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Quantum mechanics describes the microscopic properties of nature in a regime where classical mechanics no longer applies. It explains phenomena such as the wave-particle duality, quantization of energy, and the uncertainty principle and is generally used in single-body systems. Use the quantum-field-theory tag for the theory of many-body quantum-mechanical systems.

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On the "spectrum" of an operator in quantum mechanics

Very simple question, I'm new to this. I'm reading Griffiths book on QM and have a question about the "spectrum" of an observable operator. Does the spectrum of an operator require specification of a …
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Why use of creation and then annihilation operator on n state gives n+1 state when they are ...

You must have misread something. The raising $(\hat a^\dagger)$ and lowering $(\hat a)$ operators of the quantum harmonic oscillator act as follows on the $n^\text{th}$ state: $$\hat a|n\rangle=\sqrt{ …
Charlie's user avatar
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3 votes
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What are the limits of quantum theory?

The most famous limitation of quantum theory is its failure to incorporate gravity. The two most established pillars of modern physics are quantum mechanics and general relativity. Quantum mechanics i …
Charlie's user avatar
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1 vote

What is the wave function of a quantum particle?

The wave function is a mathematical object that is physically interpreted as representing the state of a quantum system. Any information you could possibly extract about a given system is "contained", …
Charlie's user avatar
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1 vote
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Biconditionality of the compatibility theorem for commuting operators

Yes it is if and only if. Given two observables $X$ and $Y$: $[X,Y]=0$ $X$ and $Y$ have a common eigenbasis are equivalent statements.
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Exchange Degeneracy of Fermions

Remember that two states differing by an overall phase are considered physically identical. Note that the particle exchange operator has the following property here: $$P_{21}|\Psi_1\rangle=-|\Psi_1\ …
Charlie's user avatar
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Probability of Finding a Quantum System in a Specific State

You will need to evolve the state in time using the full Schrodinger equation. However, if your Hamiltonian is time-independent then your system is "stationary" and all amplitudes are time-independent …
Charlie's user avatar
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3 votes
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How to prove that $\sum_{n=1}^{\infty} | \phi_n \rangle \langle \phi_n | = \hat {I}$?

You can see that an operator of this type behaves as the identity in $\Bbb R^3$ with a basis $\{|i\rangle\}$: $$\left(\sum_{i=1}^3|i\rangle\langle i|\right)|v\rangle=\sum_{i=1}^3|i\rangle\left(\langle …
Charlie's user avatar
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2 votes

Inner product of $\langle \phi | \psi \rangle$ gives a complex value - why/meaning?

As noted in Mike Stone's answer this is explained in any quantum mechanics textbook. The inner product is a complex inner product, in general: $$\langle \phi|\psi\rangle\in \Bbb C \tag{1}.$$ In quantu …
Charlie's user avatar
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3 votes

What does it means that a particle with mass can be a wave without mass?

Your intuition about what "mass" is is what's causing your problem. If you try and take mathematical descriptions of the world too literally you will hit a dead end pretty fast. In this example you're …
Charlie's user avatar
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Expectation Value in Bra-ket notation

What you have in the case of $\langle\psi|\hat A|\psi\rangle$ is effectively a matrix sandwiched between a vector and a dual vector. If you like, it is a row vector on the left, a matrix in the middle …
Charlie's user avatar
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Eigenvalue and Amplitude

No. I think your equation is perhaps meant to be $$|\psi\rangle = \frac{1}{\sqrt{2}}|\psi_1\rangle+\frac{1}{\sqrt{2}}|\psi_2\rangle,$$ without $\hat B$ in there. This is the state of the system expan …
Charlie's user avatar
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Meaning of $\langle 0 | \hat{H} | 0 \rangle$

If what you have is a quantum system in the state $|0\rangle$, then the meaning of $\langle 0|\hat M|0\rangle$ where $\hat M$ is an operator corresponding to an arbitrary observable depends on whether …
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Probability of measuring eigen-energies?

This is taking the modulus of a complex number, not just "squaring it" in the usual sense: $$|e^{i\theta}|^2=e^{-i\theta}e^{i\theta}=e^{i\theta-i\theta}=e^0=1.$$
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Understanding some experiments based on the Stern-Gerlach experiment

Regarding the apparatus (b), you are correct that the "classically" expected result would be a roughly even distribution of measurements in a spectrum between $S_z=+\hbar/2$ and $S_z=-\hbar/2$. The su …
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