If a proton is moving at high speed, does it affect its de Broglie wave nature?

1) de Broglie wavelength (core relation)

For any particle:

Non-relativistic proton

Relativistic proton (high speed)

2) What “high speed” changes physically

• The wave nature does NOT disappear. Quantum mechanics never “turns off.”
• The wavelength becomes very small. At accelerator energies it can be far smaller than atoms or even nuclei.

3) Why fast protons often look “particle-like”

Wave behavior (diffraction/interference) is easiest to see when the wavelength is comparable to the size of
slits, gratings, or other structures:

So the proton still has a wave description, but the wave effects become harder to detect
with typical instruments.

4) Relativity does not suppress quantum mechanics

5) Phase vs group velocity (subtle but important)

For a relativistic de Broglie wave:

No causality violation: information travels with the group velocity, not the phase velocity.

6) Why high-energy experiments still reveal “wave/quantum” structure

Even when λ is tiny, quantum behavior shows up strongly in scattering and diffraction-like measurements.
Higher energies probe smaller distances, revealing internal structure (e.g., quarks and gluons in the proton).

7) One-line takeaway

The post Wave nature and speed of proton (particle) appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

The post The Universal Wave Function appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

Clarifying Angular Momentum in a Particle in a Box

The statement “a particle in a box has no angular momentum” refers specifically to the absence of orbital angular momentum as a good quantum number. This does not apply to spin, which is an intrinsic form of angular momentum independent of geometry.

1. Two Kinds of Angular Momentum

(A) Orbital Angular Momentum

The operator is:

It depends on spatial coordinates and exists only when the system has rotational symmetry.

(B) Spin Angular Momentum

Spin is an internal degree of freedom and does not depend on the potential’s shape or boundary conditions.

2. What “No Angular Momentum” Really Means

A rectangular (1D or 3D) infinite potential well has no rotational symmetry. Therefore:

This means neither nor are conserved or define good quantum numbers. The particle’s orbital motion is described only by the quantum numbers .

Thus: a particle in a box has no conserved orbital angular momentum.

3. But the Particle Can Still Have Spin

The spin degree of freedom is completely unaffected by placing the particle in a box.

The full Hilbert space becomes:

The spatial eigenfunctions are the usual box states:

The spin state is independent:

The complete state is therefore:

or

If no magnetic fields or spin–orbit coupling are present:

Spin angular momentum is fully conserved inside the box.

4. What If the Particle Is “Already Rotating”?

This phrase has two possible interpretations:

Case 1 — The particle has orbital angular momentum before entering the box

The initial state might be something like:

When placed in a rectangular box:

• rotational symmetry is lost,
• orbital angular momentum is no longer conserved,
• the state becomes a superposition of box eigenstates.

The angular momentum “scrambles” because the box does not allow rotations.

Case 2 — The particle has spin

Spin remains a well-defined, conserved quantum number. The full state can be:

Spin exists independently of the geometry and survives unchanged inside the box.

5. Summary Table

Quantity	Exists in a Box?	Depends on Geometry?	Conserved?
Orbital Angular Momentum	Not a good quantum number	Yes	No
Spin Angular Momentum	Always	No	Yes (unless external fields)
Total	Not conserved (because is not)	Yes	Only for spherical potentials

In Summary

A particle in a rectangular/1D box does not have conserved orbital angular momentum because the geometry breaks rotational symmetry.
But the particle’s spin remains fully intact and unaffected by the boundary conditions.

The post Spin versus Angular Momentum appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

https://stationarystates.com/basic-quantum-theory/angular-momentum…article-in-a-box/

Energy Levels of a Particle in a Box with Angular Momentum

1. Particle in a 1D Box

In 1D, angular momentum doesn’t exist in the usual sense because rotation requires at least two dimensions.
Energy levels remain:

2. Particle in a 2D or 3D Box

For a 2D rectangle or 3D cube, the Schrödinger equation separates in Cartesian coordinates:

Angular momentum is not conserved in a cubical box. Energy depends on quantum numbers along each axis:

3. Particle in a Spherical Box

If the box is spherically symmetric, angular momentum L is a good quantum number.
The Schrödinger equation in spherical coordinates:

Separate variables:

where Y_\ell^m are spherical harmonics and \ell is the angular momentum quantum number.
Energy levels include a centrifugal term:

Here, are the zeros of spherical Bessel functions.
Larger angular momentum () increases energy because the wavefunction is “pushed outward”.

