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1

It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...


1

Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...


5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...


1

Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?


2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged


3

Mathematical reason is that the time evolution operator is unitary, which means that $U^\dagger = U^{-1}$. Therefore $\langle \psi(t) | \psi(t) \rangle = \langle \psi(0)| U^\dagger U | \psi(0) \rangle = \langle \psi(0) | \psi(0) \rangle$. We can see that it is unitary by considering the Schrodinger equation: $$ \newcommand{\ket}[1]{| #1 \rangle} ...


1

You chose the $\lvert \pm \rangle$ to be an eigenvector of $S_z$ with eigenvalue $\pm\frac{1}{2}$ - that's what the $m_s$ is: The eigenvalue of the state w.r.t. the $z$-spin. Since $S_x$ and $S_y$ do not commute with $S_z$, $\lvert \pm \rangle$ is not an eigenvector of them, hence the state cannot stay the same after they are applied to it. That the spin ...


3

Many particle wavefunctions are generally appallingly complicated objects. One way to get a handle on them is to break them down into simpler parts, understand those parts and then put them back together again. We do this by constructing the space of many particle wavefunctions as either a tensor product space or a Fock space. An obvious way break down a ...


3

A single-particle state is a state corresponding to a single particle in isolation. In weakly-interacting translation-invariant systems, for example, a particularly useful set of single-particle states are the plane-wave states $\lvert \mathbf{k}\rangle$, corresponding to a single particle with a plane-wave wavefunction $\langle \mathbf{x} \rvert ...


0

In a physics of nuclear structure, by the term single particle state is typically understood an excitation, that can be attributed mostly to one proton or one neutron that jumped to a higher orbit. Contrary to collective excitation or collective state, which is an excited level, that many nucleons participate in.


0

A closed extension $A_c$ of an operator $A$ is an operator whose action is the same as $A$, the domains satisfy $D(A_c)\supset D(A)$ and $A_c$ is closed. Given that, the smallest closed extension of a symmetric (densely defined) operator is its double adjoint $A^{**}$. We call it the closure of $A$, and denote it by $\overline{A}$. An operator $A$ is ...


1

$\newcommand{\ket}[1]{\lvert #1 \rangle}$Recall that the vacuum is annihilated by all annihilation operators: $$ a_i\ket{\Omega} = 0$$ and that all the occupied states are created from the vacuum as $$ \ket{\chi_i} = a^\dagger_i \ket{\Omega}$$ Now, if you apply an annihilation operator to a state which doesn't have the corresponding electron in it, the ...


0

Let me first answer your questions: Q1: No, the partial transposition acts on one part of the subsystem. To make this more clear, let me give a real definition of the criterion: Let $\mathcal{H}_1,\mathcal{H}_2$ be two Hilbert spaces, then the partial transpose on the second system is defined via $$ ^{PT}: \mathcal{B}(\mathcal{H}_1\otimes ...


2

$\newcommand{\ket}[1]{\lvert #1 \rangle}$You seem to be confused about what measuring an operator means. Let $A,B$ be two commuting self-adjoint operators as in your question, and let $\{u_n\}$ be a basis of simultaneous eigenvectors, that is $$ A\ket{u_i} = a_i \ket{u_i} \ \vee \ B\ket{u_i} = b_i \ket{u_i}$$ Now, a generic state $\psi$ can be written as $$ ...


3

No, because Haag's theorem states that there is no map between the free and interacting Hilbert spaces such that the fields and their commutation relations on one space are unitarily mapped onto the fields and their commutation relations on the other space. That is, the space of states of the interacting theory is as a representation of the commutation ...


1

Well, this is a problem of linear algebra substantially. The basic idea is that a square matrix $A$ is invertible if and only if $detA\neq 0$. Then, if the matrix is diagonalizable, since the determinant is invariant under coordinate transformations $A'=C^{-1}AC^1$, when you compute the determinant you get that, you get $detA=\lambda_1\dots\lambda_n$. (Note ...


2

The unitary representation of Galileian group already includes a representation of Weyl-Heisenberg group. The boost generator $K$ and the generator of translations $P$ satisfy $[K,P]= imI$ where $m$ is the mass of the system. Therefore, the subgroup whose generators are $m^{-1}K, P, I$ is the wanted unitary representation of Weyl-Heisenberg group. As a ...


4

The equation you phrase as $$|l,m\rangle=\int_\text{all space}\psi_{lm}(r,\theta,\phi)\,\left|r,\theta,\phi\right\rangle r^2\,\mathrm dr\,\mathrm d\Omega$$ is, and must be, wrong. The reason is that $|l,m⟩$ inhabits the orbital part of Hilbert space, $\mathcal H_\Omega$, and the right-hand side is a vector in the full Hilbert space $\mathcal H$, which is the ...


3

Repeatedly applying this relation to the ground state is exactly what you need to do. There's nothing more to it.


2

You're right, there is no eigenfunction. The eigenfunctions of a self-adjoint operator form a complete basis for the Hilbert space, but this is simply not true for symmetric operators. Therefore if an operator is not self-adjoint, it may not have any eigenfunctions.


0

This doesn't directly answer your question of orthogonality, but may still address your concern. I need to point out that you seem to be working on a scattering problem and resonant states. Resonant wave functions DO NOT belong to the Hilbert space. They are not even eigenfunctions of the Hamiltonian in an usual sense. You already know that they do not ...


