New answers tagged

0 votes

Difference between real operators and Hermitian operators in quantum mechanics

In order to avoid unnecessary mathematical complications, let us consider a finite-dimensional complex Hilbert space $\mathcal H$, i.e. a complex vector space with dimension $n\lt \infty$ and a (...
Hyperon's user avatar
  • 4,592
0 votes

How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

So I started with rewriting state $|x+\rangle$ in z-representation: $$ |\psi \rangle = \frac{1}{\sqrt{2}} |+\rangle|+\rangle + a|+\rangle \Bigl( \frac{1}{\sqrt{2}} \bigl( |+\rangle + |-\rangle \bigr) \...
hft's user avatar
  • 18.9k
0 votes

How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

The best way to quantify the entanglement for a 2-qubit state is something called the "concurrence"; it goes to zero for a separable state and 1 for a maximally entangled state (like a Bell ...
Ken Wharton's user avatar
  • 1,505
2 votes
Accepted

Difference between real operators and Hermitian operators in quantum mechanics

A real linear operator is one whose matrix elements are real. Operators like the momentum operator are real, for example, when we calculate its expectation value $\langle \psi|\hat p|\psi\rangle$ we ...
Albertus Magnus's user avatar
5 votes
Accepted

Hermiticity of a projection operator

Indeed, you missed that $\sigma_{00}^0=I$. \begin{equation} \hat{U}=I+\sigma_{00}(e^{i\omega}-1), \hspace{6 mm} \hat{U}^{\dagger}=I+\sigma_{00}(e^{-i\omega}-1), \end{equation} and it is now easy to ...
Ruben Campos Delgado's user avatar
1 vote

How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

A pure bipartite state $$\left|\psi\right> = \sum_{i = 1}^{d_A} \sum_{j = 1}^{d_B} c_{ij} \left|i\right>\!\left|j\right> \in H_A \otimes H_B$$ is separable if and only if the matrix $(c_{ij})$...
Vladimir Lysikov's user avatar
0 votes

Notation for spatially superposed particle using vacuum modes - Why the state of the particle and vacuum is entangled, not separable?

Both expressions are legitimate: you have identified the difference between first quantization and second quantization. The specific state you mention was subject to a lot of debate as to whether or ...
Quantum Mechanic's user avatar
3 votes
Accepted

Unitary Representation of $\text{SO}(3)$ in Position Representation

The eigenvector $|x\rangle$ of the position operator $\hat X$ associated to the eigenvalue $x$ being defined by $$\hat X|x\rangle=x|x\rangle$$ the defining property of $U_R$ gives $$U_R^{-1}\hat XU_R|...
Christophe's user avatar
  • 3,473
0 votes
Accepted

Expansion of a wavefunction of a two-particle system in one dimension in an arbitrary basis without operations associated with the tensor product

If we fix $x_1 = \bar{x}_1$, the function $\psi(\bar{x}_1, x_2)$ describes a valid state of the second particle, so can be decomposed with respect to the eigenbasis $\psi_{\omega_2}(x_2)$. The ...
Vladimir Lysikov's user avatar
1 vote

Hermiticity of a radial momentum operator $\hat{p}_r$ and the spectral theorem

Addressing a point which is not covered in the answer by @Roland F and the two pertinent comments to it: The OP is conflating spectral theory and the general theory of operators in Quantum Mechanics ...
DanielC's user avatar
  • 4,285
0 votes

How to measure the entanglement of three entangled qubits?

The three-tangle https://arxiv.org/abs/quant-ph/9907047 is such a measure. It is in a sense a generalization of the concurrence and has nice properties.
Karl Pilkington's user avatar
2 votes
Accepted

Can the hybridization of edge states in the 1D SSH model be observed numerically?

To answer point 1, you should look at the eigenvalues for those two modes: They should be very close, but not the same. This might very likely also answer 2 -- the hybridization should go down ...
Norbert Schuch's user avatar
0 votes

A question from S. Weinberg's book (Sec. 2.7)

I am not familiar with representing a group element by an operator on hilbert space. But, when we are representing elements of a group by linear transformations on a vector space, it must be that the ...
Aravind Madhavan's user avatar
1 vote
Accepted

A question from S. Weinberg's book (Sec. 2.7)

Clearly $\phi(T,1)$ is independent of $T$. Let $\phi(T,1)=\phi_0$. You can rescale your unitary operators by a constant phase ${\tilde U}(T) = e^{-i\phi_0} U(T)$. The new set of unitary operators ...
Prahar's user avatar
  • 25.6k
0 votes

Wigner-Eckart theorem: Completeness relation

Your notation suggests that $j$ and $m$ index the rotation group representation vectors, which are of course complete. And for $\alpha$ you need to have just one value to apply the theorem, so whether ...
Jos Bergervoet's user avatar
2 votes

Coherent creation operator: unitary or not?

