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2

Restricting ourselves to just vector spaces without any extra structure, the theorem is true. One way to see this is to note that any member $f$ of the dual space is uniquely defined by the value it returns acting on the basis $\{\psi_n\}$, say $f(\psi_n) = z_n$ for complex numbers $z_n$. Then $V^*$ is isomorphic to $\mathbb{C}^\mathbb{N}$, the set of ...


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There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The ...


2

The eigenvalue is something physicists should be familiar with. For some matrix, $A$, multiplied by some vector $\mathbf x$, we get $$ A\mathbf x=\lambda\mathbf x \tag{1} $$ where $\lambda$ is the eigenvalue, a characteristic of $A$ on $\mathbf x$. An eigenfunction is related to Equation (1). Given an operator (a differential operator in the case of quantum ...


1

Consider the case that $p$ is space-like. Then, there is a frame in which $p = (0,\sqrt{p^2},0,0)$. Now, define $p_+ = (\sqrt{p^2},\sqrt{p^2/2},0,\sqrt{p^2/2})$ and $p_- = (-\sqrt{p^2},\sqrt{p^2/2},0,-\sqrt{p^2/2})$. Obviously, $p_\pm$ are light-like and $p = \frac{1}{\sqrt{2}}(p_+ + p_-)$. Proceed analogously for the time-like case.


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This answer is basically what I learned from the Lubos Motl's answer(hopefully I understand it right), with a little bit generalization. Question raised We want to solve a Hamiltonian $H=H_0+\lambda V$ perturbatively . What we know is the solution of $H_0$, whose eigenstates are $|n^0\rangle$ with eigenenergies $E_n^0$ has no degeneracy; it also has a ...


3

It seems, the state $|\psi\rangle$ is a superposition of $|\phi\rangle$ and several eigenstates of the Hamiltonian: $$ \hat{H}|\chi_n\rangle = E_n|\chi_n\rangle $$ The sigma then just denotes the sum of the eigenstates. And since the Hamilton operator is linear, you can easily apply it to each element of the sum independently. $$ \hat{H}|\psi_n\rangle = ...


1

If each leg $i=1\ldots 3$ is located at $\vec{r}_i = (x_i,y_i,0)$ and each force is $\vec{F}_i = (0,0,F_i)$ then the balance of forces and moments is $$ \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{W} $$ $$ \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 + \vec{r}_3 \times \vec{F}_3 = \vec{0} $$ where $\times$ is the vector cross product. From the two ...


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ACuriousMind's Answer pretty much summed up the reasons, which are essentially mathematical. If you want to grasp the "physical significance", then I suggest you should work through an example: think of two quantum systems, each with three base states: $\left.\left|1\right.\right>$, $\left.\left|2\right.\right>$ and $\left.\left|3\right.\right>$. ...


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The "Kronecker product", better known as the tensor product, is the natural notion of a product for spaces of states, when these are considered properly: A space of states is not a Hilbert space $\mathcal{H}$, but the projective Hilbert space $\mathbb{P}\mathcal{H}$ associated to it. This is the statement that quantum states are rays in a Hilbert space. ...



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