# Tag Info

## Hot answers tagged hilbert-space

20

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be ...

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$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction ...

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The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field $\mathbb K$ (usually $\mathbb R$ or $\mathbb C$). A formal definition is given below (there are many equivalent approaches). First, if $V$ is a vector space, $V^*$ denotes its algebraic dual space, namely the vector space of the linear ...

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A wave function is a complex-valued function $f$ defined on ${\mathbb R}^1$ (if your electron is confined to a line) or on ${\mathbb R}^2$ (if your electron is confined to a plane) or ${\mathbb R}^3$ (if your electron ranges over three-space), and satisfying $$\int |f|^2=1$$ (where the integral is defined over the entire line or plane or 3-space). Every ...

8

A Hilbert space $\cal H$ is complete which means that every Cauchy sequence of vectors admits a limit in the space itself. Under this hypothesis there exist Hilbert bases also known as complete orthonormal systems of vectors in $\cal H$. A set of vectors $\{\psi_i\}_{i\in I}\subset \cal H$ is called an orthonormal system if $\langle \psi_i |\psi_j \rangle ... 7 Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector$A$and two bases$\beta=\{e_i\}_i$and$\beta'=\{e_i'\}_i$. This means you can write the components of$A$with respect to$\beta$as $$A_i=e_i·A=\sum_j\delta_{ij}e_j·A$$ and the components with respect to$\beta'$as $$... 7 This is a supplement to freude's correct answer: Hamiltonian is the infinitesimal generator of time translation defined as$$\mathrm{\hat{U}}(\mathrm dt)= 1- \frac{i}{\hbar} \mathrm{\hat{H}}(t)\mathrm dt\;.$$Time-Evolution Operator: Let the system be at |\phi\rangle\;. Now, let's wait for some time..... What is the probability amplitude of finding ... 6 This is a scalar value that is a projection of the state H|\psi \rangle on the state |\phi \rangle. The state H|\psi \rangle results from the action of the operator H on the state |\psi \rangle. If the state |\psi \rangle is an eigenstate of the operator H, the expression can be rewritten as E \langle\phi|\psi \rangle. If the state |\phi ... 5 The wave function is the solution to the Shroedinger equation, given your experimental situation. With a classical system and Newton's equation, you would obtain a trajectory, showing the path something would follow: the equations of motion. For a quantum mechanical system you get a wave function, and the rules it obeys over time. With this you can determine ... 4 You need to use a more precise notion of the cloning process, in order to understand the general statement and its repercussions. I will give you some outline here (mainly following the explanations of B. Schumacher and M. Westmoreland given in the reference), with an emphasis on the most important aspects of it, but to fully appreciate the importance of the ... 4 Why is the vector |S⟩ represented as Ψ for both bases when working out the components for the quantum mechanics case above? The first of the final two equations is simply an expression for the sifting property of the delta 'function'.$$f(x) = \int dx' f(x')\delta(x - x') $$Let's back up just a bit and write the state (ket) as a weighted 'sum' of ... 3 At the risk of revealing that I have completely misunderstood your question, a few thoughts... People sometimes talk about regular QM as being like "zero-dimensional QFT,"* and I think that correspondence is more or less what you are getting at here. I'm not sure to what extent this viewpoint has been or can be formalized. But here is my understanding of ... 3 What does it mean for a particular mode of particles in an infinite square well to have a Fock state |0\rangle with Gaussian wave function? As far as the many-particle setting is concerned, I am tempted to say that the short answer is "Nothing, really, because the |0\rangle state is not Gaussian." :D Longer answer: The formal vacuum state of ... 3 Actually, the outcome of the experimental apparatus is an interval (x_0-\delta, x_0+\delta). \delta>0 stays for the accuracy of the instrument which can be made smaller and smaller but cannot be removed. It is therefore assumed (Luders-von Neumann's axiom) that, if the state immediately before the measurement was determined by the wavefunction ... 3 The No-Cloning Theorem means that if you have an unknown state then it is not possible to make an identical copy. The original reference is to Wooters, A single quantum cannot be cloned. Of course, if you know the state, you can manufacture duplicates; or if you have many identical copies of the unknown state, provided by some quantum machine, you could ... 