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8

A unitary operator $U$ can only take a pure state $|\psi \rangle$ to a pure state. Let $U | \psi \rangle = | \psi' \rangle$. Then acting the unitary $U$ on the pure state $| \psi \rangle$ yields $$ U \left( | \psi \rangle \langle \psi | \right) U^\dagger = | \psi' \rangle \langle \psi' |,$$ which is a pure state. Another way to see this is that a (...


7

Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$ t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of ...


6

It is actually possible to improve Mackey's approach completing a program started in the seventies by Jauch and Piron. I remind you that a lattice is a partially ordered set $(\cal L, \leq)$ such that for every pair $a,b \in \cal L$ there exist $$ a\vee b := \sup\{a,b\}\in \cal L \quad \quad \mbox{and}\quad \quad a\wedge b := \inf\{a,b\}\in \cal L$$ Let us ...


6

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$. Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for $p\not\in\{0,...


5

The answer to your question is NO. The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$. For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ can'...


5

OK, since my name was taken in vain I suppose I am obliged to clarify my comment further. My invitation was to contrast charge oscillations to strangeness oscillations in the $K^0-\bar{K}^0$ system, not to use the latter to argue for the former. The finally mutated question I am addressing is “Why are there no charge oscillations and superpositions of ...


5

No, an arbitrary operator does not represent a change of basis. And even those that can be used to perform changes of basis should not always be interpreted as such. A "change of basis" in a Hilbert space is usually meant to be a change from one orthonormal basis to another. The operators that map orthonormal systems to orthonormal systems are precisely the ...


4

Bra-ket notation is just a useful short hand for some well-defined objects in functional analysis (or linear algebra if you work in finite dimensions). To understand what is allowed and what isn't, you would better know what those concepts are, so let's quickly recap: A ket $|\psi\rangle$ is just a vector in some Hilbert space $\mathcal{H}$. A bra $\...


4

Let us write the eigenvalues as follows: $$\lambda=a+ib$$ Where $a$ and $b$ are real. By definition we must therefore have: $$\lambda^*=a-ib$$ Equating these gives us: $$a+ib=a-ib$$ $$2ib=0$$ $$b=0$$ and therefore: $$\lambda=a$$ Which is a real number.


4

Preliminaries: If you "limit" your description of quantum mechanics to $L_2$ Hilbert spaces, all your bases will be discrete, both bounded or unbounded. You can have Hilbert spaces of any cardinality, but the one in "standard" quantum mechanics is $L_2$, the space of square integrable functions, which has a countable cardinality, $\aleph_0$. In this case ...


4

(1) Yes, the point spectrum is countable in your hypotheses: otherwise the operator would have an uncountable set of pairwise orthogonal vectors since eigenvectors of a self-adjoint operator with different eigenvalues are orthogonal. This is impossible because, in every Hilbert space, every set of (normalized) orthogonal vectors can be completed into a ...


4

This follows from the cyclicity of the trace, i.e. the property that $$\mathrm{Tr}(AB)=\mathrm{Tr}(BA),$$ which extends to the cyclic permutation $\mathrm{Tr}(ABC\cdots XYZ)=\mathrm{Tr}(BC\cdots XYZA)$ for larger products. Thus, if you expand $f$ in its Taylor series, you get \begin{align} \mathrm{Tr}\left(f(G^\dagger G)\right) & = \mathrm{Tr}\left(\sum_{...


4

It is not a postulate that the Hilbert space for QFT is a Fock space. In fact, for interacting theories is often almost surely not a Fock space. The requirements for a Hilbert space to be the space of a QFT is that the Wightman axioms are satisfied. For free theories, a suitable Fock representation of the canonical commutation relations satisfies the ...


3

Well, you know that the eigenfunction of $\hat p$ is $\exp(ipx)$, so let's try to find what the expectation value of $x$ is, to begin with: $$\langle p | x | p\rangle = \int \exp(-ipx) x \exp(ipx) \, dx = \int x\, dx$$ and this integral doesn't exist. Then it shouldn't come as a surprise that trying to take the time derivative gives nonsense. The underlying ...


3

I think the particular statement of Gleason's theorem is really important in elucidating this perceived ambiguity. Wikipedia states it as: Theorem. Suppose H is a separable Hilbert space of complex dimension at least 3. Then for any quantum probability measure on the lattice Q of self-adjoint projection operators on H there exists a unique trace class ...


2

It seems that by "operator" you mean a time evolution operator $\exp\left(\frac{i}{\hbar}\,H\,t\right)$ where $H$ is a quantum system's Hamiltonian, and such an operator by definition always maps (acts on) a pure quantum state to another pure quantum state. Unitary evolution is what happens whenever quantum measurement doesn't. So your statemtent "I assume ...


2

The given ansatz includes two assumptions. (1) The approximate ground states is seeked on a subspace of the Hilbert space consisting of rotated versions of single constant vector. (This subspace is a $2$-sphere parametrized by a unit vector in $\mathbb{R}^3$ (2) The value of the spin projection in the direction of the unit vector is half of the number of ...


