# Tagged Questions

A quantum observable is a measurable operator whose corresponding property of the state can be determined by some sequence of physical operations ("observation"), such as submitting the system to various electromagnetic fields and eventually reading a value. In systems governed by classical ...

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### Superposition and simultaneous observation

Trying to understand superposition. Ok, so double slit experiment. The multiple paths the particle simultaneously travels interfere with each other but as it is absorbed, it chooses one "actual" ...
487 views

### Commutation relation of a operator with Hamiltonian [duplicate]

Given that the eigenvalues of a Hamiltonian operator $H$ are bounded below, will a Hermitian operator $T$ exist such that $[T, H] = i\hbar{\bf 1}$ identity operator?
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### Hamiltonian symmetry Lie algebra

What is the connection between complete set of commuting observables and generators of the Lie group? I have a Hamiltonian written down in second quantized formalism and I also checked that it ...
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### Why does the measurement of some observable $A$, the measured value is always an eigenvalue of the operator?

Explain why when we make a measurement of some observable $A$ in QM, the measured value is always an eigenvalue of the operator $A$.
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### On the Equivalence of Schrodinger and Heisenberg Descriptions of Quantum Mechanics and Observability

I'm not a physicist, but rather a control (feedback) systems engineer eager to understand more than just a cursory explanation of quantum mechanics. The StackExchange has been an excellent forum for ...
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### How do we physically apply the operators of quantum mechanics on a particle?

What do we have to perform physically that is equivalent to applying those quantum mechanical operators on a state $|\psi\rangle$? Edit: I have removed the part I was asking regarding measurement ...
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### Repeating a measurement vs uncertainty

The wikipedia says on measurement in quantum mechanics that: Repeating the same measurement without any evolution of the quantum state will lead to the same result. On the other hand, doesn't ...
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### Motivating Irreducibility of Hilbert Space for Quantization Axioms

In the context of geometric quantization, we usually look for a map from the Poisson algebra of classical observables to the algebra of quantum observables (or rather, a sub-algebra of the classical ...
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### Eigenvalues being physical observables

I think I'm comfortable with the PDE solutions to the Schrodinger equation. But as soon as we start putting these values in a matrix (in dirac notation), I lose my understanding and everything ...
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### Commuting operators and Direct product spaces

Under what conditions is the common eigenspace of two commuting hermitian operators isomorphic to the direct product of their individual eigenspaces? When can an eigenket $|\lambda$1$\lambda$2$\rangle$...
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### What are you studying when you study a Harmonic Oscillator in QM?

This probably is a naive question - so please forgive a self-studier. In the text I am studying, one builds a HO by placing a particle in a potential that increases quadratically from the origin. The ...
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### Is there any non-hermitian operator on Hilbert Space with all real eigenvalues?

The property of hermitian is the sufficient condition for eigenvalue being real. Is there any non-hermitian operator on Hilbert Space with all real eigenvalues? If there exist, then can all ...
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### Eigenstate of field operator in QFT

Why don't people discuss the eigenstate of the field operator? For example, the real scalar field the field operator is Hermitian, so its eigenstate is an observable quantity.
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### Understanding Well Defined States

I am self-studying from a text in QM. Well defined states are mentioned several times. By and large these are consistent and seem to be readily apparent: states of well defined energy are basis kets....
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### Quantum Mechanics - Observable

If $O$ represents an operator corresponding to an observable why does the following equality hold? $$\langle f(x)\, |\, O g(x)\rangle = \langle g(x) \,|\, O f(x) \rangle$$ It is used on the last ...
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### What happens with a tunneling particle when its momentum is imaginary in QM?

In classical mechanics the motion of a particle is bounded if it is trapped in a potential well. In quantum mechanics this is no longer the case and there is a non zero probability of the particle to ...
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### The Physical Meaning behind a Commutator [duplicate]

I've just been introduced to the idea of commutators and I'm aware that it's not a trivial thing if two operators $A$ and $B$ commute, i.e. if two Hermitian operators commute then the eigenvalues of ...
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### Is commutation relation an equivalence relation?

I'm now learning quantum mechanics with Liboff. In the book it deals with "a compete set of mutually compatible observables" in order to make a state maximally informative. How can one find such set? ...
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### Non-Hermitian operator with real eigenvalues?

So we know that in Quantum Mechanics we require the operators to be Hermitian, so that their eigenvalues are real ($\in \mathbb{R}$) because they correspond to observables. What about a non-Hermitian ...
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### Discrete Values for Observables vs Average Values (Quantum Mechanics)

When considering observables and their corresponding operators, would it be correct to believe that discerning discrete values for an observable is possible ONLY when $\psi$ is an eigenfunction of the ...
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### Is $\hat{\phi}^{-}\hat{\phi}^{+}$ a well defined observable in the Quantum field theory of a scalar field?

Is the Hermitian operator $\hat{\mathcal{O}}=\hat{\phi}^{-}(x)\hat{\phi}^{+}(x)$, where $\hat{\phi}^{+}(x)$ is positive frequency part of the scalar field operator, a well defined observable in QFT? ...
265 views

### Why does a picture of a person seem to be looking in the same direction irrespective of the angle of observation? [closed]

If you observe a picture of a person hanging on a wall who seems to be looking directly towards you always seems to be looking at you even though you change your angle of observation to the extremes. ...
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### Is kinetic energy in QM a state-property or is it distributed?

Suppose we have a quantum mechanical system, which is well described by its wave function in r-representation $\Psi$. We are interested in the properties of an observable, say the kinetic energy $T$. ...
I am trying to solve an exercise in Shankar's QM book (concretely 4.2.1), and I am asked the probability of each possible value for the operator $L_x$ when the particle is in a certain eigenstate of ...