# Arguments on Quantum Bayesianism from the Viewpoint of Quantum Logic

This was a project thesis for the mathematical logic course (2016) at Beijing Normal University.  It is notorious that quantum mechanics is full of counterintuitive ideas.  Sometimes one needs a philosophical ground to make an idea easier to accept. There are alternative theories based on different philosophical viewpoints that can be as good as the orthodox in predictions.  One of them is quantum Bayesianism, another one is Bohmian mechanics.  Here I outline a case where a generalisation of a classical probability formula lacking interpretation as probability.

This thesis introduces some basic concepts in quantum logic. Then I outline the connection between Gleason’s Theorem and the density matrices. I point out one problem in the alternative formalism of quantum Bayesianism. The meaning of making noncommutative generalisation of the classical Bayesian rule is unclear. Since the space of events is non-boolean, the term interpreting the intersection of events doesn’t have proper meaning in the quantum cases. Such generalisation of probability is just formal but doesn’t have a clear interpretation of probability.

Warning: The file is in a draft state containing errors and typos (I have lost the latex file for it and this is the only draft I can find).

This article doesn’t address any important issue and it’s no longer of any interest for me. The reason I wrote such nonsense is due to my quest for an interpretation of quantum mechanics, for which I explain on my page Thoughts on Physics, Maths and Reality

Full article- Arguments on Quantum Bayesianism from the Viewpoint of Quantum Logic

# [Reading Notes] Mechanics by Landau and Lifshitz

The following is an excerpt from [$\S 43$ The action as a function of coordinate]. I would like to comment on the formal derivation of Hamilton’s equations and the problems of independence of variations.

Note that in the derivation above, the variations $\delta p$ and $\delta q$ are regarded as independent.* Actually, $\delta q$ is arbitrary but $\delta p$ is not, even though p, q are both independent variables. Since $p$ in connected with $\dot{q}$ and  $\delta p$ and $\delta \dot{q}$ are not independent.

Notice that before (43.8) is derived, we have applied Ledrendre  Transformation which requires that

$\dot{q}=\frac{\partial H}{\partial p}$.                                               (1)

So the coefficient of $\delta p$ is  $0$,  and $\delta q$ is arbitrary, so its coefficient must be $0$, Hence we get another Hamilton’s  equation

$\dot{p}=-\frac{\partial H}{\partial q}$.                                              (2)

Notice that we only derive half of Hamilton’s equations from the procedure above.

Since we can not say that we derive Hamilton’s equations by applying Hamilton’s equations. In order to make this induction above complete, we have to give the proof of another half of Hamilton’s equations, that is (1).

(1) is related to the definition of  $p=\frac{\partial L}{\partial \dot{q}}$. From the definition of Hamiltonian

$H=\Sigma \dot{q}p-L$

and $\dot{q}=\dot{q}(p,q,t)$, then

With the definition of p, $p=\frac{\partial L}{\partial \dot{q}}$, we have

$\dot{q}=\frac{\partial H}{\partial p}$.

Hence the half part of Hamilton’sequations is derived.

In this way of deriving Hamilton’s equation, strictly, we first derive (1) from the definition of p, and then by applying (1) in $\delta S$, (2) can be derived.

*We should notice that variations here are simultaneous variations and it’s for a complete system.

# [Solutions] Ch 2 Manifolds in “Spacetime and geometry”

These are parts of the solutions to exercises from:

Carroll, S. M. (2005). Spacetime and geometry: An introduction to general relativity. Addison Wesley.

Solutions:

1. First consider mapping the infinite cylinder to a semi-infinite cylinder, then consider projecting points on it to a plane.

Let $L=1$,

$(acos\theta,asin\theta,z)\mapsto ^f (acos\theta,asin\theta,e^z)\mapsto ^g (acos\theta/e^z,asin\theta/e^z,1)$.

$g\circ f$ is the map desired whose image is an open set $R^2\backslash (0,0)$.

2. No. $\mathbb{R}^k$ must be k-dimension manifolds. Furthermore, the dimension of a manifold is unique.

Suppose that $U$ is a subset of a manifold $M$, if there are two charts with different dimension $m$ and $latex n$, then there exists $\Phi_1, \Phi_2\in C^{\infty}$, $\Phi_1:U\to \mathbb{R}^m$, $\Phi_2:U\to \mathbb{R}^n$. Then $\Phi_2\circ \Phi_1$ is a diffeomorphism from $\Phi_1(U)$ to $\Phi_2(U)$. Hence m must be equal to n.

Note: Here I use such an assertion( J. Milnor, Topology from Differentiable Viewpoint, P4).

Assertion. If $f$ is a diffeomorphism between open sets $U\subset R^k$ and $V\subset R^l$, then $k$ must equal $l$, and the linear mapping

$df_x:R^k\to R^l$

must be nonsingular.

Proof. The composition $f^{-1}\circ f$ is the identity map of U; hence $d(f^{-1})_v\circ df_x$ is the identity map of $R^k$. Similarly $df_x\circ d(f^{-1})_v$ is the identity map of $R^l$. Thus $df_x$ has a two-sided inverse, and it follows that $k=l$.

