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2

Per the linked wikipedia article, vector spaces are linear mathematical structures satisfying the axioms of a vector space. Bras and kets are vectors of a specific type of complex vector space called a Hilbert space. The Hilbert space is additionally defined by the following inner product, which must be finite: $<b|k> = \int_{-\infty}^{\infty} ... 1 Unlike in QFT where you can derive spin from more basic principles, in ordinary non-relativistic QM spin is essentially defined into existence as a group of operators$S^i = (\hbar/2) \sigma^i$that satisfy the algebra $$[\sigma^i, \sigma^j] = i \epsilon^{i j}_{\,\,k} \sigma^k.$$ The dimensions in the Hilbert space on which the Pauli operators act are ... 2 In the context of physics, there are "natural" equivalence relations motivated by the following notion: mathematical objects that determine the same physics should be considered equivalent. These equivalence relations lead to partitions of the sets on which they are defined. Equipped with this idea, let's examine the two points you mention: Let$\mathcal ...

2

Restricting ourselves to just vector spaces without any extra structure, the theorem is true. One way to see this is to note that any member $f$ of the dual space is uniquely defined by the value it returns acting on the basis $\{\psi_n\}$, say $f(\psi_n) = z_n$ for complex numbers $z_n$. Then $V^*$ is isomorphic to $\mathbb{C}^\mathbb{N}$, the set of ...

9

There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The ...

3

Your "imaginary eigenvalues" don't work, because the eigenfunctions are no eigenfunctions. They do not lie in $L^2$, as you seem to be aware of. So, let's deal with the Laplacian itself: $-\Delta=-\frac{d^2}{dx^2}$. What I want to do is, I want to calculate the Fourier transform of this operator, because the Fourier transform diagonalizes $-\Delta$, as we ...

4

First off, if $k =: i\kappa$ is imaginary, the eigenvalue (“energy”) is $-\kappa^2$, i.e. real but negative: $$-\frac{d^2}{dx^2} e^{ikx} = -\frac{d^2}{dx^2} e^{-\kappa x} = -\kappa^2 e^{-\kappa x}.$$ Physically, that is an evanescent wave in one direction, but grows without bound in the other, so if your space is all of $x\in\mathbb R$, it is not a valid ...

0

A Hermitian operator is diagonalizable and has real eigenvalues. If you have an operator with real eigenvalues and such that all its eigenstates are orthonormal and, moreover, form a complete basis, then this operator is Hermitian. This is because you can pick its eigenbasis as an orthonormal basis and in that basis it will be represented by a manifestly ...

1

Step 0: Outline. We are going to define a candidate Hamiltonian. We know that we get the right answer if it has all the right behavior. While checking everything the Hamiltonian might possibly do seems daunting, things are simplified by two important facts: The Hilbert space is small -- only 4-dimensional. In fact, we're going to group three basis ...

2

The spectral theorem only holds for normal operators. Self-adjoint operators are normal, symmetric ones not necessarily so. In physicist-speak, we want the generalized eigenvectors to from a 'complete basis' of the Hilbert space. For example, the generalized eigenvectors of the momentum operator in position representation are plane waves, and even though ...

2

I'm pretty sure there exists an answer for this here already, but I can't find it (it's always about unboundedness). For the Hamiltonian, the answer is basically given by Stone's theorem on one-parameter unitary groups. There is a one-to-one correspondence between self-adjoint operators and strongly continuous one-parameter families of unitaries. Why is ...

1

Consider an operator $A$ on a Hilbert space $\cal H$, say $L^2(\mathbb R)$ in order to deal with QM of a particle on a real axis without spin. Let $D(A) \subset \cal H$ be the domain of $A$. The spectrum $\sigma(A)\subset \mathbb C$ of $A$ is defined as the union of the following three pairwise disjoint subsets $\sigma_p(A)$, $\sigma_c(A)$, $\sigma_c(A)$. ...

2

Your unnormalised state is a superposition of 2 states, $\left |\uparrow \downarrow\right>$ and $\left |\downarrow \uparrow\right>$ with probability amplitudes equal to unity. Their probability amplitudes are the coefficients in front of the 2 states. In this case they are 1. But what does normalisation mean anyway? Currently as the state is, and since ...

3

Any Hilbert space $\mathcal{H}$ with the notion of unitary time evolution also possesses the notion of Hamiltonian. If $\mathcal{U}(t) : \mathcal{H} \to \mathcal{H}$ is the time evolution operator for every $t \in \mathbb{R}$, then it forms a one-parameter Lie subgroup of the Lie group of unitary operators, which is generated by some distinct element $H$ ...