4. Key Takeaways

• 1D box: Angular momentum is irrelevant; energy levels are unchanged.
• Rectangular/cubical box: Energy depends on quantum numbers along each axis, not angular momentum.
• Spherical box: Higher angular momentum quantum number \ell raises the energy.

Intuition: Higher angular momentum → particle “rotates” more → less probability near the center → higher energy.

The post Angular momentum for particle in a box appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

Why Wavefunctions for Different Energy Levels Are Orthogonal

1. They Come From a Hermitian Operator

The time-independent Schrödinger equation is:

Ĥ ψ = E ψ

Here, Ĥ (the Hamiltonian) is a Hermitian operator, which has two key properties:

• Its eigenvalues (energy levels) are real.
• Its eigenfunctions corresponding to different eigenvalues are orthogonal.

So if:

Ĥ ψ_n = E_n ψ_n
Ĥ ψ_m = E_m ψ_m

and E_n ≠ E_m, then:

<ψ_n | ψ_m> = 0

2. Orthogonality Prevents States From Overlapping

Different energy eigenstates are physically distinct. Orthogonality ensures:

• No energy state contains any component of another.
• Measurements of energy always yield one clear value.

3. It Comes From Conservation of Probability

Take two solutions of the Schrödinger equation, ψ_n and ψ_m. Multiply the equation for ψ_n by ψ_m* and the equation for ψ_m by ψ_n*, subtract, and integrate:

(E_n - E_m) ∫ ψ_m*(x) ψ_n(x) dx = 0

Since E_n ≠ E_m, the only solution is:

∫ ψ_m*(x) ψ_n(x) dx = 0

This is orthogonality.

4. Simple Example: Particle in a Box

Energy eigenfunctions are:

ψ_n(x) = √(2/L) sin(nπx / L)

Different sine modes are orthogonal:

∫_0^L sin(nπx / L) sin(mπx / L) dx = 0   (n ≠ m)

Like different notes on a guitar string—different vibrational modes don’t “mix”.

Summary

• The Hamiltonian is Hermitian → different eigenvalues → orthogonal eigenfunctions.
• Distinct energy states must not overlap physically.
• Orthogonality pops directly out of the integrated Schrödinger equation.
• In real systems (particle in a box, harmonic oscillator, hydrogen atom), this matches the behavior of different vibrational modes.

The post Why are wave functions orthogonal? appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

In quantum mechanics, the time dependence of operators depends on which representation (picture) we use — primarily the Schrödinger picture or the Heisenberg picture.
Both are equivalent, but they treat the time evolution of states and operators differently.

1. Schrödinger Picture

In the Schrödinger picture:

• The operators are typically time-independent (unless they explicitly depend on time, like a time-varying potential).
• The state vectors evolve with time according to the Schrödinger equation.

If an operator itself depends explicitly on time (e.g., an external driving field), then its time dependence is just that explicit one.

2. Heisenberg Picture

In the Heisenberg picture, the situation is reversed:

• The state vectors are constant in time (frozen at their initial value).
• The operators carry all the time dependence.

The operator evolution is given by the Heisenberg equation of motion:

The solution can also be expressed as a similarity transformation using the time-evolution operator :

3. Examples

(a) Free Particle Momentum and Position

For a free particle with Hamiltonian :

This is a quantum analogue of classical motion with constant momentum and linearly increasing position.

(b) Harmonic Oscillator

For a 1D harmonic oscillator with :

Combining gives:

Solution:

This again mirrors classical oscillatory motion but with operator-valued amplitudes.

(c) Spin Precession in a Magnetic Field

For a spin- particle in a magnetic field , the Hamiltonian is

Then:

So the spin components precess:

where is the Larmor frequency.

4. Summary

Picture	State	Operator	Equation of Motion
Schrödinger		Usually time-independent
Heisenberg	Time-independent

Thus, in quantum mechanics, operators evolve in time via commutators with the Hamiltonian, reflecting the deep correspondence between quantum and classical dynamics (where parallels the classical Poisson bracket ).

The post Time Dependence of Quantum Mechanical Operators appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

It took me forever to understand why we could just ‘take all possible paths’ in the double slit experiment.

It wasn’t till I re-solved (30 years after my initial graduate school QM course), the problem of the STEP Potential Barrier. This simple problem has a revealating solution – it contains a REFLECTED part of the particle.