6

If $|\phi⟩$ and $|\psi⟩$ are linearly independent, then it is always possible to assign them to the column vectors $$ |\phi⟩\mapsto\begin{pmatrix}1\\0\end{pmatrix} \text{ and } |\psi⟩\mapsto\begin{pmatrix}0\\1\end{pmatrix}, $$ but if they're not orthogonal you're obviously going to need to work harder on the representation of the inner product in this basis. ...


0

Its the first one. This is exactly what the "dagger" does. It transposes the spinor, converting it from a column spinor to a row spinor, and takes every entry to its complex conjugate, i.e: $$ \psi=\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix} \xrightarrow{\dagger} \begin{pmatrix}(\psi^T_L)^* (\psi^T_R)^*\end{pmatrix} = \begin{pmatrix}\psi_L^\dagger ...


2

$\langle a|i\hat{C}|a\rangle=\langle a|[\hat{A},\hat{B}]|a\rangle = \langle a| \hat{A}\hat{B}-\hat{B}\hat{A}|a \rangle = \langle a|\hat{A}\hat{B}|a\rangle - \langle a|\hat{B}\hat{A}|a \rangle = a^{*}\langle a|\hat{B}|a \rangle - a \langle a|\hat{B}|a\rangle = 0 $ (Since $a^{*}=a$ because $\hat{A}$ is Hermitian and Eigen Values of Hermitian operators are ...


1

$\newcommand{ket}{\left| #1 \right>}$ $\newcommand{bra}{\left< #1 \right|}$ $\newcommand{\bk}[3]{\left< #1| #2 |#3\right>}$ In a one dimensional problem $\langle \hat p \rangle$ is always zero. $$\langle \hat p \rangle = \bk{\psi}{\hat p}{\psi}=\int \mathrm{d}x \,\psi^* \hat p \, \psi \propto \int \mathrm{d}x \, \psi^* \psi' \overset{(1)}{=}-\int ...


2

A hint on this could be the fact that a superposition of stationary states of different energies is NOT a stationary state, because you can not express the wave function as the product of a single time-dependent exponential tiames a spatial function.


1

Given a densely defined pre-closed operator $T$ on a Hilbert space $H$, you can define its transpose (more properly called the adjoint) $T^*$ by requiring it to be the operator with the property that $$(\eta,T\psi)=(T^*\eta,\psi)$$ for any $\eta$ in the domain of $T^*$ (which is dense) and $\psi$ in the domain of $T$. Using Dirac's notation we can rewrite ...


3

It is invertible iff its determinate doesn't vanish $$ \det([H]_B) \ne 0 $$ Note that this property of the determinate is invariant under a change of basis since: \begin{align} \det(S^{-1} \cdot [H]_B \cdot S) & = \det(S^{-1}) \cdot \det([H]_B) \cdot \det(S) = \frac{1}{\det(S)} \cdot \det([H]_B) \cdot \det(S) \\ & = \det([H]_B) \end{align} with ...


2

I) The main point is that the half-angle $\frac{\theta}{2}$ doubles when we go from the ket $$\tag{1} |\psi\rangle~=~\begin{bmatrix}\cos\frac{\theta}{2} \cr e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix}, \qquad ||\psi||~=~1, $$ to the density matrix/operator $$\tag{2}\rho~=~| \psi\rangle \langle\psi | ~=~\frac{1}{2}\left({\bf 1}_{2\times 2}+ \vec{r}\cdot ...


3

From the way it is defined $\left| \Psi \right\rangle$ is not a vector on the sphere, but rather a vector along the z-axis between $-\hat{z}$ and $\hat{z}$, because it is a linear combination of $\left|0\right\rangle$ and $\left|1\right\rangle$ which are both vectors along the z-axis. Now we want $\left|\Psi(\theta = 0 , \phi =0)\right\rangle = ...


1

I am confused. Why would $\langle x \rangle = \langle x \rangle + \delta x$? Because you acted with the translation operator on the state. This is by definition what we want the translation operator to do. If it doesn't do this then we are in trouble. Shouldn't it equal $\langle x \rangle?$ Nope. Since, $\langle x\rangle = \langle ...


1

In quantum field theory the vacuum is the vector with no particles. It is defined starting from the Fock space, that is the sum of spaces with any number $n\in \mathbb{N}$ of (identical particles). The space with zero particles is, roughly speaking, one-dimensional, and its basis vector is the vacuum. In quantum mechanics, one can use the Fock space ...


4

I believe the difficulty stems from the archaic notation. What he is trying to do is show that $-\mathrm{i}\mathrm{D}$ is Hermitian, where $\mathrm{D}=\mathrm{d}/\mathrm{d}q$. Note the first equation you wrote. It is just saying that $\mathrm{D}$ acting "backwards" on the bra is equivalent to $\mathrm{D}$ acting "forwards" on the ket. He is trying to derive ...


0

Preliminaries Recall that a representation of an algebra on a Hilbert space is a map from the algebra to the bounded operators on a certain Hilbert space. Also recall the Heisenberg canonical commutation relations $$[q_i,p_k]=i\delta_{ik}I$$ A representation of such relations is a set of operators on some Hilbert space that satisfy to the same commutation ...


7

I will try to make it as simple and intuitive as possible. In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$): $$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$ Which ...


1

$\newcommand{ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 \big| #2 \right> }$ Though I am not sure 100% if what I am going to do is legitimate I would suggest the following (I am about 90% sure that it is legitimate): The confusion arises because the author has used $x'$ for two distinct ...



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