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Questions tagged [eigenvalue]

A linear operator (including a matrix) acting on a non-zero eigenvector preserves its direction but, in general, scales its magnitude by a scalar quantity λ called the eigenvalue or characteristic value associated with that eigenvector. Even though it is normally used for linear operators, it may also extend to nonlinear operations, such as Schroeder functional composition, which evoke linear operations.

A linear operator (including a matrix) acting on a non-zero eigenvector preserves its direction but, in general, scales its magnitude by a scalar quantity $\lambda$ called the eigenvalue or characteristic value associated with that eigenvector. An operator's eigenvalues provide an invariant characterization of it through its trace, determinant, and characteristic polynomial.

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In quantum mechanics, given certain energy spectrum can one generate the corresponding potential?

A typical problem in quantum mechanics is to calculate the spectrum that corresponds to a given potential. Is there a one to one correspondence between the potential and its spectrum? If the answer to the previous question is yes, then given the…
Revo
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Why are Only Real Things Measurable?

Why can't we measure imaginary numbers? I mean, we can take the projection of a complex wave to be the "viewable" part, so why are imaginary numbers given this immeasurable descriptor? Namely with operators in quantum mechanics, why must measurable…
user24082
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Eigenfunctions of the Runge-Lenz vector

The hamiltonian for the hydrogen atom, $$ H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r} $$ is spherically symmetric and it therefore commutes with the angular momentum $\mathbf{L}$; this causes all its eigenfunctions with equal angular momentum number…
Emilio Pisanty
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Why do we use Hermitian operators in QM?

Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian ('self-adjoint') operators. The eigenvalues of the operator are…
Benjamin Hodgson
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Why do we use Eigenvalues to represent Observed Values in Quantum Mechanics?

One of the postulates of quantum mechanics is that for every observable $A$, there corresponds a linear Hermitian operator $\hat A$, and when we measure the observable $A$, we get an eigenvalue of $\hat A$ as the result. To me, this result seemed to…
Ishan Deo
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Discreteness of set of energy eigenvalues

Given some potential $V$, we have the eigenvalue problem $$ -\frac{\hbar^2}{2m}\Delta \psi + V\psi = E\psi $$ with the boundary condition $$ \lim_{|x|\rightarrow \infty} \psi(x) = 0 $$ If we wish to seek solutions for $$ E <…
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Schrödinger wavefunctional quantum-field eigenstates

The reason that I have this problem is that I'm trying to solve problem 14.4 of Schwartz's QFT book, which've confused me for a long time. The problem is to construct the eigenstates of a quantum field $\hat{\phi}(\vec{x})$, such that…
Nahc
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How to tackle 'dot' product for spin matrices

I read a textbook today on quantum mechanics regarding the Pauli spin matrices for two particles, it gives the Hamiltonian as $$ H = \alpha[\sigma_z^1 + \sigma_z^2] + \gamma\vec{\sigma}^1\cdot\vec{\sigma}^2 $$ where $\vec{\sigma}^1$ and…
user1285419
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Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be normalizable/non-normalizable?

These facts are taken for granted in a QM text I read. The purportedly guaranteed non-normalizability of eigenfunctions which correspond to a continuous eigenvalue spectrum is only partly justified by the author, who merely states that the…
Sean Allen
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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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How do we know that we have captured the entire spectrum of the Harmonic Oscillator by using ladder operators?

Consider standard quantum harmonic oscillator, $H = \frac{1}{2m}P^2 + \frac{1}{2}m\omega^2Q^2$. We can solve this problem by defining the ladder operators $a$ and $a^{\dagger}$. One can show that there is a unique "ground state" eigenvector $\psi_0$…
user140223
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Numerical solution to Schrödinger equation - eigenvalues

This is my first question on here. I'm trying to numerically solve the Schrödinger equation for the Woods-Saxon Potential and find the energy eigenvalues and eigenfunctions but I am confused about how exactly this should be done. I've solved some…
CINA
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Can a normalizable function *always* be decompose into the discrete Hydrogen spectrum?

This question has been bothering me for a while now: can one reconstruct an arbitrary (normalizable) function $\phi(\mathbf r)$ in $\mathbb R^3$, with only the (discrete) set of Hydrogen wavefunctions (or, e.g., eigenfunctions of the 3D Harmonic…
anon01
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Why is the measured value of some observable $A$, always an eigenvalue of the corresponding 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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Bounded and Unbounded (Scattering) States in Quantum Mechanics

I understand that bounded states in quantum mechanics imply that the total energy of the state, $E$, is less than the potential $V_0$ at + or - spatial infinity. Similarly, the scattering state implies that $E > V_0$. But I do not understand why for…
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