Classically, reflection should not be possible, since the energy of the particle is greater than the energy of the step. Still, reflection does occur. This means, the particle follows a path that isn’t classically allowed. This also starts to explain Feynman’s ALL POSSIBLE PATHs approach to the double slit experiment.

1) Step Potential with : Partial Reflection is Real

Consider a particle of mass incident from the left on a potential step:

Define region-dependent wavenumbers:

Time-independent Schrödinger equation solutions:

Applying continuity of and at :

Solving for reflection and transmission coefficients:

The probability current for a plane wave is . Hence,

and .

Why reflection without a barrier?
Classically, with there’s no turning point, so reflection is impossible. Quantum mechanically, the de Broglie wavelength jumps at the interface (from to ). To satisfy boundary conditions for both and , a left-moving component is required—an impedance mismatch effect analogous to Fresnel reflection at an optical interface. Reflection is thus an interference requirement, not an energy shortfall.

2) From “Reflection at a Step” to “All Paths” (Feynman’s Picture)

Feynman’s path integral expresses the amplitude for a particle to go from at time to at time as a coherent sum over all paths :

where .

• In the classical limit (), phases from wildly different paths cancel; only those near stationary action (classical paths) dominate.
• At finite , non-classical paths contribute with phases and interfere constructively or destructively.

How That Explains Step Reflection

At a sharp step, many paths “sample” the region (where kinetic energy is ) before returning to . Their actions differ due to the potential term. When all contributions are added coherently, two dominant families of paths appear at the detector on the left:

• Those that keep momentum (forward-going)
• Those that reverse to (reflected)

The relative phase between these families depends on the action difference tied to the mismatch—precisely what boundary conditions captured. The constructive interference of the “returning” family yields a nonzero .

This “impedance mismatch ⇒ coherent back-sum” corresponds directly to the path integral description.

And Now the Double Slit

In the double-slit experiment, the amplitude at a screen point is a sum over all paths passing through each slit:

Within each slit’s sum are countless non-classical trajectories—zigzags, edge grazes, small detours—each contributing a slightly different phase. The envelopes of these sums are dominated by near-straight (stationary-action) routes. Coherent addition across the two slits gives the interference pattern, with phase difference . Blocking one slit removes one entire family of paths, eliminating interference.

Unifying Intuition

• Step reflection with : coherence across paths entering the step forces a back-propagating component—nonzero without a classical turning point.
• Double-slit fringes: coherence across paths through both apertures yields the interference pattern.

In both cases, classically forbidden effects (reflection without a barrier, interference without waves) naturally emerge once we accept that all paths contribute with phase .
Classical motion reappears only when non-stationary paths’ phases cancel out. Wherever discontinuities (steps, slit edges) create phase mismatches between path families, quantum interference manifests macroscopically—as reflection, diffraction, or fringes.

The post The Simple Step Potential and How it Explains the All Paths Feynman Approach appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

Position Operator in Momentum Space and the Uncertainty Relation

In 1D momentum space the position and momentum operators act very differently:

• Momentum operator:

     

• Position operator:

     

That “multiply vs. differentiate” mismatch is exactly what produces uncertainty:
a state sharply localized in must vary slowly with ,
while a state sharply localized in requires to vary rapidly with .
These requirements are incompatible in the extreme, and the quantitative statement is the
Heisenberg uncertainty relation.

1. The Commutator in Momentum Space

Let be square-integrable and vanish at . Using the operator forms above:

So as operators,

2. Derivation via Cauchy–Schwarz Inequality

Define centered operators

and the vectors

Then

By Cauchy–Schwarz:

Split into commutator and anticommutator:

Taking expectation values and using :

Hence

Therefore,

This is the Heisenberg uncertainty relation.

(Equivalently, one can quote the general Robertson–Schrödinger inequality
, and insert .)

3. Why the Derivative Form Introduces Uncertainty

Since acts as , making very narrow around some
forces to be large in magnitude (sharp changes near the edges),
which inflates . Conversely, making very smooth (small derivative) spreads
it out in , increasing . The non-commutativity is the precise
algebraic statement of this trade-off.

4. State that Saturates the Bound

Equality holds when for real . In -space:

Solving gives a Gaussian

with and , so .

Summary

In momentum representation, the position operator being a derivative operator guarantees
non-commutation with the momentum operator. By applying the Cauchy–Schwarz inequality
to the commutator, we derive the uncertainty relation:

This expresses the fundamental trade-off between sharpness in position and sharpness in momentum.

