All Questions
Tagged with linear-algebraobservables
27 questions
0votes
0answers
59views
Are projectors observable quantities in QM? [duplicate]
Given a certain quantum state $|\Psi\rangle$, then we can project any other quantum state $|\Phi\rangle$ on the first one by using the projector: $P_\Psi\equiv |\Psi\rangle\langle\Psi|$ in such a way ...
- 2,164
1vote
2answers
230views
Projection operator onto support of distinct observables
Suppose $P_i$ is the projection operator onto the support of the observable $O_i$ defined on some (say, finite dimensional) Hilbert space. I'm curious as to whether we can define the projection ...
1vote
1answer
196views
Position Eigenstates
Suppose that the state of a system can be represented as a superposition of finite position eigenstates $|\Psi \rangle = \sum_{i}c_{i}|x_{i}\rangle$, where each $|x_{i}\rangle$ corresponds to a unique ...
- 1,073
3votes
2answers
600views
Choice of Basis for solution space of a Hamiltonian
I'm quite new to the field of Quantum mechanics, but I can't wrap my head around the following postulate of Quantum mechanics, stated as Postulate 3 in [1]. Although I approached quantum mechanics ...
- 133
1vote
1answer
56views
Technical question in how to correspond operators to dynamical variables [closed]
I have following question. Here is the book I use for QM. I understand so far until the step of 3.43. Can somebody tell me how he arrived at the step of 3.43? I tried to expand the eigenvector in 3.42 ...
0votes
0answers
65views
Difference between transformations written as $TOT^{\dagger}$ and as $TOT^{-1}$ in quantum matrix mechanics?
When we write quantum operators in matrix form, we perform transformations. I have seen that at some points the transformation $T$ of a matrix $O$ is written as, for example, $O\to TOT^{\dagger}$ $O\...
- 861
1vote
1answer
58views
What does it mean for two compatible observables to be a "coarse-graining" of a third?
In reading about quantum contextuality, I've encountered the statement that if [A,B] = 0, then there exists another observable C such that the spectral projections of A and B are a coarse-graining ...
- 1,425
0votes
2answers
139views
How to assign a value to an observable when the statevector is not an eigenvector of the operator?
We get the value of an observable $A$ for a given state $|\lambda\rangle$ of a system from the eigenequation $\hat{A} |\lambda\rangle = \lambda |\lambda\rangle$ where $\hat{A}$ is the operator ...
- 378
2votes
1answer
141views
Finding common eigenvectors for two commuting hermitian matrices [closed]
Let $A = \begin{bmatrix} 1 &0 &0 \\ 0& 0& 0\\ 0&0 &1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 &0 &1 \\ 0& 1& 0\\ 1&0 &0 \end{bmatrix}$ ...
0votes
2answers
163views
Eigenstates for $\vec{L}^2, L_z, L_x$ and $L_y$? [closed]
I am asked to find states $|j,m\rangle$ that are simultaneously eigenstates for $\vec{L}^2, L_z, L_x$ and $L_y$. I know that the $L_i$ operators do not commute and hence you cannot have a state $|\phi\...
- 824
5votes
6answers
2kviews
Angular momentum commutation relations [duplicate]
The operator $L^2$ commutes with each of the operators $L_x$, $L_y$ and $L_z$, yet $L_x$, $L_y$ and $L_z$ do not commute with each other. From linear algebra, we know that if two hermitian operators ...
- 914
0votes
2answers
786views
The general wavefunction can be expanded in such eigenstates
Suppose we have solved for the energy eigenstates of some Hamiltonian operator $\hat{H}$. We call the energy eigenstates $\psi_n (x)$, where: $n=1$: $\psi_1 (x)$ is the ground states $n=2$: ...
- 187
0votes
2answers
167views
Representatives in QM
I'm reading Dirac's book about QM. I reached the chapter called "representations" where Dirac introduces how can bras, kets, and observables be decomposed using a base. I have found issues ...
- 569
0votes
0answers
56views
Can QM be used to model 2 state systems with more than 4 linearly independent observables?
Suppose I have a system (e.g., a particle) and I have different physical measurement apparatus which can act on it. Each of the measurement apparatus (observables) has 2 distinct labeled outcomes, ...
1vote
1answer
703views
Conjugate complex of linear operators in quantum mechanics
I'm pretty new to quantum mechanics (I would like to understand it broadly as an hobbyist). I'm trying to reading Principles of Quantum Mechanics by Dirac. I've found difficult to understand a ...
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