4

No, I don't see why that should be the case. The notion of a Hilbertspace underlying a quantum-mechanical system is quite independant of the postulate that time evolution is generated by a Hamiltonian. The notion of a vectorspace enters QM, because fundamentaly QM should be a linear theory and thus allow for arbitrary superpositions. The more wonderous ...

3

It seems, the state $|\psi\rangle$ is a superposition of $|\phi\rangle$ and several eigenstates of the Hamiltonian: $$\hat{H}|\chi_n\rangle = E_n|\chi_n\rangle$$ The sigma then just denotes the sum of the eigenstates. And since the Hamilton operator is linear, you can easily apply it to each element of the sum independently. $$\hat{H}|\psi_n\rangle = ... 0 Ok, there are a lot of points here. 1) First of all, an operator in Hilbert spaces is not defined only by its action (e.g. the operation of derivation for the momentum), but also by the so-called domain of definition, i.e. the subspace of vectors of the Hilbert space where it can act. Unbounded operators are not defined for every vector of the Hilbert ... 2 If you have different Hilbert spaces, you cannot say it is the same operator on them, since operators are defined on the Hilbert space. The momentum operator is a tricky one for many systems, and rigor requires the discussion of concepts like rigged Hilbert spaces. A nice introductory discussion of this is "Mathematical surprises and Dirac's formalism in ... 1 (I'm omitting the bra-kets in this answer, because there will be no bras appearing. Every greek letter is a ket.) It seems you mean to take two states of different systems and combine them into a state of the composite system. If \mathcal{H}_1 and \mathcal{H}_2 are the Hilbert spaces of states for the individual systems, then the tensor product ... 2 This is quite simple. Consider the operator H on the Hilbert space \mathscr{H}, in your simple example it has a spectral resolution:$$H=\sum_{n}E_n \lvert n\rangle\langle n \rvert\; .$$Each eigenvalue has multiplicity 1. Now the operators H_1 and H_2 on \mathscr{H}\otimes\mathscr{H} have the same spectrum of H, but each eigenvalue has ... 15 Hilbert spaces of infinite dimension are necessary, in the minimal case, to describe the non-relativistic quantum mechanics of a massive particle with at least a single real degree of freedom, and they are needed to allow the theory to describe, in general, states with arbitrarily high level of detail and at arbitrarily far-away positions. However, for any ... 22 The canonical commutation relations are not well-defined on finite-dimensional Hilbert spaces. The canonical prescription is$$ [x,p] = \mathrm{i}\hbar\mathbf{1}$$and, recalling that the trace of a commutator must vanish, but the trace of the identity is the dimension of the space if it is finite-dimensional, we conclude that we have a space for which the ... 6 Hilbert spaces, in general, can have bases of arbitrarily high cardinality. But the he specific one used on QM is, by construction, isomorphic to the space L2, the space of square-integrable functions, and this space as an infinite (but discrete) number of dimensions. The reason you want squere integrable functions is that you want the probabilities ... 2 Quantum dynamics is commonly known to be generated by self-adjoint operators. Therefore in order to properly define the dynamics of a system it is necessary to introduce a suitable self-adjoint Hamiltonian operator. In quantum field theories, this task is extremely difficult, because the formal operators that emerge quantizing a classical field theory ... 2 The first equation is not normalized, which is the result of the lowering operator. While the second equation is the normalized state, you can check it easily by using \langle 10|10 \rangle. Or you can use the normalization condition \langle \psi|\psi \rangle = 1 where |\psi \rangle = Z \hbar | \uparrow \downarrow + \downarrow \uparrow \rangle. 3 Given an inner product (\dot{},\dot{}), \langle x \rvert is the linear functional defined by (\lvert x\rangle, \dot{}). 2 If one calculates expectation value \langle\Psi| M^2|\Psi\rangle of the mass-square operator M^2 of the state |\Psi\rangle~=~ \alpha^j_{-1}| 0; p\rangle , then the level-matching condition (2.25) would be violated. 3 Define$$ N \equiv \sum_{i=0}^{D-2} \alpha_{-n}^i \alpha_n^i,~~~~{\tilde N} \equiv \sum_{i=0}^{D-2} {\tilde \alpha}_{-n}^i {\tilde \alpha}_n^i  The formula you have written tells us un particular that for any state, we must have $N = {\tilde N}$. This condition is known as the level-matching condition. The state $\alpha^i_{-1} |0;k\rangle$ has $N = 1$ but ...

4

As showed by Solovay here, in a non-separable Hilbert space $H$ there may be probability measures that cannot be written, for any $M$ closed subspace of $H$, as $\mu (M)=\mathrm{Tr}[\rho \mathbb{1}_M]$, for some positive self-adjoint trace class $\rho$ with trace 1 (density matrix). Here $\mathbb{1}_M$ denotes the orthogonal projection on $M$. [The proof of ...

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