathcal H^+$ and $W=\mathcal H^-$.
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-115
Tue, 19 Apr 2011 20:59:46 +0000lee115@http://www.math.usm.edu/mathforum/<p>The analytic continuation of the heat equation $\frac{\partial \psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+x^{\nu+2}\psi$ into Minkowski space-time $\mathbb R^{3+1}$ is $i\frac{\partial \psi}{\partial t}=\frac{\partial^2\psi}{\partial x^2}-x^{\nu+2}\psi$. This is not a Schrodinger equation. So the heat equation of negative thermal expansion has no connection to quantum mechanics in Minkowski spacetime.
</p>lee on "PT-Symmetric Quantum Mechanics 2"
http://www.math.usm.edu/mathforum/topic.php?id=41#post-113
Sun, 17 Apr 2011 14:24:59 +0000lee113@http://www.math.usm.edu/mathforum/<p>Today I will <em>attempt</em> to unveil the true face of PT-symmetric quantum mechanics.</p>
<p><em>PT-Symmetric Quantum Mechanics and the Heat Equation of Negative Thermal Expansion</em></p>
<p>The PT-symmetric potential $V(x)=x^2(ix)^\nu$ is either negative potential or pure imaginary potential. Either case appear to be unphysical, however as long as the PT-symmetric Hamiltonian $H=p^2+V(x)$ has real eigenvalues (i.e. real energies), PT-symmetric quantum mechanics may offer a viable alternative quantum theory. In fact, numerical calculations show that $H$ has real eigenvalues under unbroken PT-symmetry. There is another distinctive feature of PT-symmetric quantum mechanics. There are eigenfunctions (pure states) that have negative probability density! In fact there are exactly the same number of eigenfunctions that have positive probability density and negative probability density. Note that these eigenfunctions correspond to mutually distinct eigenvalues. Here the probability density of the eigenfunction $\psi_n(t,x)$ is mathematically the square norm $|\psi_n|^2$. Physically it is the intensity (the amplitude) of a particle represented by the wave function $\psi$ that may appear with the energy $E_n$. In standard quantum mechanics, a wave function $\psi(t,x)$ takes its value in $\mathbb C$, the complex numbers, so the probability density is alway nonnegative. This indicates that the target space for wave functions in PT-symmetric quantum mechanics is not merely $\mathbb C$ but something else. I will reveal what it is to you later. Before I continue I will list a couple of articles on PT-symmetric quantum mechanics that may help you understand it better and provide more details:</p>
<p>1. <a href="http://arxiv.org/abs/quant-ph/9809072">PT-Symmetric Quantum Mechanics</a>, Carl Bender, Stefan Boettcher, Peter Meisinger</p>
<p>2. <a href="http://arxiv.org/abs/quant-ph/0501052">Introduction to PT-Symmetric Quantum Theory</a>, Carl M. Bender</p>
<p>The PT-symmetric Schrodinger equation $$i\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+x^2(ix)^\nu\psi$$ has its counterpart $$\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+x^{\nu+2}\psi$$ in Euclidean space $\mathbb R^2$. The Euclidean counterpart becomes the PT-symmetric Schrodinger equation under the Wick rotation $t\to it; x\to ix$. The Euclidean counterpart appears to be a heat equation but with a <em>wrong</em> sign. It has negative thermal diffusivity $-1$! <strike>Since materials expand upon heating, the thermal diffusivity is required to be positive. Well, at least that is what I learned. So at first I thought the equation is physically meaningless. To my surprise, there are materials, for example <em>cubic zirconium tungstate</em> (Zr${\rm W}_2{\rm O}_8$) that contract upon heating therefore the thermal diffusivity can be negative. For more details about negative thermal expansion, click <a href="http://en.wikipedia.org/wiki/Negative_thermal_expansion">here</a>.<br />
I am very curious as to whether PT-symmetric quantum mechanics may be used to study such materials. Especially, I am very curious if there is a material that contracts upon heating and also internally generates heat following the Wick rotated PT-symmetric potential $V(-ix)$. If so, PT-symmetric quantum mechanics may be a realistic model (I mean a physical model that can be used to solve real life problems) of quantum theory with interesting practical applications.</strike> Now, I am about to make a wild claim about PT-symmetric quantum mechanics.</p>
<p><em>PT-Symmetric Quantum Mechanics is a 2-Time Quantum Theory!</em></p>
