• Product state: We can assign each particle its own wavefunction and multiply them to get the joint amplitude.
• Entangled state: Only the combined system has a complete wavefunction. There is generally no separate pure wavefunction for each particle that can be multiplied to recover it.

explain what it means to multiple the wavefunctions of two separate particles that are interacting in some manner

For two distinguishable particles, “multiplying their wavefunctions” usually means building a joint wavefunction from two independent states:\[ \Psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2). \]

Here \(x_1\) is particle 1’s position and \(x_2\) is particle 2’s position. The product is an amplitude for the joint possibility that particle 1 is at \(x_1\) and particle 2 is at \(x_2\). Its squared magnitude gives the joint probability density:\[ |\Psi(x_1,x_2)|^2 =|\psi_1(x_1)|^2|\psi_2(x_2)|^2. \]

So this multiplication describes a state in which the particles’ position probabilities are independent. It is often written more precisely as a tensor product, \(|\psi_1\rangle\otimes|\psi_2\rangle\). We are not multiplying two waves at the same position; the full wavefunction has two position arguments.

Interaction changes the picture. Suppose the particles exert forces on each other. Their Hamiltonian contains an interaction term \(V(x_1,x_2)\):\[ H=H_1+H_2+V(x_1,x_2). \]

Even if the state starts as a product, the interaction generally evolves it into a joint wavefunction that cannot be written as one wavefunction for particle 1 times another for particle 2:\[ \Psi(x_1,x_2)\ne\psi_1(x_1)\psi_2(x_2). \]

That is entanglement. The amplitude for finding particle 1 at one position now depends on what position we consider for particle 2. For example, if two particles attract, configurations in which they are near each other may have greater probability than a product of two independent position distributions would predict.

A useful distinction is:

• Product state: We can assign each particle its own wavefunction and multiply them to get the joint amplitude.
• Entangled state: Only the combined system has a complete wavefunction. There is generally no separate pure wavefunction for each particle that can be multiplied to recover it.

There is one further wrinkle for identical particles. Quantum mechanics does not let us physically distinguish “particle 1 at \(x_1\)” from “particle 2 at \(x_1\).” Their joint wavefunction must be symmetric for bosons or antisymmetric for fermions. Even without a force between them, a simple labeled product \(\psi_a(x_1)\psi_b(x_2)\) is therefore generally not the complete physical state.

In short, multiplying two wavefunctions is a way to describe independent amplitudes within one joint system. Once an interaction creates correlations, the joint wavefunction usually contains more information than that product can express.

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For a free Schrödinger particle, start with the time-dependent Schrödinger equation:$$i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$iℏ∂t∂ψ​=−2mℏ2​∂x2∂2ψ​

Assume a plane-wave solution:$$\psi(x,t) = A e^{i(kx-\omega t)}$$ψ(x,t)=Aei(kx−ωt)

Step 1: Compute the derivatives

Time derivative:$$\frac{\partial \psi}{\partial t} = -i\omega \psi$$∂t∂ψ​=−iωψ

Therefore$$i\hbar\frac{\partial \psi}{\partial t} = \hbar\omega \psi$$iℏ∂t∂ψ​=ℏωψ

Spatial second derivative:$$\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi$$∂x2∂2ψ​=−k2ψ

Therefore$$-\frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} = \frac{\hbar^2k^2}{2m}\psi$$−2mℏ2​∂x2∂2ψ​=2mℏ2k2​ψ

Substituting into Schrödinger’s equation gives$$\hbar\omega\psi = \frac{\hbar^2k^2}{2m}\psi$$ℏωψ=2mℏ2k2​ψ

Cancelling $\psi$ψ,$$\boxed{ \omega=\frac{\hbar k^2}{2m} }$$ω=2mℏk2​​

This is the Schrödinger dispersion relation.

Visualizing the dispersion relation

The relation is quadratic in $k$k:$$\omega \propto k^2$$ω∝k2

$\omega=\frac{\hbar k^2}{2m}$ω=2mℏk2​

Unlike light waves, where$$\omega = ck$$ω=ck

the Schrödinger particle’s frequency grows as the square of the wavenumber.

Connection to momentum and energy

Using de Broglie’s relations:$$p=\hbar k$$p=ℏk $$E=\hbar\omega$$E=ℏω

Substituting into the dispersion relation:$$E = \hbar\left(\frac{\hbar k^2}{2m}\right) = \frac{\hbar^2k^2}{2m}$$E=ℏ(2mℏk2​)=2mℏ2k2​

Since $p=\hbar k$p=ℏk,$$\boxed{ E=\frac{p^2}{2m} }$$E=2mp2​​

which is exactly the classical nonrelativistic kinetic energy.

Thus the Schrödinger dispersion relation is simply the wave version of Newtonian mechanics.

