# Tag Info

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Right, in general you're not going to see a straightforward equivalence there. We can use Dirac notation with $\hat P_b = |b\rangle\langle b|$ to see that $\langle \hat A \hat B \rangle = \sum_{a,b} a~b~\langle \psi | a \rangle~\langle a | b \rangle~\langle b | \psi \rangle$ and even inserting an identity matrix for $b$ (call it $b'$) gives: $$... 2 Indeed using a rigged Hilbert Space is almost certainly what you should use and there are tutorials on it on arxiv (I'll add them when/if/I can find them again). However it isn't necessarily going to help you understand what an average physicist is doing on the one hand, and on the other hand there were already many things you were doing that not rigorous ... 10 What exactly does \Delta_{r_e} mean ? Your wave function isn't a field in space, it is a field on configuration space, i.e. it assigns complex numbers to a configuration. If your electron is at (x_e,y_e,z_e) and your proton is at (x_p,y_p,z_p) then the configuration is the point (x_e,y_e,z_e,x_p,y_p,z_p) in a 6d space. A point in that 6d space ... 1 \Delta_e is called the Laplacian operator and it is defined as$$\Delta_e \equiv \frac{\partial^2}{\partial e_x^2} + \frac{\partial^2}{\partial e_y^2}+ \frac{\partial^2}{\partial e_z^2}$$The Laplacian operator is related to the Del operator \nabla_e$$\nabla_e \equiv \frac{\partial}{\partial e_x}\hat e_x+ \frac{\partial}{\partial e_y}\hat e_y+ ...

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The index of the Laplacian tells you which of the coordinates it acts on, that is, if you write $r = (r_x,r_y,r_z)^T$ and $R = (R_x,R_y,R_z)^T$ as Cartesian coordinates, then \begin{align} \Delta_r & := \frac{\partial^2}{\partial {r_x}^2} + \frac{\partial^2}{\partial {r_y}^2} + \frac{\partial^2}{\partial {r_z}^2} \\ \Delta_R & := ...

2

Say the time evolution for Hamiltonian $H$ is given by $U(t) = \exp(-iHt)$ and the corresponding evolution on the support of $P$ is $$PU(t)P = P\exp(-iHt)P = \exp(-iH_\text{eff}t) \equiv U_\text{eff}(t)$$ assuming $H_\text{eff}$ exists. The desired identity follows from $$\lim_{\eta \rightarrow 0} \int_0^\infty dt \; U(t) e^{i (\epsilon + i \eta) t} = ... 1 It's a map from the quantum state space onto itself that represents how the state changes with time. Simplest example: a spin half spin state is a 1\times 2 column vector \psi holding the two probability amplitudes for the system to be in spin up and spin down. Now leave the system isolated wait for a time t. In QM, the operator is linear, so that ... 1 Comments to the question (v2): It seems that the question does not explain how a 'Hamiltonian' H differs from a self-adjoint operator A (presumably bounded from below). This would make OP's question a duplicate of the linked Phys.SE post. Perhaps a 'Hamiltonian' H is also supposed to generate 'time'-evolution for some distinguished parameter t, ... 0 I think the following explains it: $$|n_\alpha\rangle=\frac{1}{\sqrt{n_\alpha!}}(b^+_\alpha)^{n_\alpha}|0\rangle$$ $$b^+_\alpha|n_\alpha\rangle=\frac{1}{\sqrt{n_\alpha!}}b^+_\alpha(b^+_\alpha)^{n_\alpha}|0\rangle = \frac{1}{\sqrt{n_\alpha!}}(b^+_\alpha)^{n_\alpha+1}|0\rangle$$ and since ... 2 Let$$|n\rangle = \frac{(a^{\dagger})^n}{\sqrt{n!}} |0\rangle$$be the n'th normalized eigenstate and recall the commutation relation for bosonic creation/annihilation operators, [a, a^{\dagger}] = 1. Then we have$$a|n\rangle = \frac{ a \,(a^{\dagger})^n}{\sqrt{n!}} |0\rangle = \frac{1}{\sqrt{n!}}\big([a, (a^{\dagger})^n] - (a^{\dagger})^n ...

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Hints: $v(a)$ is identified with a left multiplication operator $L_{v(a)}f:= v(a)f$. $L_{v(a)}^{\dagger}=L_{\overline{v(a)}}$; $\left(\frac{\partial}{\partial a}\right)^{\dagger}=-\frac{\partial}{\partial a}$; and $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$.

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The matrix elements $D_{PQ}$ can either be zero if the orbital $P$ or $Q$ are not occupied -- then $a_Q\vert x\rangle=0$ or $\langle x \vert a_P^\dagger = 0$ -- and otherwise, they are overlaps of $N-1$ electron states $a_P\vert x\rangle$. If you define a matrix $X$ whose columns are $a_P\vert x\rangle$, then you can write $D=X^\dagger X$, and a matrix of ...

