New answers tagged

1

Yes. We can do this for any number of qubits using N dimensional spherical coordinates. For two qubits if we can write a general density matrix as a linear combination of direct products of Pauli matrices and identity , $$ \rho = \sum_{ij=0}^3 a_{ij}~ \sigma_{i} \otimes \sigma_{j}$$. Here $\sigma_{0} = I$, and the rest are the usual Pauli matrices. For ...


1

A valid density operator is any Hermitian, trace 1, matrix (with complex entries) and all eigenvalues between 0 and 1. Any two qubit system may be represented therefore by a Hermitian, trace 1 4x4 matrix. Your qubit representation could be rewritten, more suggestively as: \begin{align} \rho &= \frac{1}{2}\left(\operatorname{I} + a_1 \sigma_x + a_2 \...


-1

A one qubit state may be written in general as \begin{equation} |\psi_1\rangle=\alpha|0\rangle+\beta|1\rangle \end{equation} where $\alpha,\beta \in \mathbb C$ and there is the further restriction that $\langle \psi_1|\psi_1\rangle=|\alpha|^2+|\beta|^2=1$. However, only the relative phase between $|0\rangle$ and $|1\rangle$ is physically meaningful (quantum ...


1

The word "coherent" is used in Physics in a rather sloppy way. Your first state is a linear combination of harmonic oscillator eigenvectors that turns into a gaussian in momentum/position representations. In a more general background, a coherent state is just a state where coherences (off-diagonal terms in the density matrix) are non-zero, which means the ...


0

You can model the wave function as collapsing into either $|slit1\rangle$ or $slit2\rangle$ when the electron passes through the slits, or when the electron hits the detector, or when a human experimenter comes along and examines the results. The corresponding theories make all the same predictions (this is a theorem of von Neumann), so it's entirely up to ...


1

Let's imagine the double slit experiment you proposed as something simpler, yet equivalent. In the electron beam experiment you have a free particle's wavefunction that suddenly faces a decision: go left or right, and then collapses at a screen, giving you a result. I propose finding an analogous example with spin: suppose we have a hydrogen atom trapped ...


0

Maybe it is not relevant to you now, but it will be to someone. $[x,p]$ is Heisenberg algebra and you can easily see that this algebra is solvable. You have Lie theorem that says that every finite-dimensional irreducible representation of solvable algebra has to be one dimensional. So if you have one dimensional representation everything would commute and ...


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 ...


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.


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 ...


-1

It is not required to put the state in density matrix form prior to performing the unitary operation $U$. However, you want to learn properties about the reduced density matrix after tracing out some of the system post-unitary, so you need a density matrix eventually. So you could do the unitary, and then construct the density matrix: $$ \newcommand{\ket}[1]{...


0

There is no derivation of the probabilistic nature of the wavefunction: it is an interpretation (postulate), the only one that makes the theory consistent. From Sakurai, Modern Quantum Mechanics: Schrödinger published his famous wave equation in February 1926 in the famous paper Quantisierung als Eigenwertproblem (Quantization as an Eigenvalue Problem),...


0

In the last integral $ x^2 -y^2$ will integrate to zero. You can show that by splitting the integral into two pieces and changing variables in one of the integrals as $ x\leftright y $


-1

From the list of Spherical Harmonics here, I can rewrite your wavefunctions as : $$ \psi_1(\vec r) \propto Y_1^{-1}(\theta,\phi) r g(r) \\ \psi_2(\vec r) \propto Y_1^{0}(\theta,\phi) r g(r) \\ \psi_3(\vec r) \propto Y_1^{1}(\theta,\phi) r g(r) \\ $$ Use the fact that $Y_{lm}$'s are a orthogonal set of functions. $$ \int Y_l^m \ Y_{l'}^{m'} \text d \Omega = \...


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 ...


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 ...


0

It's an axiom which leads to a formalism that has great agreement with experiment. Asking why an axiom is what it is isn't really useful.


0

The so-called rigged spaces are made with a triple $(S,\mathscr{H},S')$; where $\mathscr{H}$ is the usual Hilbert space, $S$ is a dense vector subspace of $\mathscr{H}$, and $S'$ the dual of $S$. Usually when $\mathscr{H}=L^2(\mathbb{R}^d$, then $S$ is taken to be the rapidly decrasing functions, and $S'$ the tempered distributions. If this is the case, ...


0

They kets represent a configuration of the field (and the terminology is usually upgraded from "Hilbert space" to "Fock space"). $\lvert 0 1 0 \cdots 0\rangle$ is that there is 1 excitation (a particle) in some place specified by the second place in that vector. To explain, let's take a simple example: A particle in a box. Normally, in undergrad quantum you ...


0

I mean, suppose Ψ is a Quantum Field, in that case Ψ(r) is one observable. Ψ(r) is the operator field, not an observable. The observable is what appears after the operation on the ground state. Take the electron field. It is described mathematically over all space time, but obviously all of space time is not filled with electrons! If that's true, Ψ(r)...


0

Hilbert spaces are needed because there exists physical systems that possess infinitely many eigenstates.


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 ...


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_{...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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\...


0

It's both - those two "long form" expressions you wrote are equivalent. Matrix multiplication distributes over tensor products, so when you multiply them together the $i$th factor in the tensor product just becomes $\sigma^{(i)}_x$ times a bunch of identity matrices.


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 ...


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.


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: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\sigma_{1x}|+\rangle_1\otimes \sigma_{2x}|+\rangle_2+\sigma_{1y}|+\rangle_1\otimes \sigma_{2y}|+\rangle_2+$$ $$\...


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 ...


0

Update: Comments by Robin pointed out my confusion. Consider the following: $$ [x, p^2] = x p p - p p x = x p p - p x p + p x p - p p x = [x,p] p + p [x,p] = 2 i h p $$ If you plug this in the initial expression, you will get exactly what you would expect from this observable: $p/m$. But formally correct manipulations that you performed provide a ...


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.


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 ...


0

Were you bothered by having the differential $dx$ and the ket $|x\rangle$ in the 1D example? If not, what makes this different? I'll point out that the position operator is not $\hat{r}_n$; the position operator is a vector of operators, $\hat{\vec{r}}=(\hat{r}_1,...,\hat{r}_n)$. Maybe you realize this and your notation is just different than what I'm used ...


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 $\...


0

Others have pointed out that time is not an operator in the Schrodinger equation (I'll link to posts when I find them), but this isn't the end of the story. For example, how should one denote the monentum operator acting on a bra? $\partial_x \psi(x,t)$ could be denoted as $\langle \psi | \hat p^\dagger$ (preferred) but I've also seen variants like $\...


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: $\...


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 ...


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 (...


0

I consider a related question but where I consider a pair of Schrodinger equations for unperturbed and perturbed states. The occurrence of time is an inconvenience, so instead work with the Hamiltonians and Green' function. This is much the same, in that the S-matrix is concerned with the asymptotic scattering states and their differences in eigenspectrum. ...


0

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ The original post leaves out the strength of the Hamiltonian, which is confusing, so I'm going to put it back in. We write $H/\hbar = \Omega \sigma_x$. Following the original post we have \begin{align} \exp(-i H t / \hbar) = \exp \left( -i \Omega t \sigma_x \right) &= \left( \begin{array}{cc} \...


1

Essentially, separation of variables in the time-independent Schroedinger equation amounts to diagonalizing the Hamiltonian. One can see this easily by considering the case where the Hilbert space is finite-dimensional, and the Hamiltonian is a Hermitian matrix. In case of a partially continuous spectrum one gets the same, except that the sum must be ...



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