17 votes
Accepted

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 ...
  • 111k
15 votes

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 ...
  • 174k
14 votes
Accepted

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 ...
11 votes
Accepted

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 ...
  • 111k
10 votes
Accepted

What is the matrix representation of a function of an quantum operator?

It's not clear what you mean by "the function acting on the matrix of the operator." If you mean $$f\left(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right) = \begin{pmatrix}f(a) & ...
9 votes
Accepted

Do non-commuting Hamiltonians have non-commuting time evolution operator?

The short answer is yes the unitary evolution will not commute if considered in full generality of two parameter dependence but not under certain special circumstances (and also single parameter ...
9 votes
Accepted

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

It is just matter of math. Even if one may also try to find physical interpretations of this result a posteriori. Let us focus on the identity $$\frac{d^2 \psi}{dx^2}=\frac{2m}{\hbar^2}[V(x)-E]\psi\:.$...
9 votes

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

Expectation values of quantum mechanics correspond to classically measured values by construction. Indeed, if the classical theory were not a limiting case of the quantum one, this latter would be ...
  • 42.2k
8 votes
Accepted

Separable Hilbert space in quantum mechanics

Separability is not a necessary physical requirement, at least in modern approaches. First of all, all mathematical technology, as the spectral theory, is valid both for separable or non-separable ...
5 votes

Can the wavefunction be inferred from the expectation values of operators?

Intuitively, if you know the expectation values of all possible observables, that should be enough to fix the state of the system. This almost sounds tautological, since the state is just a ...
  • 6,111
5 votes
Accepted

Tensor products in second quantization

If these were bosonic operators $b_{\alpha}^{\dagger}$, you could rightly regard the product $b_{\alpha}^{\dagger} b_{\beta}^{\dagger}$ as a tensor product. This is because you can think of the ...
  • 2,404
5 votes

Defining propagators as vacuum expectation values of products of field operators

It's simply because that commutator is proportional to identity operator, as your result imply. Considering a normalized vacuum state then $$ \left[ \hat{\phi}^+(x), \hat{\phi}^-(y) \right] = \text{...
  • 699
5 votes

What to understand by $\langle \phi | \hat{A}|\psi \rangle$?

Both. Recall that a linear map $A:V\to W$ automatically induces a map $A^*:W^*\to V^*$ where $V^*$ is the dual space of $V$. Given $f\in W^*$ we can evaluate $f(A(x))$ for any $x$ in $V$, and so $f(...
  • 43.3k
4 votes
Accepted

Reality of Hermitian Operators

"Real" for an operator does not mean anything. At most you can wonder if the matrix of the operator is real. It depends on the used basis. If the basis is made of eigenvectors and the ...
4 votes
Accepted

Momentum operator's position when we calculate the expectation value of momentum

When we calculate the expectation value of the momentum operator, we use $$⟨p⟩ = \int ψ^*\left(−iℏ\frac{∂}{∂x}\right)ψ\ dx\tag{1}.$$ I'm wondering if we can get $⟨p⟩$ by $$⟨p⟩ = \int \left(iℏ\frac{∂}{∂...
  • 9,375
4 votes

Momentum operator's position when we calculate the expectation value of momentum

The momentum operator is hermitian ($\widehat{p}^\dagger=\widehat{p}$), therefore the expectation value is a real number, which is of course important for physical interpretation: $$\langle\widehat{p}\...
3 votes

What is the matrix representation of a function of an quantum operator?

Formally/heuristically, for sufficiently nice operator $$\begin{align}\hat{A}~=~&\sum_{i,j\in I}|i\rangle A^i{}_j \langle j|\cr ~=~&\begin{bmatrix}|1\rangle & |2\rangle & \cdots \end{...
  • 174k
3 votes

Can the wavefunction be inferred from the expectation values of operators?

Just a generalization to Meng Cheng's answer and theoretically how you can construct $O_{mn}$ given known operators like $x$ and $p$. Given some Hermitian operator $H$ which we have full knowledge ...
  • 861
3 votes
Accepted

Spectral theorem in QFT

I am not sure to understand the nature of the problem. The spectral theorem, as it is a mathematical fact, holds also in QFT. It does not matter if we do not know how the Hilbert space is made, it is ...
3 votes
Accepted

Quantum Hamiltonian and Classical Hamitonian Rotationally Invariant

You have hit upon a recondite problem of quantization, which rarely crops up in practical problems: A classical $\cal H$ has several, indeed, many different quantum Hs which have it as their common ...
3 votes

Is this proof that $a|0\rangle=0$ wrong (ladder operators and number operator)?

The proof is correct, just it is a proof by contradiction. Suppose that $a|0\rangle\neq 0$, then we can normalise it and we call $|-1\rangle$ a unit vector parallel to it. Therefore $$a|0\rangle = c_0|...
3 votes
Accepted

How to make this matrix unitary?

I think requiring Hermiticity for X, and hence real, instead of pure imaginary, a dooms you. Let me review the mainstream representation of the conventional SU(2) group element, instead, $$U =a_0 + i ...
3 votes

For $n\in\mathbb{N}$ is the $n$th power of a hermitian operator always hermitian?

And since two different hermitian operators will not necessarily commute they might not be hermitian in that case. That may be the case, but an operator commutes with all its powers, i.e. $[A^m,A^n] =...
  • 111k
3 votes

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)
3 votes

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 ...
  • 28.2k
2 votes

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 $\langle x, Ax\rangle \geq 0$...
2 votes

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 ...
  • 3,318
2 votes
Accepted

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 $J$ ...
  • 174k
2 votes
Accepted

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^\...
2 votes

Can the wavefunction be inferred from the expectation values of operators?

Can the wavefunction be inferred from expectation value of operators? The answer is yes. If we consider "the wavefunction" to be the ground state wavefunction then the first Hohenberg-Kohn ...
  • 9,375

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