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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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Time-ordering and Minkowski metric's negative sign

There are lots of answers already on the site for example here that discuss in varying levels of detail/nuance the ways in which the metric signature changes things. The common theme among all of them …
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2 votes

Interpretation of Sakurai's ambiguous statement about wavefunctions and overlap

Consider two functions, $$f(x):\Bbb R^3\rightarrow \Bbb R$$ $$g(x):\Bbb R^3\rightarrow \Bbb R$$ They may separately have non-zero integral on all of $\Bbb R^3$, i.e.: $$\int_{\Bbb R^3}f(x)\mathrm dx\n …
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4 votes

Hamiltonian Operator

It is not correct that applying the Hamiltonian gives you the energy of the system if the system is in a general state, $|\psi\rangle$. Upon measurement (in an experiment for example) the system "jump …
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0 votes
1 answer
54 views

Extra second-order term in Dyson Formula Expansion in David Tong's Notes

Right at the bottom of page 52 of David Tong's QFT notes we have just defined the time ordered Dyson formula, David Tong then shows the expansion of $(3.20)$ however an extra second-order term has app …
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10 votes
2 answers
619 views

Time evolution in quantum mechanics of states not contained in the Hilbert space

Eigenstates of, for example, $\hat p$, are not elements of the standard quantum mechanical Hilbert space, i.e. $\psi(x)=e^{ipx}\notin\mathcal L^2(\Bbb R)$. This prompts the question of - given that af …
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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 …
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1 vote
Accepted

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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4 votes
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Are particles literally waves or just abstract probability waves?

It's tempting to try to interpret the mathematics of QM or QFT too literally, however this can be a dangerous game to play. Quite often you will cause more problems for yourself than you will solve by …
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7 votes
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Correct interpretation of $\langle x | \psi \rangle$?

The first of your two suggestions doesn't make sense, "the probability amplitude of finding the particle at position $x$ in the state $|\psi\rangle$". The particle is either in the state $|\psi\rangle …
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3 votes
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What physical quantity does the Hamiltonian operator represent?

The Hamiltonian corresponds to the energy of the system. The equation you have written is the Schrodinger equation and it tells you that the Hamiltonian is a special observable operator that dictates …
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0 votes

Is $(L^2, L_z)$ a complete set of commuting observables?

Regarding the Hydrogen atom, being in an eigenspace of both $L^2$ and $L_z$ means knowing the type of orbital the electron is in ($s$, $p$, $d$, etc.) - this gives the $l$ label - and also which speci …
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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 …
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17 votes
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Quantum mechanics and rigorous math

Kets In quantum mechanics the possible states of the system are elements of a separable, projective Hilbert space (i.e. two states differing by an overall complex constant are equivalent). The kets e. …
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1 vote

Time dependent Schrödinger equation with time independent potential and separation of variables

Yes, see for example this small set of notes that outlines how the Schrodinger equation comes apart into two separate equations: $$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x) \tag{ …
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2 votes

Is there a difference between a quantum state and a quantum wave function?

Remember that - because our observables correspond to Hermitian operators acting on the Hilbert space, $\mathcal H$, containing the state vectors, $|\psi\rangle$ - the set of all eigenvectors of each …
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