3 I could give you an answer by barking up a very different tree indeed! In phase space QM, and not, repeat not geometric quantization, you may work on flat phase spaces and forfeit spheres altogether, the way you actually do in Hilbert spaces. If you can stomach that, read on, otherwise not, lest you feel your expectations betrayed. Plain vanilla ... 3 This completeness relation of the basis means that you can reach all possible directions in the Hilbert space. It means that any |\psi \rangle can be made up from these basis vectors. If the sum of the projectors (the ket-bras) would not be the unit matrix, the vector |\psi\rangle could have components which cannot be represented within your basis. ... 3 The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows:$$ \left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle $$where the second factor is a Clebsch-Gordan coefficient and q=-1,0,1 indicates the type of transition. For ... 2 The delta function is not really a function, it is a distribution, In the strict sense both \delta (x) and e^{ikx} are not normalizable when n=m One way to prove your equations is to use fourier transforms Using Placherels theorem the fourier transform F([f(x)]k) for the function f(x) is given by ... 2 Everything you write is correct and there is no inconsistency. When you write \left<x\right|Hf(x)g(x′)\left|x′\right> you just, indeed, put the numbers on the right. But numbers commute so there is nothign wrong with it. Note that you have really no conclusion from it. You can't even transform back to operators on the form ... 2 Yes, their outer product is defined as you said. Further, the product of operators is given by$$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$Therefore,$$ [A \otimes B, F \otimes 1_B] = [A,F] \otimes [B,I_B] = [A,F]\otimes {\bf 0} = 0 $$2 The point is that the domain D(P) of P must be such that P is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that D(X)= L^2([0,L],dx) instead, where X is automatically self-adjoint. The vector \psi you use to prove Heisenberg inequality has to belong to D(PX) \cap D(XP) as you see by direct ... 2 Schmidt decomposition is in general a singular value decomposition (SVD) and it is applied on wave vectors and not on density matrices. While dealing with bi-partite wave vectors we use SVD because there is no restriction that the size of the two systems in question are the same. So the matrix of the wave vector coefficients can be rectangular and SVD can ... 2 A crucial hypothesis is missed in your construction. Each \phi_i must also satisfy \phi_i \not \perp \psi_i, otherwise \langle \psi_i |E_i \psi_i\rangle >0 is false. This point provides an answer to your last question as well. If \psi is an added further vector, linearly dependent on the vectors \psi_i, the construction you made cannot be ... 2 The Noether charge is the generator of the symmetry it belongs to, see e.g. this answer by Qmechanic. This relationship is also preserved in the quantum theory, see this question, in the sense that the quantum Noether charge Q must commute with the Hamiltonian H, at least in the absence of anomalies and if we do not run into "quantization issues" when ... 2 Supersymmetry generators are not always Hermitian. If you impose SUSY, and then compute de corresponding Noether's currents, and then you calculate the conserved charge, i.e., the fermionic Lorentz generators, you will get two non-Hermitian conserved currents. (By the way, the relation Q^\dagger=\bar Q is only valid in Lorentzian signature, in Euclidean ... 2 It's simply the most general kind of interaction Hamiltonian you can write down in this simplified two-level system. On the 2D Hilbert space spanned by \lvert R\rangle,\lvert L \rangle, the most general linear operator is written as$$ A = a_\text{RR}\lvert R\rangle\langle R\rvert + a_\text{RL}\lvert R\rangle\langle L \rvert + a_\text{LR}\lvert ... 2 That's a rather delicate topic. I suggest you to begin with Section 2.1 ''Quantum Mechanics'' of Weinberg's ''The Quantum Theory of Fields'', Volume I. All (normalized) wave vectors in the Hilbert space which only differ by phase represent the same physical state of the system. All such wave vectors (corresponding to the same state) can be united into a ... 2 Hint: The linear Hilbert space$H$of$n$qubits has$2^n$complex dimensions. The set of density operators on$H$, which by definition are Hermitian (actually semipositive) and have trace$=1$must then have real dimension$(2^n)^2-1$. 2 I'm having a hard time totally understanding the question here but I think the resolution might be to think of different modes as different spatial dimensions. Recall that a three dimensional particle in box has three quantum numbers ($n_x$,$n_y$, and$n_z\$). In terms of quantum information content (at a logical level), is there a difference between a 2D ...

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