12 votes
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Property of a wavefunction

The only constraint on a particle's wavefunction $\psi$ is that it must be square-integrable - i.e. $\int_{-\infty}^\infty \overline{\psi(x)} \psi(x) \mathrm dx < \infty$. So it's perfectly ...
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  • 51.9k
9 votes

How to write completeness of wavefunctions without bra ket notation?

One way you can show the completeness relation without bra-ket notation is just $$\sum_{i} \langle \psi_i , v \rangle \psi_i = v \qquad \forall v\in\mathcal{H},$$ where $\mathcal{H}$ is the Hilbert ...
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6 votes

How to write completeness of wavefunctions without bra ket notation?

The completeness relation reads: $$\sum_{I=1}^{\infty} \psi^{*}_{I}(x') \psi_{I}(x)=\delta(x'-x ).$$ Proof: Suppose any wave function $\Psi$ can be expanded using the $\psi_i$: $$ \Psi(x)=\sum_{I=1}^{\...
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  • 317
4 votes

Can we have an eigenfunction that belongs to $L^{2}$ ($L=$ angular momentum operator) but not to $ L_{x}, L_{y}, or L_{z} $?

Can we have an eigenfunction that belongs to $L^{2}$ (L= angular momentum operator) but not to $ L_{x}, L_{y}, or L_{z} $? Not sure what "belongs to... " means here, but if it means "...
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  • 7,613
4 votes
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Why this Non-Hermitian Operation is non-existential?

This successively measuring or operation is denoted by a multiplication of operators in linear algebra. Like if we measure A first then B, total operation on your state is BA. Good. No. You're ...
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  • 38.8k
3 votes

Are Grassmann numbers always "under the hood" if we deal with fermionic ladder / field operators?

Supernumbers are needed & used in (among other things): the path/functional integral formalism of a QFT with fermionic fields, cf. e.g. this Phys.SE post. fermionic coherent states, cf. e.g. ...
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  • 170k
3 votes

Are Grassmann numbers always "under the hood" if we deal with fermionic ladder / field operators?

It is possible to define the antisymmetric Fock space, and related fermionic creation and annihilation operators, without any need to introduce Grassmann numbers. Grassmann numbers are only necessary ...
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  • 11.4k
3 votes

How to compute the bracket of two different angular momentum eigenvectors?

$\vert j~ m \rangle$ is a shared eigenvector of two Hermitian operators ($J^2$ and $J_z$). Eigenvectors of Hermitian operators vanish if they don't correspond to the same eigenvalue, so $$\langle j_1 ...
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3 votes
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Is $[\hat x, \hat p_x] = i\hbar\, \mathbb{I}$ contradicting a fact about commutators?

Just to develop what people already said in the comment: in fact, you put your finger on the proof that position and momentum operators can't have bounded spectra. Since those operators have an ...
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  • 2,256
3 votes
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How to prove that a $d$-dimensional Hilbert space can only have $d^2$ equiangular vectors (i.e. that a SIC is a maximal collection of that kind)?

I found one answer in this paper. Let $S$ be the $d \times N$ matrix whose columns are the SIC vectors, and let $G = S^\dagger S$ be their Gram matrix. The rank of $G$ is equal to $d$ unless the ...
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2 votes

Is there a difference between a Hermitian operator and an observable?

First, a mathematical subtlety. In finite dimensional Hilbert spaces spaces, every self-adjoint operator can be represented by a Hermitian matrix, i.e. one that is equal to its conjugate transpose, ...
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  • 137
2 votes
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Wavefunction properties tunnel effect

The eigenfunction of the 1D Schrödinger equation satisfy a second order linear differential equation. If there is a point $x$ where $\psi(x) = \psi'(x) = 0$, then $\psi$ is zero everywhere, which is ...
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  • 3,981
2 votes

Property of a wavefunction

If $\psi(x)$ is a "completely real" wavefunction, then $\phi(x) \equiv i \psi(x)$ is an "completely imaginary" wavefunction that is experimentally indistinguishable from $\psi(x)$. ...
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2 votes
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Why is time-evolution unitary? - The Heisenberg-picture Version

It depends on the hypotheses, in particular on the set of observables, you assume. If assuming that elementary YES-NO observables (also known as tests) are the orthogonal projectors on a Hilbert ...
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2 votes
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Regarding Page 224 Grififth Quantum Mechanics 3rd Edition

The states $|\uparrow\rangle,|\downarrow\rangle$ aren't eigenvectors of $S_x,S_y$ and the coefficients $\pm\tfrac{i\hbar}{2}$ aren't eigenvalues. My suggestion is that you should learn the basics of ...
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2 votes
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Basis for the two particle Hilbert space

A two-particle Hilbert space is just a tensor product $H\otimes H$ for the one-particle Hilbert space $H$. Essentially by definition of the tensor product, given any two bases $\{b_i\}_{i\in I}$ and $\...
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  • 107k
2 votes

Why this Non-Hermitian Operation is non-existential?

