# Tag Info

## Hot answers tagged hilbert-space

Accepted

### 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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### 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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### 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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### 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 ...

### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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Accepted

### 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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### 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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### 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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### 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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### 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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### 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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