Why on earth do you want NY to do anything about CO2 emissions?
Global warming is an overexaggerated LIE to rip off people's money. If I knew that taxes like the congestion charge really make a difference, I wouldnt mind but global warming is one BIG lie to tax people and every government knows that no amount of tax and charges will help them to make a difference. So you should be happy that NY does not waste any of taxpayers money on this rubbish.
PS even if there is global warming, all that the humanity could do is make a difference that is insignificant, thus not worth any money, time or effort.
Nothing, in fact, from classical probability theory, we know that the expectation of a random variable X is completely determined by its distribution DX by
\mathbb{E}(X) = \int_\mathbb{R} \lambda \, d \, \operatorname{D}_X(\lambda)
assuming, of course, that the random variable is integrable or that the random variable is non-negative. Similarly, let A be an observable of a quantum mechanical system. A is given by a densely defined self-adjoint operator on H. The spectral measure of A defined by
\operatorname{E}_A(U) = \int_U \lambda d \operatorname{E}(\lambda),
uniquely determines A and conversely, is uniquely determined by A. EA is a boolean homomorphism from the Borel subsets of R into the lattice Q of self-adjoint projections of H. In analogy with probability theory, given a state S, we introduce the distribution of A under S which is the probability measure defined on the Borel subsets of R by
\operatorname{D}_A(U) = \operatorname{Tr}(\operatorname{E}_A(U) S).
Similarly, the expected value of A is defined in terms of the probability distribution DA by
\mathbb{E}(A) = \int_\mathbb{R} \lambda \, d \, \operatorname{D}_A(\lambda).
Note that this expectation is relative to the mixed state S which is used in the definition of DA.
Remark. For technical reasons, one needs to consider separately the positive and negative parts of A defined by the Borel functional calculus for unbounded operators.
One can easily show:
\mathbb{E}(A) = \operatorname{Tr}(A S) = \operatorname{Tr}(S A).
Note that if S is a pure state corresponding to the vector ψ,
\mathbb{E}(A) = \langle \psi | A | \psi \rangle.
Von Neumann entropy
Main article: Von Neumann entropy
Of particular significance for describing randomness of a state is the von Neumann entropy of S formally defined by
\operatorname{H}(S) = -\operatorname{Tr}(S \log_2 S) .
Actually, the operator S log2 S is not necessarily trace-class. However, if S is a non-negative self-adjoint operator not of trace class we define Tr(S) = +∞. Also note that any density operator S can be diagonalized, that it can be represented in some orthonormal basis by a (possibly infinite) matrix of the form
\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 & \cdots \\ 0 & \lambda_2 & \cdots & 0 & \cdots\\ & & \cdots & \\ 0 & 0 & \cdots & \lambda_n & \cdots \\ & & \cdots & \cdots \end{bmatrix}
and we define
\operatorname{H}(S) = - \sum_i \lambda_i \log_2 \lambda_i.
The convention is that \; 0 \log_2 0 = 0, since an event with probability zero should not contribute to the entropy. This value is an extended real number (that is in [0, ∞]) and this is clearly a unitary invariant of S.
Remark. It is indeed possible that H(S) = +∞ for some density operator S. In fact T be the diagonal matrix
T = \begin{bmatrix} \frac{1}{2 (\log_2 2)^2 }& 0 & \cdots & 0 & \cdots \\ 0 & \frac{1}{3 (\log_2 3)^2 } & \cdots & 0 & \cdots\\ & & \cdots & \\ 0 & 0 & \cdots & \frac{1}{n (\log_2 n)^2 } & \cdots \\ & & \cdots & \cdots \end{bmatrix}
T is non-negative trace class and one can show T log2 T is not trace-class.
Theorem. Entropy is a unitary invariant.
In analogy with classical entropy (notice the similarity in the definitions), H(S) measures the amount of randomness in the state S. The more dispersed the eigenvalues are, the larger the system entropy. For a system in which the space H is finite-dimensional, entropy is maximized for the states S which in diagonal form have the representation
\begin{bmatrix} \frac{1}{n} & 0 & \cdots & 0 \\ 0 & \frac{1}{n} & \dots & 0 \\ & & \cdots & \\ 0 & 0 & \cdots & \frac{1}{n} \end{bmatrix}
For such an S, H(S) = log2 n. The state S is called the maximally mixed state.
Recall that a pure state is one of the form
S = | \psi \rangle \langle \psi |,
for ψ a vector of norm 1.
Theorem. H(S) = 0 if and only if S is a pure state.
For S is a pure state if and only if its diagonal form has exactly one non-zero entry which is a 1.
Entropy can be used as a measure of quantum entanglement.
Hold on to your pants if you are shocked about New York City why don't you take a look at China top 100 Cities. They have worst smog than LA ever did and we have to develop the air cleaning technology. China just refuse's to spend the money. The United States pollution number have constantly gotten better for the last 30 years.
The Lear Jet Leftists intend to form the core of the global Green Inner Party. They will continue to enjoy the finest foods and most advanced technology the world has to offer -- while telling YOU how to live.
If you work hard and show unwavering loyalty they may allow you to join the Green Outer Party. There you will be allowed occasional servings of meat -- and air conditioning on a rationed basis.
The rest of us will eat beans and swelter -- for the good of the planet.
Sadly, the US is doing very little about CO2 emissions, since a large subset of our population keeps loudly proclaiming that there is no problem. It makes it hard to make any real progress on the issue...
God only knows the question to that answer.
I was shocked to read that a city such as New York City doesn't even have a low emission zone. It begs the question how long do the US have to lag behind the rest of Europe?