<p>I mentioned earlier that there may be a physically realistic counterpart of the PT-symmetric quantum mechanics in Euclidean space. Although the corresponding heat conduction problem is one-dimensional, it may be considered as a heat conduction problem in Euclidean 4-space $\mathbb R^4$ with condition that heat conduction occurs along only one coordinate $x$-direction. Let $(t,x,y,z)$ be the coordinates of $\mathbb R^4$. Then the Euclidean metric on $\mathbb R^4$ is given by the quadratic form $(dt)^2+(dx)^2+(dy)^2+(dz)^2$. While the heat equation turns into the PT-symmetric Schrodinger equation under the Wick rotation $t\to it; x\to ix$ the Euclidean space $\mathbb R^4$ turns into $\mathbb R^{2+2}$ with indefinite metric $-(dt)^2-(dx)^2+(dy)^2+(dz)^2$. This is a space-time with two time coordinates. So PT-symmetric quantum mechanics is a quantum mechanics in $\mathbb R^{2+2}$, not in $\mathbb R^{3+1}$ as people have thought. It is well-known that $\mathbb R^{2+2}$ can be identified with the space of <em>paraquaternions</em> by the identification $$(t,x,y,z)\longleftrightarrow \xi=t{\mathbf 1}+x{\mathbf i}+y{\mathbf j}+z{\mathbf k},$$ where ${\mathbf 1},{\mathbf i}.{\mathbf j},{\mathbf k}$ satisfy the multiplication law: \begin{eqnarray*}{\mathbf i}^2=-{\mathbf 1},\ {\mathbf 1}^2={\mathbf j}^2={\mathbf k}^2={\mathbf 1},\\{\mathbf i\mathbf j}=-{\mathbf j\mathbf i}={\mathbf k},\ {\mathbf j\mathbf k}=-{\mathbf k\mathbf j}=-{\mathbf i},\ {\mathbf k\mathbf i}=-{\mathbf i\mathbf k}={\mathbf j}.\end{eqnarray*} That is, $\mathbb R^{2+2}$ may be regarded as a 4-dimensional real algebra $\mathbb R{\mathbf 1}\oplus\mathbb R{\mathbf i}\oplus\mathbb R{\mathbf j}\oplus\mathbb R{\mathbf k}$. We may identify the field of complex number $\mathbb C$ with $\mathbb R{\mathbf 1}\oplus\mathbb R{\mathbf i}$ and thereby we can identify $\mathbb R^{2+2}$ with the right complex linear space $\mathbb C\oplus\mathbb C{\mathbf j}$. If we define the conjugate paraquaternion $\bar\xi=t{\mathbf 1}-x{\mathbf i}-y{\mathbf j}-z{\mathbf k}$, then the square norm $|\xi|^2$ in $\mathbb R^{2+2}$ is given by $|\xi|^2=\xi\bar\xi$. If we write $\xi=z+w{\mathbf j}$ where $z,w\in\mathbb C$, then $|\xi|^2=|z|^2-|w|^2$. Here $|z|^2$ denotes the square modulus $|z|^2=z\bar z$ of the complex number $z$.</p>
<p>In quantum mechanics, the wave function takes its value in the complex plane $\mathbb C$, but no one knows why. Furthermore, no one knows what the wave function really is. It is just that that is the way quantum mechanics was built and it worked! Often in physics why it works matters not much as long as it works. de Broglie hypothesized in his Ph.D. thesis that not only light but other matters would exhibit both aspects of particles and waves. (I could be wrong but if I remember correctly, de Broglie was the first person of two physicists who won the Noble physics prize from their Ph.D. theses. The other person was Sheldon Lee Glashow.) In quantum mechanics, a free particle is modeled by a de Broglie wave $\psi(x,t)=Ae^{ikx-\omega t}$, a complex-valued plane wave function satisfying the dispersion relation $p=\hbar k$, where $A$ is the amplitude factor, $k$ a wave number, $p$ momentum, $\hbar$ the reduced Planck constant, and $\omega$ the frequency. The complex-valued plane wave was well known even before the birth of quantum mechanics, because it is also a solution of the Maxwell's equations in electromagnetism. Since the wave function $\psi$ describes a particle in space-time, wouldn't it be more reasonable to assume that the wave function $\psi$ takes its value in space-time, not in the complex plane $\mathbb C$? In PT-symmetric quantum mechanics case, $\psi$ should take its value in $\mathbb R^{2+2}$ or equivalently in $\mathbb C\oplus\mathbb C{\mathbf j}$. If we denote the linear space of all states by $\mathcal H$, then $\mathcal H$ may be written as $$\mathcal H=\mathcal H^+\oplus\mathcal H^-,$$ where $\mathcal H^+$ is the linear subspace of states of positive square norm (i.e. positive probability density) with the null state $0$, and $\mathcal H^-$ is the linear subspace of states of negative square norm (i.e. negative probability density) with the null state $0$. The pure states $|\psi_n^+\rangle$ with positive probability density form an orthonormal system of $\mathcal H^+$ and the pure states $|\psi_n^-\rangle$ with negative probability density form an orthonormal system of $\mathcal H^-$. Since the inner product $\langle\ |\ \rangle$ on $\mathcal H$ is indefinte, $\mathcal H$ is not a Hilbert space. However we can somehow give a Hilbert space structure. Define an endomorphism $\mathcal C: \mathcal H\longrightarrow\mathcal H$ by $$\mathcal C=\sum_{n=1}^\infty\{|\psi_n^+\rangle\langle\psi_n^+|+|\psi_n^-\rangle\langle\psi_n^-|\}.$$ Then one can easily check that $$\mathcal C|\psi_m^+\rangle=|\psi_m^+\rangle,\ \mathcal C|\psi_m^-\rangle=-|\psi_m^-\rangle,\ m=1,2,\cdots.$$ Hence, $\mathcal C^2=\imath_{\mathcal H}$, i.e. $\mathcal C$ is an involution. We now define a new inner product $(\ ,\ )$ on $\mathcal H$ by $$(\phi,\psi):=\langle\phi|\mathcal C\psi\rangle.$$ Then $(\ ,\ )$ is a positive definite inner product. The indefinite inner product space $\mathcal H$ with the involution $\mathcal C$ is a <a href="http://en.wikipedia.org/wiki/Indefinite_inner_product_space"><em>Krein space</em></a>. With the inner product $(\ ,\ )$, all states have positive probability density and hence we no longer have to deal with physically awkward negative probability.</p>
<p><em>To be continued</em>.