Using the Baker-Campbell-Hausdorff formula, the unitary operator $$\hat{D}(z):= e^{za^\dagger-z^\ast a}, \qquad \quad [a,a^\dagger]=\mathbf{1},$$ can also be written as $$\hat{D}(z)=\underbrace{e^{-|...
Hyperon's user avatar
  • 4,592
3 votes

Coherent creation operator: unitary or not?

Unitarity is not defined by the action on a single specific state any more than hermiticity is defined by the action on a single state. $L_x$ is hermitian but $L_x\vert \ell,\ell\rangle=\frac12 L_-\...
ZeroTheHero's user avatar
  • 45.3k
1 vote
Accepted

How does an operator in the denominator act on a state?

The comment by Filippo tells you everything you need to know, but I will explain a little. When we write a quantity such as $\exp(i \hat{\phi})$ we are giving ourselves the 'right' to evaluate a ...
Andrew Steane's user avatar
1 vote
Accepted

Schmidt decomposition of density operators

The density matrix can be written as $$ \rho = \sum_{ijkl} \rho_{ijkl} |i\rangle_A|j\rangle_B \langle k|_A\langle l |_B\;, $$ where $|i\rangle_A,|j\rangle_B, \langle k|_A$ and $\langle l |_B$ are ...
By Symmetry's user avatar
  • 9,091
2 votes

Hermiticity of a radial momentum operator $\hat{p}_r$ and the spectral theorem

The operator $$\hat p_r \ (...) = -i \hbar \frac{1}{r}\partial_r \left[\ r \ (...)\right] $$ is obvoiusly hermitean/symmetric on $\mathit L^2(\mathbb R_+ , r^2 dr)$: $$\int_0^\infty \ r^2 \ dr \ \psi(...
Roland F's user avatar
2 votes
Accepted

How does one rigorously define two-point functions?

Firstly as quadratic forms, $a^\ast(x)$ and $a(x)$ make perfect sense on the form domain generated by finite linear combinations of n-particle wavefunctions lying in Schwartz space. In fact any normal ...
Prox's user avatar
  • 457
3 votes

How to actually implement a global phase gate?

Quantum states are only defined up to a global phase. Thus, it is not meaningful to implement a "global phase gate", in fact, quantum gates are, for the same reason, also only defined up to ...
Norbert Schuch's user avatar
0 votes

Rotational states in higher dimensions: multiple magnetic quantum numbers

Thanks to LPZ's answer, I just want to graphically show the restrictions on the magnetic quantum numbers $m_{12}$ and $m_{34}$ at least in 4 dimensions, to show how wrong my guesses were: $m_{12}$ ...
Nanite's user avatar
  • 3,390
5 votes
Accepted

Outer product as an operator in an infinite dimensional Hilbert space

For any $\psi\in H$, you can define an operator $P_\psi$ by $P_\psi \phi:=\langle \psi, \phi\rangle_H \psi$ for all $\phi\in H$. It is easily verified that $P_\psi$ is a linear bounded operator, which ...
Tobias Fünke's user avatar
1 vote
Accepted

In trouble with QFT field operators

You are quantizing inside a box of volume $V$. Therefore in this context, $$\int d^3x \neq \int_{\textrm{all space}}d^3x = \infty\,.$$ Instead, $$\int d^3x = \int_{\textrm{box of volume } V}d^3x = V\,....
march's user avatar
  • 7,409
2 votes

Interpretation of the vector space in notation bra ket

The bra, ket notation for quantum states is a way of expressing the state in the abstract. The quantum state ket $|\alpha\rangle$ is a vector in Hilbert space. Here the concept of vector is to be ...
Albertus Magnus's user avatar
2 votes
Accepted

Interpretation of the vector space in notation bra ket

You can think of $\psi(\vec x, t)$ as being a vector in an infinite dimensional complex vector space (a Hilbert space), where $|\psi(\vec x, t)|^2$ is the probability that the particle will be found ...
gandalf61's user avatar
  • 50.7k
0 votes

How do we know the creation and annihilation operators for angular momentum give rise to a complete basis?