Phase velocity

The phase velocity is$$v_p=\frac{\omega}{k}$$vp​=kω​

Substituting the dispersion relation:$$v_p = \frac{\hbar k}{2m} = \frac{p}{2m}$$vp​=2mℏk​=2mp​

Since$$v=\frac{p}{m}$$v=mp​

we obtain$$\boxed{ v_p=\frac{v}{2} }$$vp​=2v​​

The phase of the wave moves at half the particle velocity.

Group velocity

A particle is represented by a wave packet, not a single plane wave.

The packet moves at the group velocity:$$v_g = \frac{d\omega}{dk}$$vg​=dkdω​

Differentiating,$$v_g = \frac{\hbar k}{m}$$vg​=mℏk​

Using $p=\hbar k$p=ℏk,$$v_g = \frac{p}{m}$$vg​=mp​

Therefore$$\boxed{ v_g=v }$$vg​=v​

The group velocity equals the particle’s classical velocity.

Why wave packets spread

Because$$\omega \propto k^2$$ω∝k2

different Fourier components travel at different group velocities:$$v_g=\frac{\hbar k}{m}$$vg​=mℏk​

Large-$k$k components move faster than small-$k$k components.

As time passes, the packet spreads out.

This is called dispersion.

For light in vacuum,$$\omega=ck$$ω=ck

and$$\frac{d\omega}{dk}=c$$dkdω​=c

for every $k$k, so no spreading occurs.

For Schrödinger waves,$$\frac{d^2\omega}{dk^2} = \frac{\hbar}{m} \neq 0$$dk2d2ω​=mℏ​=0

and the packet inevitably disperses.

Why Lorentz invariance fails

The Schrödinger equation assumes$$E=\frac{p^2}{2m}$$E=2mp2​

which leads directly to$$\omega=\frac{\hbar k^2}{2m}.$$ω=2mℏk2​.

A Lorentz transformation mixes energy and momentum:$$E’=\gamma(E-vp)$$E′=γ(E−vp) $$p’=\gamma\left(p-\frac{vE}{c^2}\right)$$p′=γ(p−c2vE​)

and the transformed quantities no longer satisfy$$E’=\frac{p’^2}{2m}.$$E′=2mp′2​.

Thus the Schrödinger dispersion relation is preserved only under Galilean transformations, not Lorentz transformations.

That is why relativistic quantum theory replaces the Schrödinger relation$$E=\frac{p^2}{2m}$$E=2mp2​

with$$E^2=p^2c^2+m^2c^4,$$E2=p2c2+m2c4,

leading to the relativistic wave equations such as the Klein-Gordon and Dirac equations.

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Key Result

The phase of a Schrödinger plane wave can be rewritten in Lorentz-transformed coordinates. However, the transformed energy and momentum do not satisfy the nonrelativistic Schrödinger dispersion relation.

Therefore, the Schrödinger equation is not Lorentz invariant.

The Free-Particle Plane Wave

A free particle in nonrelativistic quantum mechanics can be represented by the plane wave:

ψ(x,t) = A exp[i(kx − ωt)]

Using the de Broglie relations:

p = ℏk

E = ℏω

the wavefunction can also be written as:

ψ(x,t) = A exp[(i/ℏ)(px − Et)]

For a free particle governed by the Schrödinger equation, energy and momentum are related by:

E = p²/(2m)

Equivalently, frequency and wave number satisfy:

ω = ℏk²/(2m)

This is the nonrelativistic Schrödinger dispersion relation.

Applying a Lorentz Transformation

Consider two inertial reference frames moving relative to one another along the x-axis.

The inverse Lorentz transformation is:

x = γ(x′ + vt′)

t = γ(t′ + vx′/c²)

where:

γ = 1/√(1 − v²/c²)

Substitute these expressions into the phase px − Et:

px − Et = pγ(x′ + vt′) − Eγ(t′ + vx′/c²)

Collecting the coefficients of x′ and t′ gives:

px − Et = γ(p − vE/c²)x′ − γ(E − vp)t′

This can be rewritten as:

px − Et = p′x′ − E′t′

provided that:

p′ = γ(p − vE/c²)

and:

E′ = γ(E − vp)

These are the Lorentz transformation rules for relativistic momentum and energy.

The Transformed Wave

The transformed wave can now be written as:

ψ′(x′,t′) = A exp[(i/ℏ)(p′x′ − E′t′)]

Equivalently:

ψ′(x′,t′) = A exp[i(k′x′ − ω′t′)]

where:

k′ = γ(k − vω/c²)

and:

ω′ = γ(ω − vk)

The phase therefore retains the same form:

kx − ωt = k′x′ − ω′t′

At first glance, this may appear to suggest that the Schrödinger plane wave is compatible with special relativity. The problem becomes apparent when we examine the transformed dispersion relation.