</p>lee on "PT-Symmetric Quantum Mechanics 1"
http://www.math.usm.edu/mathforum/topic.php?id=36#post-110
Thu, 07 Apr 2011 23:43:21 +0000lee110@http://www.math.usm.edu/mathforum/<p>Thanks, Larry. I was paying attention to only the hamiltonian but not to the Schrodinger's equation. I think I can come up with another explanation (but in a sense equivalent). The hamiltonian $H=-\frac{\partial^2}{\partial x^2}+V(x)$ is PT-symmetric, where $V(x)$ a PT-symmetric potential. So the energy operator $i\frac{\partial}{\partial t}$ must be PT-symmetric too. In order for the energy operator to be PT-symmetric, the time-reversal operator $\mathcal T$ must map $t\to -t;\ i\to -i$. This way the motion of a particle (wave function) evolves according to Schrodinger's equation regardless of time-orientation, forward in time or backward in time.
</p>lrmead on "PT-Symmetric Quantum Mechanics 1"
http://www.math.usm.edu/mathforum/topic.php?id=36#post-104
Thu, 07 Apr 2011 08:41:38 +0000lrmead104@http://www.math.usm.edu/mathforum/<p>One additional comment: All PT symmetric H's seem to have all real eigenvalues<br />
provided the V(x) is differentiable. For example, in the<br />
$V=(ix)^\epsilon |x|^n$ potential, there are all real eigenvalues for n an EVEN<br />
integer, but only a finite number of real eigenvalues otherwise. Also, $\epsilon$<br />
must be an integer greater than or equal to 1. Both conditions are needed for<br />
the differentiability in x.</p>
<p>This has been verified numerically in a recent paper of mine with C. Bender.
</p>lrmead on "PT-Symmetric Quantum Mechanics 1"
http://www.math.usm.edu/mathforum/topic.php?id=36#post-103
Thu, 07 Apr 2011 08:37:30 +0000lrmead103@http://www.math.usm.edu/mathforum/<p>The garbled equation in the last post should read,<br />
-${d^2\over dx^2}\psi^{*}(x,-t)+v(x)\psi^{*}(x,-t)=+i{d\over dt}\psi^{*}(x,-t).$
</p>lrmead on "PT-Symmetric Quantum Mechanics 1"
http://www.math.usm.edu/mathforum/topic.php?id=36#post-102
Thu, 07 Apr 2011 08:34:18 +0000lrmead102@http://www.math.usm.edu/mathforum/<p>I can answer the question about time reversal being t -> -t and i -> -i.<br />
The Schroedinger equation (sans all the constants) is,<br />
$-\psi'' + v(x) \psi(x) = i {d\over dt} \psi$.<br />
We believe that (like Newtons law) physics is independent of time; that is, if a movie<br />
of a physical event is played backwards, then the backwards event is physically allowed by<br />
the laws of nature (initial conditions for the "backward" event may be tough to achieve).<br />
Ok, suppose $\psi(x,t)$ is a solution of the above. Can we generate a solution from it<br />
that is time reversed? Well, taking t -> -t, gives<br />
$-\psi''(x,-t) +v(x)\psi(x,-t) = -i {d\over dt} \psi(x,-t)$, which *isn't* the Schrodinger<br />
equation! However, if one takes the complex conjugate,<br />
$-{d^2\over dx^2}\psi^{*} +v(x) \psi^{*}(x,-t)=+i{d\over dt}\psi^{*}(x,-t)$, which *is*<br />
the schrodinger equation again with solution<br />
$\psi(x,t) \to \psi^{*}(x,-t)$.