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I define the set of projectors $P_{a} = |a\rangle\langle a|$, what is the physical observable (like 'mopentum') that I need to measure to perform the operation? Even if you have some projections, even if they are each self adjoint, and even if they are mutually orthogonal and their range spans the whole space they still don't linearly combine into a ...

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It is so, because Dyson formula was designed for the interaction picture in Quantum Mechanics. Interaction picture may seem strange, especially in comparison with the 'natural' Heisenberg picture, but has been proven useful, especially in perturbative computations. The basic idea is to split the total Hamiltonian into two parts: $$H = H _0 + H _{int}$$ ...

1

Observables (I refer here to Hermitian operators) are confusingly named since they are not observable. What is observable, as you noted, is the eigenvalues that represent the outcomes. So what are observables for? The clearest way to understand the issue is to look at Heisenberg picture observables. These observables change over time but the state does not ...

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Comments to the questions (v2): The Hamiltonian $H(t)$ in QFT often depends on time $t$, e.g. if there are interactions in the interaction picture, or if there are external sources. In particular, Hamiltonians at different times need not commute. For an explicit example see e.g. my Phys.SE answer here. Well, if the Hamiltonian $H(t_1)=H(t_2)$ is the same ...

1

Observables are measurable quantities represented by certain mathematical structures dependent on the theory. In classical mechanics, measurable quantities are represented by functions on a phase space. In quantum mechanics, they are represented by operators on a Hilbert space. In classical mechanics, a measurement is equivalent to evaluating the function ...

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In non relativistic QM observables are represented by hermitian operators $\hat{O}$ acting on the states $\psi$, they represent measurements of some physical quantity ($\hat{O}$ represents say angular momentum, or spin). The result of the measurement will be an eigenvalue $o$ of the operator $\hat{O}$. The eigenvalues represent the actual measured value.

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Yes, they can. It's not great form, but they can. For a simple example, consider the hamiltonian for a 1D particle in a potential $V(x)$, $$\hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}).$$ This holds equally well if the potential is a delta function at $x_0$, $V(x)=\kappa \delta(x-x_0)$, in which case $$\hat H=\frac1{2m}\hat{ p}^2+\kappa\delta(\hat{x}-x_0).$$ ...

3

Let $\mathcal{H}$ be the space of states of our theory. Then, time evolution is given by a unitary operator $U(t_2,t_1) : \mathcal{H}\to\mathcal{H}$ that evolves "stuff" from time $t_1$ to time $t_2$. For time-independent Hamiltonians it is just $\mathrm{e}^{\mathrm{i}H(t_2 - t_1)}$. If we are in the Schrödinger picture, we say states "carry the time ...

3

Observe that the number operator $N := a^\dagger a$ is positive-semidefinite because $$\langle \psi \vert N \psi \rangle = \lvert a \lvert \psi \rangle \rvert^2 \ge 0$$ and hence has no negative eigenvalues. Since $a\lvert n \rangle = \sqrt{n}\lvert n - 1 \rangle$ for a normalized eigenstate of $N$ with eigenvalue $n$ (which one can derive from the ...

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If you want to define your $J_\pm$ in terms of $J_x$ and $J_y$ then you'll need to know those matrices. If you consider $J_x=\frac{\hbar}{\sqrt 2}\left(\begin{matrix} 0&1&0\\ 1&0&1\\ 0&1&0\end{matrix}\right),$ and $J_y=\frac{\hbar}{\sqrt 2}\left(\begin{matrix} 0&-i&0\\ i&0&-i\\ 0&i&0\end{matrix}\right)$ then ...

0

Once you admit the tunnelling the "topology" of the problem changes to "particle in a circle with an infinite barrier". In fact, to avoid arguments with energy when crossing the barrier, it can be described as "particle in a circle with an infinite barrier in one point". So physically it can be argued to be a different problem.

2

For an analytic function $f$, $f(A)$ can be defined as the Maclaurin series, as described in the OP. However, when $A$ is an element of a C*-algebra rather than just a Banach algebra, one can also consider continuous functions over the spectrum of $A$, which can still be defined as limits of certain polynomials of $A$ (as a consequence of the ...

0

There really is not a unique way to define the ladder operators, especially with the constants in front. If you change them from what one book has, you might match another book. Those differences will show up in different normalization constants for "next" states that the operators create. Eventually, you will use your operators to find the algebraic form ...

2

The term to look for is Coulomb wave. These wavefunctions are well explained in the corresponding Wikipedia article. Depending on your mathematical background, you should be ready for a bit of a formula jolt, as these wavefunctions rely very intimately on the confluent hypergeometric function. If you want the short of it, then I can tell you that the ...

2

Consider a molecule of oxygen in a balloon. You know that at nonzero temperature all those molecules are bouncing around in all directions. Of course, the mass of air doesn't have any net motion in any direction. Indeed, the average velocity of each molecule is zero: $$\langle v \rangle = 0 \, .$$ Of course, the average energy of any particular molecule is ...

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