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5

The sort of trick involved in removing the $|P\rangle$ on both sides to get the conjugate imaginary equation $$\langle P|\xi|P\rangle = \langle P|a|P\rangle \tag1 $$ is quite common but it is indeed nontrivial to grasp the first time. In essence, you leverage the fact that in an equation of the form $$ ⟨\psi|\hat A|\phi⟩=⟨\psi|\hat B|\phi⟩\tag2 ...


-1

The Hilbert space of an electron is the electron itself ! Even in classical mechanics we can replace the subjective notion of a system by your own phase space, this is not different in quantum mechanics. The phase space in classical mechanics tells us how to build physical quantities: as a function defined in the phase space to the real line. In quantum ...


1

Think of $\lvert \psi \rangle$ as being written $a \lvert n \rangle + b \lvert n{+}1 \rangle$ -- it is just a linear combination of $\lvert n \rangle$ and $\lvert n{+}1 \rangle$ with (possibly complex) coefficients $a$ and $b$. Converting from a ket to a bra (i.e., finding the dual) distributes over addition and scalar multiplication, and it ...


3

Expanding my comments: Quantum mechanics is formulated in separable Hilbert spaces, i.e. Hilbert spaces with a countable orthonormal basis. The usual space describing a three dimensional particle is the space of square integrable functions $L^2(\mathbb{R}^3)$. This is separable, as well. On separable Hilbert spaces, the spectrum of a self-adjoint operator ...


0

The Hamiltonian as given in http://en.wikipedia.org/wiki/Landau_quantization is\ (note that $x$ and $y$ are interchanged) is \begin{equation*} H=\frac{p_{x}^{2}}{2m}+\frac{1}{2}m\omega _{c}^{2}(x-\frac{\hbar p_{y}}{% m\omega _{c}})^{2}, \end{equation*} acting in $\mathcal{H}=L^{2}(\mathbb{R}^{2},dxdy)$.Then, since \begin{equation*} x-\frac{\hbar ...


0

Doesnt the system collapse into ζ1 given we know this is the state at time t=0? True, but after the measurement at $t_0$ the system follows the unitary dynamics given by $H$. As you can easily check, $\zeta_1$ is not an eigenstate of the Hamiltonian, iff $E_1\neq E_2$. Hence the system will not remain in the state $\zeta_1$ during time-evolution. If ...


1

It is a distinction corresponding to different types of spectral measures. The absolutely continuous spectrum corresponds to absolutely continuous measures, singular spectrum to continuous singular measures (both with respect to Lebesgue measure). Refer e.g. to Reed-Simon Chapter VII for a more detailed description.


2

Hints: Assume that $H$ is a complex Hilbert space. Assume that $A:H\to H$ is a normal operator$^1$. Then a version of the Spectral Theorem says that $A$ is orthonormally diagonalizable. Let $(\lambda_i)_{i\in I}$ denote the set of different eigenvalues of $A$ with corresponding multiplicities $(m_i)_{i\in I}$. Let $P_i$ be the orthogonal projection ...


2

This is because there are just two possible values to the spin in any direction, $-\frac{\hbar}{2}$ and $\frac{\hbar}{2}$, the just differ in a sign, so when you square it you get a single value $\frac{\hbar^2}{4}$. Think about this, the only possible value when you measure the square of $S_z$ is $\frac{\hbar^2}{4}$ for any state, so $$ ...


8

OP asks: Is there any physical meaning to this? Yes, the Pauli matrix $\sigma_j$ represents (up to a proportionality factor) the spin in the $j$th direction of a spin $\frac{1}{2}$ system. Such system has only two spin states: $\uparrow$ and $\downarrow$, with opposite eigenvalues. The square $\sigma_j^2$ can no longer see the sign, so it only has one ...


1

It is better seen in the Heisenberg representation. Physical quantities, Observables, are represented by hermitian linear operators. Equation of movement is then (for a non-relativistic massive particle) : $$ m \dfrac{d^2\hat X(t)}{dt^2} = - \dfrac{\partial V(\hat X)}{\partial \hat X}(t) \tag{1}$$ with the quantization conditions : $[\hat X(t),m ...


4

It should be understood that physics - at least in it's current form - does not provide answers to "Why these laws?" questions. It can only describe an emergent law from a deeper and more fundamental one. Quantum theory is so far the most fundamental framework we have, so there is no more fundamental "reason" to describe it's structure aside from finding ...


0

$\langle \phi_{k'} \rvert$ is not an eigenfunction of $G_k$, hence the second line of your derivation is not correct. The eigenfunctions are $\langle \psi_{k} \rvert$, and that is exactly what we usually solve Lippmann-Schwinger equation for.


2

First of all, it is not always true. Most of the time, people work with positive definite Hilbert spaces. Second, the physical subspace of the Hilbert space has to be positive definite because the (squared) norms appear as probabilities and those can't be negative. Third, one often works with an extended Hilbert space that also contains unphysical states ...


