# Tag Info

## Hot answers tagged operators

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### Why does the expectation value in quantum mechanics correspond to the classically measured value?

In general, there is no such thing as a "classically measured position" for a generic quantum system/state. Some situations are simply not well-modeled by classical physics, and Ehrenfest's ...
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### Do non-commuting Hamiltonians have non-commuting time evolution operator?

Here is a counterexample to the first part of OP's question: Imagine $K$ is diagonalizable with eigenvalue spectrum $\subseteq \mathbb{Z}$ within the integers. Then $e^{i2\pi K}={\bf 1}$ is the ...
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### Can the wavefunction be inferred from the expectation values of operators?

It actually suffices to know the expectation values of all projection operators of the form $P_\psi:=|\psi\rangle\langle \psi|$ for $\psi \in H$ (which are of course observables). Indeed, suppose we ...
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### Where does it become apparent in real scalar QFT that the field has to be an operator-valued distribution, as opposed to an operator-valued function?

Since the commutation relations are $$[\phi(x,t),\pi(y,t)] = \mathrm{i}\delta(x-y)$$ at least one of $\phi$ and $\pi$ must be a distribution, too, since functions are closed under multiplication and ...
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### Momentum operator's position when we calculate the expectation value of momentum

(2) does work to get $\langle P \rangle$ but acting $\hat{P}$ on the bra rather than the ket introduces an extra minus sign which might cause trouble. So it is just easier to use (1)

### Schrödinger Equation Energy Requirement $E \geq V_{\min}$

Think about a similar classical problem to understand this. Suppose you have a mass m that can be sitting on the ground anywhere along the $x$ axis. Suppose each point along the axis is at a different ...
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### Square root of number operator for quantum harmonic oscillator

The square root exists and it is defined by standard functional calculus for every selfadjoint operator $A: D(A) \to H$ with non negative spectrum (which is equivalent to $\langle x, Ax\rangle \geq 0$...
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### Why rotation is not an observable?

An operator $A$ acting over a state $|\psi\rangle$ is an observable if $A=A^{\dagger}$ (it is its own self-adjoint or transpose conjugate). What this means is that the eigenvalues of the operator are ...
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### Question on ordered exponential explanation in Wikipedia

OP is presumably missing that Wikipedia mentions that $\gamma$ is an infinitesimal rectangle. Each of the 4 contour integrals of the 4 sides of $OE[-J]$ are replaced to with the initial value of $J$ ...
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### Expressing the four-momentum operator in terms of field operators

Simplicity is subjective here. You are meant to do something with your expression. Assuming your expression is correct, interchange the k and x,x' integrations, and trade the k for an x gradient, P^\...
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