2

Notation-wise, you can use either of \begin{align} A\rho & = \sum_{ijk} p_k \Big(|i \rangle \langle j | \otimes |i \rangle \langle j |\Big) \Big(| k \rangle \langle k | \otimes |k\rangle \langle k |\Big) \\ & = \sum_{ijk} p_k |i \rangle \langle j |k \rangle \langle k | \otimes |i \rangle \langle j |k \rangle \langle k | . \end{align} As you well ...


2

$\hat U$ is an operator, and an operator is very different from a scalar. Just think about this: every operator can be expressed as a matrix in some basis and every state as a vector. So the difference between $$\exp \left(-\frac{i \hat H t}{\hbar} \right) \mid \psi \rangle$$ and $$\exp(-i \phi) \mid \psi \rangle$$ where $\phi$ is a real number, is the ...


2

$G^\dagger G$ and $G G^\dagger$ are unitary equivalent in the finite dimensional case : There exists the polar decomposition $G = R U$ where $R \geq 0$ and $U$ unitary. Then $G^\dagger G = U^\dagger R^2 U = U^\dagger (G G^\dagger) U$. But from this it follows that $G^\dagger G$ and $G G^\dagger$ have the same eigenvalues. In the infinite dimensional case ...


2

They are not orthogonal because planar waves do not belong to hilbert space. It is true that you can have hilbert spaces of any cardinality, but the one in quantum mechanics is $L_2$, the space of square integrable functions, which has a countable cardinality, $\aleph_0$. A consistent way to do the math would be to have atoms in a box of length $l$, where $l$...


2

Your expression for: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\vec \sigma_1 |+\rangle\otimes \vec \sigma_2 |+\rangle_2$$ Is wrong. It sould read: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\sigma_{1x}|+\rangle_1\otimes \sigma_{2x}|+\rangle_2+\sigma_{1y}|+\rangle_1\otimes \sigma_{2y}|+\rangle_2+$$ $$\...


2

First, let's answer the questions precisely as you worded them: The point spectrum is always discrete in the sense that it consists of at most countably many points. This is true by proving the following results: a) the space spanned by all eigenvectors is a closed subspace of the Hilbert space, hence we have an orthonormal system of eigenvectors, b) two ...


2

Let us label the state spaces clearly as $\mathcal{H}_1$ and $\mathcal{H}_2$ for the first and second particle respectively and denote the canonical isomorphism sending a state in $\mathcal{H}_1$ of the first particle to the exact same state of the second particle by $\phi : \mathcal{H}_1\to\mathcal{H}_2$. Let us further denote the canonical "flip ...


2

You did all right, but you forgot about delta function's property. $$ \delta(x)f(x) = \delta(x)f(0) \, . $$ That's correct for every smooth function (it is hard to tell what $\theta(x)\delta(x)$ is, but there is a way to define it too). Physicists usually explain it superficially. Therefore $$ \delta(r'-r) \exp \left[-\frac{\gamma}{2}(r'-r)^2 \right] = \...


1

To normalize this function you have $$ 1~=~\int_0^{2\pi}\Phi^*\Phi d\phi ~=~|C|^2\int_0^{2\pi} e^{-im\phi}e^{im\phi}d\phi~=~|C|^2\int_0^{2\pi}d\phi~=~2\pi|C|^2. $$ The result is then obvious.


1

is it possible, based on some principal physical argument, to derive it? No, not really. This has been one of the driving goals of the field of quantum foundations for multiple decades, but as yet we don't know of any substantially simpler, more intuitive, or even different principle from which to derive the linearity axiom of quantum mechanics. We do ...


1

Both of the forms you propose, $$\sigma^{(1)}_x \otimes \sigma^{(2)}_x \ldots \otimes \sigma^{(n)}_x \tag1$$ and $$\left(\sigma_x \otimes \mathbb{I}_{n-1}\right)\cdot \left(\mathbb{I}\otimes \sigma_x\otimes \mathbb{I}_{n-2}\right) \cdot \left(\mathbb{I}_2\otimes \sigma_x\otimes \mathbb{I}_{n-3}\right)\cdot \ldots \cdot \left(\mathbb{I}_{n-1}\otimes \sigma_x\...


1

The unbound free states are not plane waves. And, in the extended (rigged) Hilbert space that allows for un-normalizable states, they are orthogonal to all the bound states.


1

I think the proof can actually work, but it needs to be formulated a bit better and it needs a bit more of explaining. Take $|\psi_n \rangle$ as a non-degenerate Eigenstate of $\hat{A}$. Then for all other Eigenvectors $|\psi_m \rangle$: $$\langle \psi_m | [\hat{A},\hat{B}]| \psi_n \rangle = (a_m - a_n)\langle \psi_m | \hat{B} | \psi_n \rangle = 0$$ So: $\...



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