By the way, $k=l$ can also be seen from the $max\{rank(BA), rank(AB)\}=min\{rank(A),rank(B)\}$, $A,B$ denote the linear map $d(f^{-1})_v$ and $df_x$ respectively.

3. Trivial.

4. First two can be verified directly.

Composition formula:

$[X,Y]^{\mu}=X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu}$.

$[X,Y]^{\mu}=[X,Y](x^{\mu})=X(Y(x^{\mu}))-Y(X(x^{\mu}))=X(Y^{\mu})-Y(X^{\mu})\\ =X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu}$.

Transformation:

$[X,Y]^{\mu'}=X^{\mu'}\partial_{\lambda'}Y^{\mu'}-Y^{\lambda'}\partial_{\lambda'}X^{\mu'}\\ =X^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}}Y^{\mu})-Y^{\lambda}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}}X^{\mu})\\ =X^{\lambda}Y^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}})+\frac{\partial x^{\mu'}}{\partial x^{\mu}}x^{\lambda}\partial_{\lambda}Y^{\mu}-X^{\lambda}Y^{\mu}\partial_{\lambda}(\frac{\partial x^{\mu'}}{\partial x^{\mu}})-\frac{\partial x^{\mu'}}{\partial x^{\mu}}Y^{\lambda}\partial_{\lambda}X^{\mu} \\ = \frac{\partial x^{\mu'}}{\partial x^{\mu}}(X^{\mu}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu})= \frac{\partial x^{\mu'}}{\partial x^{\mu}}([X,Y]^{\mu})$.

5. Set  $V_{1}=\partial _{x}$, $V_{2} = A(x, y)\partial _{x} + B(x, y)\partial _{y}$. The commutator will be given by

$[V_{1} , V_{2} ] = [\partial _{x} A(x, y)]\partial _{x} + [\partial _{x} B(x, y)]\partial _{y}$.

Set $V_{2} =x \partial _{x} + \partial _{y}$. Then $V_1$, $V_2$ are nowhere vanishing, and their commutator is nowhere vanishing as well.

6.

(a)

$r=\sqrt{x^2+y^2+z^2}=\sqrt{1+\lambda^2}$

$\theta=arccos \frac{z}{\sqrt{x^2+y^2+z^2}}=arccos\frac{\lambda}{\sqrt{1+\lambda^2}}$

$\phi=\lambda$

($\lambda\in [0,2\pi)$)

(b)

Cartesian: $(-sin\lambda, cos\lambda,1)$,

Polar:$(\frac{\lambda}{\sqrt{1+\lambda^2}},\frac{-1}{1+\lambda^2},1)$.

7.

(a) $\frac{\partial x^{\mu}}{\partial x^{\nu'}}=\frac{\partial(x,z)}{\partial(\chi,\theta)}=\left(\begin{array}{cc} {cosh\chi sin\theta}&{sinh\chi cos\theta}\\{sinh\chi cos\theta}&{-cosh\chi sin\theta} \end{array}\right)$
(b)$dx=cosh\chi sin\theta dx+sinh\chi cos\theta d\theta$,
$dz=sinh\chi cos\theta dx-cosh\chi sin\theta d\theta$,
$ds^2=(sin^2\theta+sinh^2\chi)(dx^2+d\theta^2)$.

8.
$d(\omega\wedge\eta)_{\mu_1\dots\mu_{p+q+1}}\\=\frac{(p+q)!(p+q+1)}{p!q!}\partial_{[\mu_1}\omega_{[\mu_2\dots\mu_{p+1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]}\\=\frac{(p+q)!(p+q+1)}{p!q!}[(\partial_{[\mu_1}\omega_{[\mu_2\dots\mu_{p+1}}).\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]}+\omega_{[\mu_{2}\dots\mu_{p+1}}.(\partial_{[\mu_1}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]})]\\=\frac{(p+q)!(p+q+1)}{p!q!}[(\partial_{[\mu_1}\omega_{[\mu_2\dots\mu_{p+1}}).\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]}+(-1)^p\omega_{[\mu_{1}\dots\mu_{p}}.(\partial_{[\mu_{p+1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]})]\\=(d\omega)\wedge\eta+(-1)^p\omega\wedge(d\eta)$

9.

(a)$d*F=3\partial_{[\mu}F_{\nu\rho ]}=1/2[\partial_{\mu}(\epsilon^{\nu_1\nu_2}_{\nu\rho}F_{\nu_1\nu_2})+\partial_{\nu}(\epsilon^{\nu_1\nu_2}_{\rho\mu}F_{\nu_1\nu_2})+\partial_{\rho}(\epsilon^{\nu_1\nu_2}_{\mu\nu}F_{\nu_1\nu_2})]=\epsilon^{\sigma}_{\mu\nu\rho}J_{\rho}=*J$
($\partial_{\mu}F^{\nu\mu}=J^{nu}$)
(b)$F=-(**F)$
$*F=\left( \begin{array}{cccc} 0&0 &0 &0 \\0 &0 &0 &0 \\0 &0 &0 &qsin\theta\\ 0&0 &-qsin\theta&0 \end{array}\right)$
$F=(- **F)=\left( \begin{array}{cccc} 0&-q/r^2 &0 &0 \\q/r^2 &0 &0 &0 \\0 &0 &0 &0\\ 0&0 &0&0 \end{array}\right)$
(c)$E_r=q/r^2$,$E_{\theta}=E_{\phi}=0$,
$B_{\mu}=0, \mu=r,\theta,\phi$
(d)$0$