2

i think you should read about abstract vector spaces with inner products. An Hermitian linear transformation is one that satisfy $$ \langle u,H \, v\rangle = \langle H\,u,v\rangle $$ For arbitrary vectors $u,v$. It happens that when you express this linear transformation in a orthonormal basis (with respect to the given inner product) satisfy $$ H^\dagger = ...


1

The relation $$\langle a|b\rangle\propto\delta(a-b)$$ is nothing unusual, it is simply an orthogonality condition. If the proportionality was an equality, and in addition we had completeness, the set of states would form an orthonormal basis. The reason why the delta function shows up is that you assume your operator to have a continuous spectrum of ...


1

The steps that you wrote down till eq. 1 is in fact a simple proof of the following theorem (which can be looked up in elementary text books of quantum mechanics): Eigen functions (of a Hermitian operator or more generally a symmetric operator on a separable Hilbert space) belonging to distinct eigenvalues are orthogonal. This is always true for separable ...


2

While i was typing, two good answers were posted. Since I don't want to delete everything, I'll leave this here nontheless. Without appealing to Lie theory, one might argue by physical reasoning. The unitary operators your book has in mind depend on a continous parameter $\alpha$. They describe continous transformations of the quantum mechanical state ...


3

Well, quantum mechanics is famous for not being intuitive for earthlings like us, but the following couple of facts might help: Observables in quantum mechanics are Hermitian/selfadjoint operators. The spectrum ${\rm Spec}(\hat{A}) \subseteq \mathbb{R}$ of a Hermitian/self-adjoint operator $\hat{A}$ belongs to the real axis $\mathbb{R}\subseteq ...


11

There's no escaping the Lie theory if you want to understand what is going on mathematically. I'll try to provide some intuitive pictures for what is going on in footnotes, I'm not sure if it will be what you are looking for, though: On any (finite-dimensional, for simplicity) vector space, the group of unitary operators is the Lie group $\mathrm{U}(N)$, ...


0

There is a nice and introductory presentation on the first book of Reed and Simon, Section VII.4, and a detailed bibliography at the end of the section.


1

See for example Topics in Koopman-von Neumann Theory by D. Mauro. This should be one of the most extensive overviews of KvN Theory, it also contains some examples of applying this theory to some well known problems such as Aharonov-Bohm Effect.


0

Koopman’s foundational paper is freely available on pubmed (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1076052/) and is an excellent resource by itself. Beyond that, the usual approach to teaching KvN is through analogy with the modern forulation quantum mechanics in complex separable Hilbert space, i.e the state space (see, for example, ...


1

As other answers mention, it was originally (in QED) about getting a neutral vacuum. It is useful to go back to Schwinger's old version of QED, before Dyson's approach became accepted. See Pauli: Selected topics in field quantization. Pauli presents both ways of looking at it: 1) define the electric current as sum of two terms (p.20 [6.4]), such that the ...


3

Note that: \begin{equation} \begin{aligned} \langle j | A \rangle & = \langle j |\left( \sum\limits_{i} a_i |i \rangle \right) \\& = \langle j| \left(a_1 |1 \rangle + a_2 |2 \rangle + a_3 |3 \rangle + \cdots + a_j |j \rangle + \cdots \right) \\& = a_1 \underbrace{\langle j | 1 \rangle}_{=\delta_{j1} = 0} + a_2 \underbrace{\langle j | 2 ...


2

If you think in terms of vectors and matrices with indices, $a_i$ is a vector and $\delta_{ij}$ is the unit matrix. The summation on the right hand side of your equation represents nothing but a multiplication of the vector $a_i$ with the latter. Since free indices are preserved, the remaining object carries the index $j$. You can write down an arbitrary ...


2

The choice of normal ordering prescription $:~:$ is typically adjusted to the choice of vacuum state $|\Omega\rangle$ so that the bra-ket-sandwich of normal-ordered operators $$\langle \Omega|:\hat{\cal O}_1\ldots \hat{\cal O}_n : |\Omega\rangle~=~0 $$ vanishes. The relation of normal ordering prescription to Wick theorem and other operator ordering ...


2

In classical physics, quantities are ordinary, commuting $c$-numbers. The order in which we write terms in expressions is of no consequence. In quantum field theory (QFT), on the other hand, quantities are described by operators that, in general, don't commute. Classical physics is a low-energy approximation of quantum physics - the road from quantum to ...


5

This is the same notation that you'll find in Weinberg's books. $$(\psi, \chi)$$ is the inner product of the two states $\psi$ and $\chi$, and corresponds to $$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to $$ \frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$ This new object is just the normalized ...


4

Your intuition that If, instead, my measurement is only partly accurate and says that the momentum of the particle is in a set $\Delta =(a_x,b_x)\times(a_y,b_y)\times(a_z,b_z)$, will the measurement collapse the wave function into $P\Psi$ (where $P$ is the spectral projector of the momentum operator on the set $\Delta$)? is exactly correct. ...



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