Tag Info

Hot answers tagged

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 ...


5

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 & := ...


4

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

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 ...


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

I) Let us for clarity use a subscript "$S$" (and "$H$") to denote the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively. Moreover, let us assume that the two pictures coincide at the instant $t_0$ (which Ref. 1 assumes is $t_0=0$). II) Recall first of all the possibly confusing ...


3

Functions of an operator are (or, can be) defined by their power series: $$f(\hat{A}) = f_0 + f_1 \hat{A} + f_2 \hat{A}^2 + \cdots$$ It's easy to prove that $$\hat{H}\lvert a_n\rangle = a_n\lvert a_n\rangle \quad\implies\quad \hat{H}^k\lvert a_n\rangle = a_n^k\lvert a_n\rangle$$ and if you plug that into the power series definition of the function, it will ...


2

Your final expression is correct except that $V$ should be in front since it does not commute with $Θ_0$ in general. In potential scattering ${Θ_0}V$ is often a compact operator and for large positive imaginary part of $z=E+iℏε$ its norm becomes arbitrarily small so the series converges. In certain cases one can do things a little differently by ...


2

The simplest way is to exploit the symmetries here. So, instead of mindlessly going through the algebra, solving equations etc. you just use the fact that you are free to call any direction the x-direction, and set up a right handed coordinate system. In particular, this means that you are free to cyclically permute x, y, z in the equations. The problem of ...


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 ...


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} = ...


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 ...


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 ...


2

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 ...


1

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 $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$.


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 ...


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 ...


1

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 ...


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 $t$, ...


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

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: $$ ...


1

Question 1 I'm not sure one can make this idea "rigorous". This is essentially the process of "quantization of a classical theory". It's actually an experimental result: if we take classical expressions defining relationships between classically measurable quantities and replace them by observables and, if further, we replace Poisson brackets in the ...


1

we need to assume f can be Taylor expanded, first calculate the power of H, H can commute with itself, so that the eigenvalue is just the power of a



Only top voted, non community-wiki answers of a minimum length are eligible