The post Deriving Uncertainty from the Position Operator in p-space appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

Spin-½ measurement along a rotated axis (Pauli-matrix derivation)

We will show explicitly—using Pauli matrices—why a spin-½ measurement along any direction still yields exactly two outcomes, and compute the probabilities when the measurement axis is rotated by from the original -axis.

1) Pauli matrices and spin operators

The spin operator along a unit direction is

For any direction , the eigenvalues of are always . Thus there are always two outcomes, independent of the axis.

2) Choose a new axis rotated by from the -axis

Let the original measurement axis be . Rotate the apparatus by an angle toward about the -axis. The new unit vector is

Then

For with and ,

3) Eigenvalues (two outcomes only) and eigenvectors

The characteristic equation of is

Therefore the measurement outcomes for are always .

A corresponding eigenvector (for general ) can be taken as

when we work in the (i.e., ) basis and choose azimuth (the plane).

4) Start in and measure along the rotated axis

In the basis, the eigenstates are

If the system is prepared in and we measure , the probability of obtaining is

and . This can be derived either by the explicit overlap of the spinors above, or by the Bloch-vector identity

5) Plug in

Therefore, for a rotation of the measurement axis relative to , a state prepared as yields

• Outcome along with probability .
• Outcome along with probability .

6) What is special about the chosen direction?

Physically, no direction is special: quantum spin for a spin-½ particle always has two eigenvalues along any axis. What is special is that the experimentalist’s chosen direction defines the observable:

and distinct directions correspond to non-commuting operators (e.g. ). Changing the apparatus orientation rotates the measurement basis; it does not change the two-valued spectrum.

7) One-line summary

Rotate the experiment by any angle you like (e.g. ): the spin-½ measurement still has outcomes ; only the probabilities change according to and .

The post Choosing the Axis to Measure Spin appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

Is There a Difference Between a Slow-Moving and a Fast-Moving Electron Interacting with a Photon?

Yes — the interaction depends strongly on the electron’s speed (equivalently its kinetic energy and Lorentz factor). Below is a structured comparison.

1. Energy and Momentum Transfer

Slow (Nonrelativistic) Electron

• Kinetic energy is small compared to typical photon energies in many scenarios.
• Thomson scattering applies (elastic; negligible photon frequency change).

where is the classical electron radius.

Fast (Relativistic) Electron

• Significant Doppler shifts and energy exchange with photons.
• Compton scattering with electron recoil; frequency can change substantially.
• In electron’s rest frame, incident photons are blue-shifted; after scattering, lab-frame photons can be strongly boosted (inverse Compton).

2. Reference Frame Effects

• Slow electron: Electron frame lab frame; photon field nearly unchanged under transformation.
• Fast electron: Incident radiation is Lorentz-transformed; photons are aberrated into a forward cone and blue-shifted.

3. Cross-Section Regimes

Thomson (low energy/slow electron): total cross section approximately constant for .

Klein–Nishina (relativistic/high energy): energy-dependent, decreases as photon energy rises:

with and the incident and scattered photon angular frequencies (in the same frame).

4. Physical Outcomes

• Slow electron + photon: Elastic scattering; photon energy nearly unchanged; angular pattern .
• Fast electron + photon: Large frequency shifts (Compton redshift or inverse-Compton blueshift); forward-peaked scattering; potential production of high-energy photons.

5. Examples

• Photoelectric/low-energy scattering in solids: Conduction electrons interacting with visible light Thomson limit for scattering.
• Astrophysics: Relativistic electrons in jets upscatter CMB/starlight to X-rays or -rays (inverse Compton); synchrotron plus IC spectra.
• Laboratory: High-energy electron beams colliding with lasers produce energetic backscattered photons due to strong Doppler boosting.

Summary Table

Feature	Slow Electron	Fast Electron (Relativistic)
Regime	Thomson (classical)	Compton / Klein–Nishina (relativistic)
Energy exchange	Negligible (elastic)	Significant (inelastic with recoil)
Photon frequency shift	Minimal	Doppler shift + recoil (large)
Angular distribution		Forward-peaked
Cross section	(constant for )	Energy-dependent; decreases at high energies
Dominant processes	Thomson scattering; (in media) photoelectric absorption	Compton / inverse Compton; synchrotron + IC in astrophysics

The post Slow Moving vs Fast Moving Electrons – Photon Interaction appeared first on Time Travel, Quantum Entanglement and Quantum Computing.