# Tag Info

## Hot answers tagged hilbert-space

24

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, ...

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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.

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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. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x),$$ $$\tag{2} ... 16 There is nothing to prove; this just involves making definitions as follows: Let an element |\psi\rangle of the Hilbert space \mathcal H of a particle moving in three dimensions be given. Let |\mathbf x\rangle denote a simultaneous eigenstate of the position operators X,Y,Z corresponding to eigenvalues x,y,z where \mathbf x = (x,y,z). Then for ... 16 The problem with this hamiltonian is that there is a difference between symmetric/Hermitian operators and self-adjoint operators. It looks like a nit-picky mathematician's poking holes into everything, but it is in fact important: In general, the domains of \hat{A} and \hat{A}^\dagger do not coincide. If \hat{A}=\hat{A}^\dagger on D(\hat{A}), ... 15 The expression you wrote for \sigma_z^1 + \sigma_z^2 is not quite right, but it's not surprising that you're unsure of how to proceed because the notation is somewhat obscuring the real math behind all of this. What's actually going on here is manipulations with tensor products of Hilbert spaces. The spin state of a single spin-\frac{1}{2} particle is ... 15 I) Well, one can identify a complex-valued observable with a normal operator$$\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 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 ... 13 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 ... 12 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 ... 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 First, let me state that the notion of a Hilbert space is not fundamental in quantum theory. One can realize the same quantum physical system using different Hilbert spaces. This is because quantum states (which are really the objects which physically matter) are only weakly connected to vectors on a Hilbert space. It is true that pure states correspond to ... 11 From a pure mathematical point of view the answer is negative. As you probably know, wavefunctions are all of the functions \psi from, say, R to C such that |\psi(x)|^2 has finite (Lebesgue) integral, namely \psi belongs to the Hilbert space L^2(R). One can simply construct functions that belong to L^2(R) and that oscillate with larger and ... 10 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 ... 10 Your doubt is not ridiculous, it is probably simply due to the confused way often mathematics is taught in physics. (I am a physicist too and, during my career, I had to bear ridiculous misconceptions, wasting lot of time in tackling non-existent pseudo-mathematical problems instead of focusing on genuine physical issues). There are sensible mathematical ... 10 On the actual Hilbert space of a consistent relativistic quantum mechanical system, the Lorentz transformations including boosts actually are unitary – which also means that the generators J_{0i} are as Hermitian as the generators of rotations J_{ij}. We say that the Hilbert space forms a unitary representation of the Lorentz group. What the OP must be ... 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 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 ... 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 ... 9 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 ... 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 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 ... 8 Although I think user Siva's argument is nice for intuition, I feel that the key mathetmatical point is being obscured; you just need to be careful about what you mean by the "size" of a vector space. The dimension theorem for vector spaces tells us that any two bases for a vector space must have the same cardinality. This allows us to define the dimension ... 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 $a^\dagger$.

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