# [Solutions]Ch 1 Special Relativity and Flat Spacetime

1-6

https://petraaxolotl.wordpress.com/chapter-1-special-relativity-and-flat-spacetime/

9. For a system of discrete point particles the energy-momentum tensor take the form

$T^{\mu \nu}=\Sigma_a \frac{p_{\mu}^{(a)} p_{\nu}^{(a)} }{p^{(a)}}\delta^{(3)}(x-x^{(a)})$,

where the index a labels the different particles. Show that, for a dense collection of particles with isotopically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor

$T^{\mu\nu}=(\rho +p)U^{\mu}U^{\nu} + p\eta^{\mu\nu}$.

Doing average over 4-volume,

${\bf T}^{\mu\nu}=\dfrac{1}{\Delta V_4}\int_{\Delta V_4}T^{\mu\nu}d V=\dfrac{1}{\sqrt{-g} d^3x^i dx^0}\int_{\Delta V_4}T^{\mu\nu} \sqrt{-g} d^3x^i dx^0$.

Then a) in ${\bf T}^{\mu\nu}$ only delta-functions depend on x, b) metric determinant g is a macroscopic quantity, is constant over selected volume and can also be taken away from the integral.

${\bf T}^{\mu\nu}=\dfrac{1}{d^3x^i dx^0}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}\int_{\Delta V_4} \delta^{(3)}({\bf x}-{\bf x}^{(a)}) d^3 x^i dx^0 = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}$

In the last expression the sum is taken over the particles which have worldlines passing through $\Delta V_4$.

Consider symmetry,

${\bf T}^{0 0} = \dfrac{1}{d^3x^i}\sum_a p_a^0 \equiv \rho$

${\bf T}^{i 0} = \dfrac{1}{d^3x^i}\sum_a p_a^i$. Because of isotropy, ${\bf T}^{i 0}= 0$.

${\bf T}^{i j} = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^i p_a^j}{p_a^0}$. The sum should produce a symmetric macroscopic 3-tensor of second order. But all 3-tensors are defined by 3 eigenvectors. Because no preferred direction exits and no preferred directions correspond to the case when the matrix has all eigenvalues equal, that is when the matrix is proportional to kronecker delta. The coefficient of proportionality is the pressure: ${\bf T}^{i j} \equiv P \delta^{ij}$.

Replace ：

$\delta ^{ij}\to \eta^{ij}+U^i U^j$

$T^{00}\to \rho U^0U^0$,

then we get the result.

10. Using the tensor transformation law applied to $F_{\mu\nu}$, show how the electric and magnetic field 3-vectors E and B transform under

(a) a rotation about the y-axis;

(b) a boost along the z-axis.

(a) $\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} 1&0&0&0 \\ 0&cos\theta&0&sin\theta\\ 0&0&1&0\\ 0&-sin\theta&0&cos\theta \end{array}\right)$

(b)$\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} cosh\theta&0&0&-sinh\theta \\ 0&1&0&0\\ 0&0&1&0\\ -sinh\theta&0&0&cosh\theta \end{array}\right)$

$\overline{F}^{\mu\nu}=\Lambda F \Lambda^T=\left(\begin{array}{cccc} 0& - E1*cosh(x) - B2*sinh(x)& B1*sinh(x) - E2*cosh(x)& E3*sinh(x)^2 - E3*cosh(x)^2\\ E1*cosh(x) + B2*sinh(x)& 0& B3&- B2*cosh(x) - E1*sinh(x)\\ E2*cosh(x) - B1*sinh(x)& -B3& 0& B1*cosh(x) - E2*sinh(x)\\ E3*cosh(x)^2 - E3*sinh(x)^2& B2*cosh(x) + E1*sinh(x)&E2*sinh(x) - B1*cosh(x)& 0 \end{array}\right)$.

13. Consider adding to the Lagrangian for electromagnetism an additional term of the form $\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}$

(a) Express $\mathcal{L}'$ in terms of E and B.

(b) Show that including $\mathcal{L}'$  does not affect Maxwell’s equations. Can you think of a deep reason for this?

(a) $\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=\tilde{\epsilon}_{ijk}(F^i F^{jk}-F^{i0}F^{jk}+F^{ij}F^{0k}-F^{ij}F^{k0})=4F^{0i}F^{jk}\tilde{\epsilon}^{ijk}=-8\vec{E}\vec{B}$

(b) The expression in (a) can be expressed by 4-dimension divergence:

$\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=4\partial_i[\tilde{\epsilon}^{ijlm}A_k (\partial_l A_m)]$

Integrate $\mathcal{L}'$ over 4-dimension space, according to Stokes theorem, we can get the additional term in Lagrangian $L$. This term vanishes when varying action $S$