# Tag Info

## Hot answers tagged hilbert-space

23

The differential operator itself (defined on some domain) encodes local information about the dynamics of the quantum system . Its self-adjoint extensions depend precisely on choices of boundary conditions of the states that the operator acts on, hence on global information about the kinematics of the physical system. This is even true fully abstractly, ...

16

You need nothing more than your understanding of $$\int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a)$$ Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this: $$\int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda)$$ So there you go.

16

There is nothing to prove; this just involves making definitions as follows: Let an element $|\psi\rangle$ of the Hilbert space $\mathcal H$ of a particle moving in three dimensions be given. Let $|\mathbf x\rangle$ denote a simultaneous eigenstate of the position operators $X,Y,Z$ corresponding to eigenvalues $x,y,z$ where $\mathbf x = (x,y,z)$. Then for ...

15

Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function. One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x),$$ $$\tag{2} ... 15 The problem with this hamiltonian is that there is a difference between symmetric/Hermitian operators and self-adjoint operators. It looks like a nit-picky mathematician's poking holes into everything, but it is in fact important: In general, the domains of \hat{A} and \hat{A}^\dagger do not coincide. If \hat{A}=\hat{A}^\dagger on D(\hat{A}), ... 15 I) Well, one can identify a complex-valued observable with a normal operator$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$A version^1 of the spectral theorem states that an operator A is orthonormally diagonalizable iff A is a normal operator. Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. ... 14 The expression you wrote for \sigma_z^1 + \sigma_z^2 is not quite right, but it's not surprising that you're unsure of how to proceed because the notation is somewhat obscuring the real math behind all of this. What's actually going on here is manipulations with tensor products of Hilbert spaces. The spin state of a single spin-\frac{1}{2} particle is ... 12 I don't know of any books which use this language exclusively, but the basic idea is pretty straightforward: All Hilbert spaces are isomorphic (if their dimensions match). This would present conceptual problems in quantum mechanics if we ever talked about the Hilbert space alone; how could we distinguish them? But it's OK because we are actually ... 12 The proof is probably not the right word since the expression \Psi(x,t) = \langle{x}|{\Psi(t)}\rangle is actually the definition of position space wave function. Basis in finite dimensional vector space Any vector |v\rangle from some finite dimensional vector space V(F) can be written as a linear combination of basis vectors |e_{i}\rangle from an ... 12 As a mathematical structure, the field of complex numbers does not admit an order relation which is an extension of the order we have in \mathbb{R}. This means that there is absolutely no way of saying if 5+3i is bigger or smaller than 5+6i for example. We just know it is not equal and we have to stop here. Therefore it is physically really hard ... 9 Aram's answer seems perfect, but since you are also asking about the case for higher dimensional systems, let me add that there is a simply way to get somewhat non-trivial upper and lower bounds on C(j_S,j_L). As a lower bound, you can simply synthesize an arbitrary gate which implements communication between the quantum systems (for an explicit algorithm ... 9 The wording used in your textbook was sloppy. A acts as A^* on a bra, as \langle u\rvert A\lvert v\rangle:=\langle u\lvert Av\rangle~ is the same as \langle u\rvert A\lvert v\rangle=\langle A^*u\lvert v\rangle~, by definition of the adjoint. The latter formula also shows that \langle A^*u\rvert=\langle u\rvert A. Everything becomes very simple ... 9 This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, \delta-functions are not valid \mathcal{L}^2(\mathbb{R}) functions, so we need to define a kosher function which gives the \delta-function as a ... 9 There is no particularly interesting new physical significance to such a state vector. As you already stated, it represents exactly the same physical state. The only difference is that, on taking the modulus squared, the new state gives an unnormalised probability distribution over possible measurement outcomes. You can easily extract the probability of ... 9 Several reasons: Orthogonal functions arise naturally in the study of Sturm-Liouville theory which includes many classical and quantum system mathematical models; More generally, it is the class of normal operators (and an important special case self adjoint operators) which the spectral theorem most readily works and is most complete for. The eigenvectors ... 8 I'll try my best to explain to you what I've learned about this. Well, as you probably know, the Dual Space of a Vector Space V is the space of all linear functions on the space V. This is one abstract mathematical concept, however it can give to us very nice ways of representing things on physics. Think about it for a while: the dual space is composed ... 8 The ground state of the harmonic oscillator |0\rangle obeys$$a|0\rangle = 0$$which means that the action of a can't be undone: once you act with it on a state, you set to zero the coefficient in front of |0\rangle in the decomposition into occupation eigenstates. Any candidate inverse operator a^{-1} acting on zero will give you zero again; you ... 8 OK, here you go: No. Consider the one-particle sector, with states |n\rangle having occupation number n. Wlog we can suppose that the states are normalized. Then$$a|n\rangle =\sqrt{n}|n−1\rangle,$$so$$\|a|n\rangle\| = \sqrt n.$$This means that a is not bounded. The same goes for a^\dagger. 8 Although Emilio's answer is insightful, I don't think it directly answers your question. I'll attempt to do that here. This answer proceeds in two parts: We'll show that the operator you try to write down is hermitian with appropriate domain, but that it is not self-adjoint and has no self-adjoint extensions. We'll show that your formal manipulations ... 8 Actually, what your professor isn't telling you - what we always gloss over in intro quantum mechanics for simplicity, but I might as well give it away now because everything will make so much more sense once you get this - is that kets aren't wavefunctions at all. Forget that you ever learned about wavefunctions for a minute. Kets are a form of notation ... 7 I believe there is such a representation, as follows: You need only consider operators which can be written as \Omega = \sum_k \alpha_k \gamma_k, where \gamma_k = \sum_\ell P_\ell \big( \sigma_{k_1} \otimes \cdots \otimes \sigma_{k_N} \big)P_\ell^\dagger with P_\ell being the operator corresponding to the \ell^{\text{th}} permutation of qubits. Note ... 7 Suppose I want to show$$\int \delta(x-a)\delta(x-b) dx = \delta(a-b) $$To do that , I need to show$$\int g(a)\int \delta(x-a)\delta(x-b) dx da = \int g(a)\delta(a-b) da$$for any function g(a).$$LHS = \int \int g(a) \delta(x-a)da \ \delta(x-b) dx =\int g(x)\delta(x-b)dx =g(b) $$But RHS clearly = g(b) too. The result follows putting ... 7 This is often confusing to people getting acquainted to QM and you need to stare at it for a while and convince yourself about how it works. Firstly \sigma^{x,y,z} are the Pauli spin matrices and \vec{\sigma}_1 \cdot \vec{\sigma}_2 \equiv \sigma_1^x \otimes \sigma_2^x + \sigma_1^y \otimes \sigma_2^y + \sigma_1^z \otimes \sigma_2^z The \sigma_z ... 7 The Hilbert space {\cal H} of the one-dimensional harmonic oscillator in the position representation is the set L^{2}(\mathbb{R}) of square integrable functions \psi:\mathbb{R}\to\mathbb{C} on the real line. The Dirac delta distribution \delta(x-x_{0}) is not a function. In particular, it is not square integrable, cf. this Phys.SE post. One may ... 7 Starting with:$$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$If t_i=0:$$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$Using the identity: \sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right| Since the exponential of an operator is (by ...

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