</p>lee on "PT-Symmetric Quantum Mechanics 1"
http://www.math.usm.edu/mathforum/topic.php?id=36#post-100
Mon, 04 Apr 2011 12:49:15 +0000lee100@http://www.math.usm.edu/mathforum/<p>I came across PT-symmetric quantum mechanics years ago when my friend Larry gave a talk on it in physics department. If I remember correctly PT-symmetric quantum mechanics was first studied by Carl M. Bender, a well-known physicist and a professor in physics at Washington University, St. Louis. If I remember correctly, Carl was a former research mentor (not his thesis adviser though) of Larry back when he was a graduate student there. He also has been Larry's research collaborator.</p>
<p>When I first heared about PT-symmetric quantum mechanics it just did not make sense to me and appeared to be totally unphysical. Part of reason was that I was familiar with the standard hermitian quantum mechanics like many others who were exposed to quantum mechanics and I had (probably still have) my own prejudices. Many years since then I have somewhat changed and became more open then I used to be. Accordingly my view of (or attitude toward) PT-symmetric quantum mechanics is different now from before. I still do not fully understand PT-symmetric quantum mechanics and my impression is no one does, especially the mathematics of it. I still am not convinced that this is truly a physical theory but I will leave that judgement to physicists. It is however a very peculiar and intriguing theory. My hope is that at some point I can have a better understanding on the physics as well as the mathematics of PT-symmetric quantum mechanics throughout discussions here. Before this becomes some sort of my own confession, I had better get on the subject. A friendly warning is that what I am going to write is based on my own perspective (especially as a mathematician) and it may not be coherent with a physicist's perspective, especially that of Carl Bender or Larry Mead.</p>
<p><em>So what is PT-symmetric quantum mechanics anyway?</em></p>
<p>In PT-symmetric quantum mechanics the hamiltonian or the energy operator $H=T+V$ is PT-symmetric in contrast to the hamiltonian being hermitian in the standard quantum mechanics. Before continue, I need to explain what $\mathcal P$ and $\mathcal T$ stand for. $\mathcal P$ stands for the parity operator and $\mathcal T$ stands for the time reversal operator that are Lorentz transformations of space-time. In 1+1-dimensional case, $${\mathcal P}: x\to -x;\ {\mathcal T}:t\to -t.$$ In addtion, $\mathcal T$ is also required to map $i$ to $-i$. As a result, the momentum operator $\hat p=-i\frac{d}{dx}$ is PT-symmetric. (Here the reduced Planck constant $\hbar=1$ for simplicity.) The additional requirement of the time-reversal operator may be in fact understood as a time-reverse that occurs in the Euclidan counter part $\mathbb R^2$ of the space-time $\mathbb R^{1+1}$. That is, the time-reversal operator $\mathcal T$ flips time-orientation in both Lorentzian space-time and Euclidean space-time. I do not understand yet why the additional requirement is needed for the time-reversal operator $\mathcal T$ or why it is important to make the momentum operator PT-symmetric. Maybe Larry can answer that. Another important thing to mention is that the position operator $\hat x$ is not recognized as an operator in PT-symmetric quantum mechanics because it is not PT-symmetric. $\hat x$ is hermitian so it is an operator in the standard quantum mechanics. PT-symmetric quantum mechanics is a quantum theory of bound state particles, i.e. particles whose motion is restricted by potential energy $V(x)$. Free particle motion is governed only by the kinetic energy $T=p^2$ and the kinetic energy is hermitian (and also PT-symmetric) and so it is not considered in PT-symmetric quantum mechanics.</p>
<p><em>Non-Hermitian PT-Symmetric Hamiltonians</em></p>
<p>The following class of hamiltonians provides typical examples of non-hermitian PT-symmetric hamiltonians. $$H=p^2+x^2(ix)^\nu,$$ where $\nu> 0$ is a positive integer. (Here the mass $m=\frac{1}{2}$ for simplicity.) Clearly $H$ is PT-symmetric and what makes the hamiltonian peculiar is the non-hermitian PT-symmetric potential energy $V(x)=x^2(ix)^\nu$. Note that $\nu=0$ is the hamiltonian of harmonic oscillator which is hermitian. Unlike hermitian hamiltonian case, it is not clear at all that the PT-symmetric hamiltonians would have real eigenvalues (i.e. real energies) and that the minimum energy. Some numerical calculations show that they indeed have real eigenvalues but so far no one has been able to prove that PT-symmetric hamiltonians do have real eigenvalues. One difficulty with attempt to prove it is that it depends on each hamiltonian. In other words, depending on the hamiltonian one needs to impose additional condition in order to make sure that PT-symmetry is not broken. PT-symmetric hamiltonians often fail to have real eigenvalues under broken PT-symmetry.</p>
<p><em>To be continued</em>.
</p>