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New answers tagged hilbert-space

1

Hints: Whenever you have a summation that goes over a Kronecker delta, you can just cancel out one of the indices, so that e.g. $$\sum_{ij}a_i b_j \delta_{ij} = \sum_{i}a_i b_i.$$ For your specific case, you can do this term-wise, so that e.g. \begin{align} \sum_{ablm} \epsilon_{iab} \epsilon_{jlm} \delta_{bl} |{a}\rangle\langle{m}| = \sum_{alm} \...

1

The problem is a rather fundamental one in fact, and it cannot be overcome so easily, and it is not per se about having different Hilbert spaces, but rather inequivalent representations. All separable, infinite dimensional, Hilbert spaces are isomorphic (they are thus the same mathematical structure essentially). Nonetheless, they accommodate infinitely ...

0

Yes, the non-unitary Lindbladian time-evolution of an open quantum system can be formulated in the Heisenberg picture: https://en.wikipedia.org/wiki/Lindbladian#Heisenberg_picture.

0

$\newcommand{\bra} {\left< #1 \right|} \newcommand{\ket} {\left| #1 \right>} \newcommand{\bracket} {\left< #1 \vert #2 \right>}$ Your calculation of the Hamiltonian $$\hat{H} = \hbar^2 \ket{-z}\bra{-z}$$ is correct so far. One eigenvector is $\ket{-z}$, with eigenvalue $\hbar^2$. But you missed (as Dani already wrote in his comment): ...

-1

My quantum mechanics is a little rusty but I will attempt an answer. The problem is that the Hamiltonian is not Hermitean. A Hermitean operator in this case would have two eigenvectors. You need to add the hermitean conjugate of the S_{-}S_{+} term to the hamiltonian. From what I remember this is common practice in condensed matter physics.

4

Absolutely. The Hermite polynomials $$H_n(x) = (-1)^n e^{x^2} \partial_x^n e^{-x^2} = \left(2x - \partial_x\right)^n \cdot 1$$ are orthogonalized by $$\int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n! ~ \delta_{nm} ~,$$ whereas the (nonpolynomial) Hermite functions $$\psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{... 0 The book is not so clear. I believe what he means is the following: Given a density operator$$\sum_k p_k \left| \psi^{(k)} \right>\left< \psi^{(k)} \right|$$we can always find a basis in which it is diagonal. Let's say$$\left\{ \left| \phi_i \right> \right\}_{i=1}^{N},$$where N is the dimensionality of the problem. In this basis the ... 1 The energy does not depend on m, which is why all m values have the same energy. Basically the eigenvalues of your Hamiltonian depend only \ell, in the same way that all the Y_{\ell,m}(\theta,\varphi) for fixed \ell are eigenstates of \vec L\cdot\vec L with eigenvalue \ell(\ell+1). The connection between m and \ell is found in any ... 1 Given an arbitrary state \rho\in\mathcal B(\mathbb C^n) whose eigendecomposition reads$$\rho=\sum_{k=1}^{\mathrm{rank}(\rho)} p_k|\psi_k\rangle\!\langle \psi_k|,\quad p_k>0,\tag A$$The set of its purifications is the set of vector states |\Psi\rangle\in\mathbb C^{n}\otimes\mathbb C^{m} for m\ge \mathrm{rank}(\rho) whose SVD (called Schmidt in this ... 1 The direct product space is spanned by the 4 states \vert{s_1m_1}\rangle\vert{s_2,m_2}\rangle. This Hilbert space carries a representations of \frac{1}{2}\otimes\frac{1}{2}, and group theory tells you this representation is reducible as 1\oplus 0. Moreover, group theory explicitly helps in constructing states which reduce this representation since ... 4 The first thing to know is how to do matrix multiplication, in your case for example:$$\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} =\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$The second thing to know is the convention for ket vectors:$$ \begin{bmatrix} 1 \\ 0 \end{bmatrix} = |0\rangle \quad \begin{bmatrix} 0 \\ ...