In this specific case BA = XY ~ Z, the Pauli Z matrix. There is a global phase $i$ on this term, but the product is still Hermitian. Yes, measuring X and Y with the Stern-Gerlach (SG) experiment is ...
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  • 121
2 votes

Where does the definition of partial trace as ${\rm Tr}_b(O)=\sum_j(I\otimes \langle b_j|)O(I\otimes|b_j\rangle)$ come from?

I'm not sure exactly what you're looking for, but I can try to provide some intuition for the formula you're citing: Recall that the standard trace takes a linear operator $O$ and spits out a number $\...
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  • 56
1 vote

Interpreting the Time Evolution Operator on a hands-on example

It is indeed good your intuition that the time evolution is a rotation but to get the general sense of what is happening you have to think in terms of Hilbert spaces and basis on such spaces. What is ...
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1 vote

Change of basis: incomplete and overcomplete bases?

Your assertion is entirely correct; a change of basis can only happen if the number of basis elements is the same as the previous basis. The reasoning behind this is very simple: the number of basis ...
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  • 779
1 vote

What is the geometry of a quantum system?

The physical position of the particle in real space (let's say Euclidean as you said, assuming non-relativistic quantum mechanics) is simply the eigenvalue of the position operator $\hat{\vec{r}}$ ...
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  • 779
1 vote
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Why does Identity operator work in general in quantum mechanics?

Since the set of states $\{\left | n \right>\}$ forms a complete basis, you can always write $A\left | n \right>=\sum_kc'_k\left | k \right>$ for some coefficients $c'_k$. If you plug this ...
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  • 561
1 vote
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Bit flip superoperator for three qubits

The operator $\mathcal{E}_{BF}$ is by definition a single-qubit operator, and so we construct an operator that flips the bit (with noise) of each qubit in a three-qubit system by taking the tensor ...
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  • 4,062
1 vote
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Combination of 3 spin $\frac{1}{2}$ particles to yield a state of net spin $\frac{1}{2}$

I guess the question I ask can be explained in a simpler manner eh? The narrow question can be answered by elementary linear algebra, of course, but the point of such questions is that you understand ...
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1 vote

Diagonalizing Operators Simultaneously

Any linear combination of degenerate eigenvectors is also an eigenvector, corresponding to the same value. Thus, we can first find the eigenvectors of $H$ and then look for eigenvectors of $A$ as ...
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  • 39.2k
1 vote

Does Schwarz Inequality guarantee that the expectation value for an operator exists and is finite?

Not for an arbitrary observable. For an explicit counter example, see this question. It is, however, easy to see that this is true when the observable is a bounded operator, i.e. one in which for any $...
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1 vote

Equation of motion for space and momentum of a $\textbf{coherent state}$

Given any sufficiently regular time dependent quantity (or set of quantities) $\vec u_0(t)$, it is possible to interpret the dynamics as a solution to differential (or integral, or integro-...
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  • 2,883
1 vote

Relation between the bosonic harmonic oscillator and the 'ordinary' harmonic oscillator

It's the same thing. Think of a vibrating string with modes of frequency $\omega_n$. Each of those modes is a harmonic oscillator, so when we quantize the string each mode can have energy $(N+1/2)\...
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  • 41.8k
1 vote

Quantum state from probability

Assuming you are talking about pure state, and assuming you mean, that you have lots of identical states which you can try against x and y Pauli matrices as much times as you want, and those are the ...
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  • 551
1 vote
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Action of translation operator on ket in momentum representation

The translation opertor is $T_a= \exp\{- a\partial_x\}= \exp\{-ia\hat p\}$ acts on a momentum eigenstate $|p\rangle$ by $$ T_a |p\rangle= e^{-iap}|p\rangle, \quad \langle p|T_a= e^{iap}\langle p| $$ ...
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  • 41.8k

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