The Schrödinger Dispersion Relation Is Not Preserved

The original wave satisfies:

ω = ℏk²/(2m)

After the Lorentz transformation:

ω′ = γ(ω − vk)

and:

k′ = γ(k − vω/c²)

In general, these transformed quantities do not satisfy:

ω′ = ℏk′²/(2m)

Instead:

ω′ ≠ ℏk′²/(2m)

Therefore, although the transformed expression still looks like a plane wave, it is not generally a solution of the same free-particle Schrödinger equation.

The Central Result

The plane-wave phase px − Et can retain its form under a Lorentz transformation, but the Schrödinger energy–momentum relation is not preserved.

Lorentz-transforming the phase does not make the Schrödinger equation Lorentz invariant.

Why the Conflict Occurs

Special relativity requires energy and momentum to satisfy:

E² = p²c² + m²c⁴

This relation is Lorentz invariant. If it holds in one inertial frame, it holds in every inertial frame.

The Schrödinger equation instead uses the nonrelativistic relation:

E = p²/(2m)

This is only the low-speed approximation for a particle’s kinetic energy. It excludes the rest energy mc² and is not preserved by Lorentz transformations.

The conflict is therefore not with the plane-wave form itself. The conflict arises from combining a Lorentz transformation with a nonrelativistic energy–momentum relation.

Relativistic Quantum Equations

A Lorentz-invariant quantum theory must begin with the relativistic energy–momentum relation:

E² = p²c² + m²c⁴

For spin-0 particles, this leads to the Klein–Gordon equation.

For spin-½ particles, it leads to the Dirac equation.

The Schrödinger equation remains extremely successful in the nonrelativistic domain, where particle speeds are much smaller than the speed of light. Its natural spacetime symmetry is Galilean rather than Lorentzian.

Final Conclusion

The phase:

px − Et

can be rewritten in Lorentz-transformed coordinates by treating energy and momentum as components of a relativistic four-vector.

However, the transformed quantities E′ and p′ do not satisfy the Schrödinger relation:

E′ = p′²/(2m)

Therefore:

The phase can be Lorentz-transformed, but the Schrödinger equation itself is not Lorentz invariant.

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In quantum mechanics, observables (like position, momentum, energy) are represented by operators. Requiring those operators to be Hermitian (more precisely, self-adjoint) is not arbitrary—it follows from a few fundamental physical requirements.

1. Measurement outcomes must be real numbers

A physical measurement always gives a real value.

If an operator $\hat{A}$A^ represents an observable, its possible measurement outcomes are its eigenvalues.

A Hermitian operator guarantees:$$\text{All eigenvalues of } \hat{A} \text{ are real}$$All eigenvalues of A^ are real

$\hat{A} = \hat{A}^\dagger$A^=A^†

If the operator were not Hermitian, you could get complex eigenvalues like $3 + 2i$3+2i, which have no physical meaning as measurement results.

2. Expectation values must be real

Even before measurement, we often compute the expectation value:$$\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle$$⟨A⟩=⟨ψ∣A^∣ψ⟩

For a Hermitian operator:$$\langle \psi | \hat{A} | \psi \rangle \in \mathbb{R}$$⟨ψ∣A^∣ψ⟩∈R

If $\hat{A}$A^ were not Hermitian, the expectation value could be complex—which would make no physical sense as an “average measurement.”

3. Orthogonality of eigenstates (clean measurement structure)

Hermitian operators have a powerful property:

• Eigenstates corresponding to different eigenvalues are orthogonal

This gives us a clean decomposition:$$|\psi\rangle = \sum_i c_i |a_i\rangle$$∣ψ⟩=i∑​ci​∣ai​⟩

Where:

• $|a_i\rangle$∣ai​⟩ are eigenstates of the observable
• $|c_i|^2$∣ci​∣2 are probabilities

Without Hermiticity, this orthogonal structure breaks down → probabilities become ambiguous.

4. Probability interpretation requires it

Quantum mechanics relies on:$$P(a_i) = |\langle a_i | \psi \rangle|^2$$P(ai​)=∣⟨ai​∣ψ⟩∣2

This only works cleanly if:

• Eigenstates form an orthonormal basis
• The operator is Hermitian

Otherwise, you lose a consistent probability framework.

5. Connection to unitary time evolution

Hermitian operators also generate unitary transformations.

Example: the Hamiltonian $\hat{H}$H^$$U(t) = e^{-i \hat{H} t / \hbar}$$U(t)=e−iH^t/ℏ

If $\hat{H}$H^ is Hermitian:

• $U(t)$U(t) is unitary
• Total probability is conserved

If not:

• Probability could grow or decay → physically unacceptable

6. Deeper insight (physics intuition)

You can think of Hermitian operators as enforcing:

• Reality → measurements are real
• Stability → probabilities don’t explode
• Consistency → repeatable measurements give structured outcomes

In a deeper sense:

Hermiticity ensures that the mathematical structure of quantum mechanics aligns with the physical requirement that observations are real, probabilistic, and consistent.

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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

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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.

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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.

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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.

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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 ).

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