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Assuming the functions are well-behaved (continuous and differentiable), you can change the order of differentiation. $$\left(\frac{\partial T}{\partial V}\right)_S=\frac{\partial}{\partial V}\left(\frac{\partial E}{\partial S}\right) = \frac{\partial}{\partial S}\left(\frac{\partial E}{\partial V} \right) = -\left(\frac{\partial P}{\partial S}\right)_V$$ ...

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Generically, any square-integrable function is an admissible wave function, and the space of square-integrable complex functions indeed has uncountable dimension as a vector space over $\mathbb{C}$. And it is also true that the eigenstates of the Hamiltonian span the space of states, and that they are countably many. This is the content of the spectral ...

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The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$. The definition of the exponential for an ...

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You misunderstood the meaning of $H^{\bullet}_{S^1}(U_1) = \mathbb{C}[\Omega]$. This does not mean $H^i_{S^1}(U_1) = \mathbb{C}[\Omega]\forall i$, it means that the cohomology ring (with multiplication given by the cup product) is given by the polynomial algebra in one element that has degree 2. Translated back into the individual degrees $H^i$ this ...

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The phase potrait for this physical system describes the oscillation of the particle described above to be a Homoclinic orbit-that is, the particle oscillating between extreme ends of a double well. No, not even close. I assume that your misunderstanding originates from confusing phase space and geometrical space. Suppose, we describe the dynamical ...

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Let $H_1$ be the Hamiltonian of a Harmonic oscillator, and let $m=\hbar=\omega=1$, that is, $$H_1=\frac{1}{2}P_1^2+\frac{1}{2}X_1^2-\frac{1}{2}$$ Let $|n_1\rangle$ be the eigenvectors of $H_1$, i.e., $$H_1|n_1\rangle=n_1|n_1\rangle$$ If we define $H=H_1 H_2$ with$^1$ $[H_1,H_2]=0$ we get multiplicative eigenvalues:  ...

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Yes. That's how the tensor calculus is formulated. For two observers on different frame of reference, all they have to see is the same physical laws. On this basis we formulate our physical laws so that it will be valid for any observers irrespective of their reference frames. Let's take a tensor of rank one (a vector) as example. Suppose two observers on ...

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