The irreducible representations of the Lie algebra of $\rm SU(2)$ (defined by $[T_k,T_\ell]= i \varepsilon_{k \ell m}T_m$) can be classified by the "weights" $j=0,\, 1/2,\, 1,\, 3/2, \,2, \...
Hyperon's user avatar
  • 4,592
4 votes
Accepted

Poincaré invariance and uniqueness of vacuum state

The equation $$ U^\dagger(\Lambda) \phi(x) U(\Lambda) = S(\Lambda) \phi(\Lambda x)$$ is essentially the consistency requirement that applying a Lorentz transformation commutes with quantization. The l....
ACuriousMind's user avatar
  • 124k
1 vote

What kind of physical process would correspond to an operator that doesn’t result in an eigenvalue equation: $ \hat{A}ψ=a ψ$?

The wavefunction, $\psi(x)$, is the position representation of the more abstract vector (or "ket"), $| \psi \rangle$, which is an element of the Hilbert space (a complex vector space with ...
Ben H's user avatar
  • 920
1 vote

Wave functions and Schwartz spaces (QM)

It is possible for a function to go to zero at infinity without "getting all flat". For example, imagine a function which oscillates in such a way that its amplitude goes to zero at infinity,...
Albertus Magnus's user avatar
1 vote
Accepted

Rotational states in higher dimensions: multiple magnetic quantum numbers

I’ll start by a quick refresher on $\mathfrak{so}(d)$. More generally, the Lie bracket of two basis vectors is: $$ [L_{ij},L_{kl}]=i(\delta_{ik}L_{jl}+\delta_{jl}L_{ik}-\delta_{il}L_{jk}-\delta_{jk}L_{...
LPZ's user avatar
  • 10.9k
1 vote
Accepted

On the Wigner symmetry representation theorem

I suppose that by the brackets you mean the equivalence class of the state, that is, $[\psi]\in\mathbb{PH}$ is the ray to which $\psi\in\mathbb{H}$ belongs. Now, if your question is whether every ...
Albert's user avatar
  • 292
3 votes
Accepted

What is the difference between $(\mathcal{H}\setminus \{ 0\})/\mathbb{C}^*$ and $\mathcal{H}_1/U(1)$?

The text mentions two definitions of the space of physical states. One is $\mathcal{H} \setminus \{0\}$ modulo $\mathbb{C}^*$, and the other is the unit sphere in $\mathcal{H}$ modulo $U(1)$. They are ...
Javier's user avatar
  • 28k
1 vote
Accepted

Change of basis in bra-ket notation

State vectors in quantum mechanics are transformed via some unitary operator, for example, $|\psi\rangle_{new}=U|\psi\rangle_{old}$. Since the action of a linear operator on a state vector is enirely ...
Albertus Magnus's user avatar
4 votes

Change of basis in bra-ket notation

A generic quantum state, $|\psi\rangle$ can be expanded in terms of any complete set of eigenstates. If $|u\rangle$ are the eigenstates of an observable $\hat u$, then one can use the completeness of ...
Albertus Magnus's user avatar
1 vote

Translation of Jordan and Wigner's original paper?

I finally found the translation in a book. Duck I, Pauli W, Sudarshan E C G. Pauli and the spin-statistics theorem[M]. World Scientific, 1997. Have fun reading it!
1 vote
Accepted

Transition from "spin down" state to "spin up" state for an electron

It depends on what you mean by intrinsic. It's all linear algebra. The generic spin state for the electron is $$ |\psi\rangle= \cos\theta|\uparrow\rangle +e^{i\phi}\sin\theta |\downarrow\rangle $$ and ...
Cosmas Zachos's user avatar
0 votes

Algebraic definition of ground state

Perhaps a concrete example would help. Consider the $*$-algebra of complex $2\times 2$ matrices $\mathscr A:= \mathrm{Mat}_{2\times 2}(\mathbb C)$, with the star operation given by the conjugate ...
J. Murray's user avatar
  • 68.7k
-2 votes

Algebraic definition of ground state

A bound state has negative eigenvalue with potential term, but a strict positive expectation in the free algebra of positive observables, so its not the ground state of the free Heisenberg algebra of ...
Roland F's user avatar
1 vote

Scattering, Perturbation and asymptotic states in LSZ reduction formula

When I first faced the issue of in-out asymptotic states reading Bjorken-Drell I've had your same doubt 2) and I've worked out some details trying to disentangle it. Before writing them down, I should ...
dallla's user avatar
  • 59
1 vote

Relationship between position kets in different topologies

As you've written yourself, $S^1 = \mathbb R/L\mathbb Z$. This quotient extends to functions on the space. Start with a wave function $\psi: \mathbb R\to\mathbb C$, you can send it to a wave function ...
LPZ's user avatar
  • 10.9k

Top 50 recent answers are included