# Tagged Questions

In physics, an operator is almost always either a square matrix or a linear mapping from one space of functions (often on $\mathbb{R}^N$ or $\mathbb{C}^N$) to the same or other like space of functions. Operators serve as *observables* and as *time evolution operators* in Quantum Mechanics. This tag ...

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### Countable Matrix Representation

In my quantum mechanics class, my professor explained that the Hamiltonian along with position and momentum operators can be represented by matrices of countable dimension. This is especially usefull ...
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### Eigenstates of a Hermitian field operator

Consider a Hermitian field operator $\phi(x)$ with eigenstates satisfying $$\phi(x) |\alpha\rangle = \alpha(x) | \alpha \rangle$$ I'm trying to determine the inner product between the eigenstates. ...
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### Tricky operator identity: $[L^2,[L^2,\vec{r}]]=2 \hbar ^2 \{ L^2, \vec{r}\}$?

This operator identity showed up in a course I was taking, and it was given without proof. $$[L^2,[L^2,\vec{r}]]=2 \hbar ^2 \{ L^2, \vec{r}\}$$ The curly brackets denote the anticommutator, $AB+BA$. ...
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### The Momentum Operator in QM

I've seen the 'derivation' as to why momentum is an operator, but I still don't buy it. Momentum has always been just a product $m{\bf v}$. Why should it now be an operator. Why can't we just multiply ...
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### Linearizing Quantum Operators

I was reading an article on harmonic generation and came across the following way of decomposing the photon field operator. $$\hat{A}={\langle}\hat{A}{\rangle}I+ \Delta\hat{a}$$ The right hand side ...
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### Is the energy always discrete?

In the von Neumann axioms for quantum mechanics, the first postulate states that a quantum state is a vector in a separable Hilbert space. It means it is assumed the Hilbert space has a basis with at ...
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### Self-adjoint and unbounded operators in QM

An operator $A$ is said to be self-adjoint if $(\chi,A\psi)=(A\chi,\psi)$ for $\psi, \chi \in D_A$ and $D_A=D_{A^\dagger}$. But for the free particle momentum operator $\hat{p}$ these inner products ...
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### Applying an operator to a function vs. a (ket) vector

I have a question regarding the effect of quantum mechanical operators. The definition that I'm familiar with says that an operator $A$ acts on a vector from a Hilbert space, $|\psi\rangle$, and the ...
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I am a physicist. I always heard physicists used the terminology "symmetric", "Hermitian", "self-adjoint", and "essentially self-adjoint" operators interchangeably. Actually what is the difference ...
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### Regularisation of infinite-dimensional determinants

Can a regularisation of the determinant be used to find the eigenvalues of the Hamiltonian in the normal infinite dimensional setting of QM? Edit: I failed to make myself clear. In finite dimensions,...
### The cleverest way to calculate $\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]$ with $\left[\hat{a},\hat{a}^{\dagger}\right]=1$
Who can provide me some elegant solution for $$\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]\qquad\text{with} \qquad\left[\hat{a},\hat{a}^{\dagger}\right]~=~1$$ other than brute force calculation? =...