2

Given a linear (but possibly non-unitary) time-evolution operator $\hat O(t)$, "manual normalization" would mean to consider the time evolution $$|\psi(t)\rangle = \frac{\sqrt{\langle \psi_0 | \psi_0 \rangle}}{\sqrt{\langle \psi_0 | \hat O(t)^\dagger \hat O(t) |\psi_0\rangle}} \, \hat O(t) |\psi_0\rangle .$$ It is clear that this map $|\psi_0\rangle \... -1 The total angular momentum, involved in the spin-orbit interaction: \begin{equation} {\bf J} = {\bf L} \oplus {\bf S} \end{equation} Cannot be expressed as a function of$X$and$P$(because of the spin part${\bf S}$), and it doesn't commute with them either (because${\bf L}$doesn't). 3 No. For any particular initial state$|\psi_0\rangle$, we can manually normalize the hypothetical non-unitary but linear time-evolution operator$\hat{O}(t)$in such a way that the manually normalized operator$\hat{O}_n(t) \equiv N_{\psi_0}(t)\, \hat{O}(t)$produces a time-evolved trajectory$|\psi(t)\rangle = \hat{O}_n(t) |\psi_0\rangle$with constant norm.... 2 (NB: Some of this isn't completely rigorous, particularly regarding infinite dimensional vector spaces, but hopefully it's not misleading.) Broadly speaking, the state of a quantum system can be described as being a superposition of the possible classical configurations. So a classical bit is a system with two possible states$\{A,B\}$and the quantum ... -1 Yes, there are plenty of such operators. For example, define a nonlinear operator $$\hat O \psi(x) \equiv \sin(x).$$ It does not commute with$x$or$p$, nor can it be expressed as a function of them. 6 In classical mechanics, any "observable" has to be a function of$x,p$because we define$x,p$to be the degrees of freedom of the system. If we experimentally find an observable that does not depend on$x,p$, it means there were extra degrees of freedom that we forgot to include. If so, we just enlarge our phase space, i.e., we include these extra degrees ... 1 An operator can be defined by the comutators it has with respect to x and p. Suppose we have an operator$y$that has a non-trivial commutation relation with$x$, i.e. $$[x,y]=f(x,p,y),\qquad [p,y]=g(x,p,y)$$ Note that ordering is determined by$f$and$g$. We can always solve this equation by the ansartz$y(x,p)$, up to some integrability condition. 2 Let's say we're working in a Hilbert space of square-integrable functions$f(x)$. Indeed it's natural to talk about the operators$X$and$P$which act in a simple way on$f(x)$. In principle, you can talk about any linear operator that acts on$f$. For instance, you could try to solve the Schrödinger equation $$(P^2 + X^2 + T)\psi(x) = E \psi(x)$$ where $$T\... 2 Hint: A_1^{\uparrow} = \begin{pmatrix}\cos\left(\frac{\theta_1}{2}\right)\end{pmatrix}, A_1^{\downarrow} = \begin{pmatrix}e^{i\phi_1}\sin\left(\frac{\theta_1}{2}\right)\end{pmatrix} 9 In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (... 1 I don't think this question has really anything to do with a specific physical system. A "wavefunction" is really just one way to describe a state, in which we characterise the state via the set of amplitudes in some continuous basis (such as the position basis). However, there isn't really anything special in this choice of basis, from a fundamental point ... 2 Take a Hydrogen atom (simpler than a Lithium one) made of just one electron and one proton. The complete orthonormal bases that span the Hilbert for the electron and for the proton are \{|\phi_{\mathrm{el}}^i\rangle} and \{|\phi_{\mathrm{nuc}}^j\rangle\} respectively. The state of the atom |\Psi\rangle can then be expressed as$$ |\Psi\rangle = \... 0 You are really asking a question about conventions of the bra-ket notation,$\psi(x)=\langle x| \psi\rangle. $$\hat{E}(\psi(x))=\alpha\psi(x)$$ is a synonym for $$\langle x| \hat{E} |\psi\rangle= \langle x| \alpha |\psi\rangle = \alpha \langle x| \psi\rangle ,$$ so that, likewise, \langle x| \hat{X} |\psi\rangle= \langle x| x |\psi\rangle = x \... 1 It has to be a constant, because, by definition, an eigenvalue belongs to the field of scalars \mathbb{K} underlying the vector space which is the domain and codomain of the linear operator in question, i.e. the one whose eigenvector and corresponding eigenvalue we are talking about. In quantum mechanics of the kind you are describing, the vector space is ... 1 It has to be constant. Otherwise essentially every function would be an eigenfunction of the x operator (except perhaps where x=0 or \psi=0). 1 Here's an almost solution, based on symmetry relations between the CGs. One has quite genrally \begin{align} \langle \ell_1m_1;\ell_2 m_2\vert LM\rangle = \sqrt{\frac{2L+1}{2\ell_1+1}}(-1)^{\ell_2+m_2} \langle L,-M ;\ell_2 m_2\vert \ell_1 -m_1\rangle \end{align} so set \ell_2=1 and m_2=0. Now, if m_1=m_2=M=0, then one obtains, with \ell_1=L: \... 0 In quantum mechanics, a and a^\dagger are ladder operators, in that they raise and lower the number of excitations presence in the harmonic oscillator: a^\dagger |0\rangle \propto |1\rangle \quad \Rightarrow \quad a^\dagger |n\rangle = \sqrt{n+1}|n+1\rangle, \\ a|1\rangle \propto |0\rangle \quad \Rightarrow \quad a|n\rangle = \sqrt{n}|n-1\rangle.$$... 1 You want to prove that given an arbitrary matrix A, we can write A as a linear combination of positive, unit-trace matrices. To do this, you start by writing A in terms of its Hermitian and skew-Hermitian components (see also this post about this decomposition):$$A=\underbrace{\frac{A+A^\dagger}{2}}_{\equiv A_1}+i\underbrace{\frac{A-A^\dagger}{2i}}_{ ... 0 This equation $$H|\psi(0)\rangle=E(a|1,1\rangle+b|1,2\rangle)$$ is not correct. The energy is not the same for both states, so you can't factor it out. 1 The equations you write are correct, now use the fact that the initial energy is 3, i.e.\langle \psi(0)|H|\psi(0)\rangle = 3$and that your state is normalized, i.e.$\langle \psi(0)|\psi(0)\rangle = 1$. This gives you a system of two linear equations for$|a|^2$and$|b|^2$from which you can determine a and b up to some phase factors. 0 In general you can't use fewer than three CNOTs for this task. There are special cases where you can use fewer, but they have measure zero. Consider the operation$U(a, b) = e^{ia(X\otimes X+Y\otimes Y)} \cdot e^{ibZ\otimes Z}$This operation preserves$Z_2$symmetry. In fact it preserves the total number of set bits, so it's even more restrictive. When ... 1 Well... you can take$-\vert G\rangle$and then all your coefficients would be negative. The point is: the signs of the coefficients in your eigenvector are not related to the signs of the eigenvalues or the signs of the entries in the matrix. For instance, consider the matrix$\$ M(\lambda)=\lambda \left(\begin{array}{cc} 1&0\\ 0&1\end{array}\right)+...

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