Chapter 1: Starting from Classical Mechanics
1.1 Fundamentals of Newtonian Mechanics
Introduction to Equations of Motion
Newton's Second Law, as a fundamental law of motion, expresses the relationship between force and acceleration as follows:
$$
\mathbf{F} = m\mathbf{a} = m\frac{d^2 \mathbf{r}}{dt^2}
$$
Here, $\mathbf{r}$ represents the position vector, $m$ is mass, and $t$ is time. For one-dimensional motion, this simplifies to:
$$
F = m \frac{d^2x }{dt^2 }
$$
Properties of the Equation of Motion
This differential equation is a second-order ordinary differential equation, which has a unique solution given the following initial conditions:
- Initial position: $x(t_0) = x_0$
- Initial velocity: $\dot{x}(t_0) = v_0$
Physically, this means that once the initial state is given, the subsequent motion is completely determined.
1.2 Derivation of Energy Conservation
1. Equation of Motion in a Potential Field
The equation of motion for a particle in a conservative force field, where force is a function of position only, can be written as:
$$
m\frac{d^2x }{dt^2 } = F(x) = -\frac{dV}{dx}
$$
Here, $V(x)$ is the potential energy.
2. Transformation of the Equation of Motion
Multiplying both sides by velocity $dx/dt$:
$$
m\frac{d^2x }{dt^2 }\frac{dx}{dt} = -\frac{dV}{dx} \frac{dx}{dt}
$$
The left side relates to the time derivative of kinetic energy:
$$
m\frac{d^2x }{dt^2 }\frac{dx}{dt} = \frac{d}{dt}( \frac{1}{2}m( \frac{dx}{dt})^2 )
$$
The right side is the time derivative of potential energy:
$$
-\frac{dV}{dx}\frac{dx}{dt} = -\frac{dV}{dt}
$$
3. Energy as a Primitive Function
Therefore, the equation can be rewritten as:
$$
\frac{d}{dt}( \frac{1}{2} m(\frac{dx}{dt})^2 + V(x)) = 0
$$
This shows that the following quantity is time-independent:
$$
E = \frac{1}{2}m(\frac{dx}{dt})^2 + V(x) = \text{constant}
$$
4. Energy as a Conserved Quantity
$E$ is the total energy of the system:
- $T = \frac{1}{2}m(\frac{dx}{dt})^2$: Kinetic energy
- $V(x)$: Potential energy
Constant determined by initial conditions: - Initial position $x_0$ and velocity $v_0$ at $t = 0$ determine
- $E = \frac{1}{2}mv_0^2 + V(x_0)$
5. Kinetic and Potential Energy
The total energy $E$ of the system is expressed as the sum of kinetic energy $T$ and potential energy $V$:
$$
E = T + V = \frac{1}{2}mv^2 + V(x)
$$
Using momentum $p = mv$:
$$
E = \frac{p^2}{2m} + V(x)
$$
6. Hamiltonian and Canonical Equations
Defining this energy as the Hamiltonian $H$:
$$
H(x,p) = \frac{p^2 }{2m} + V(x)
$$
The Hamiltonian leads to the canonical equations:
$$
\frac{dx}{dt} = \frac{\partial H}{\partial p} = \frac{p}{m}
$$
$$
\frac{dp}{dt} = -\frac{\partial H}{\partial x} = -\frac{dV}{dx}
$$
The canonical equations provide a system of differential equations for the two variables $(x,p)$ - position and momentum. While Newtonian mechanics starts from the equation of motion, modern formalism finds it more insightful to begin with these canonical equations to describe various physical phenomena.
7. Derivation of Energy Conservation
The total time derivative of the Hamiltonian, using the above canonical equations, transforms as follows:
$$
\frac{dH}{dt} = \frac{\partial H}{\partial x} \frac{dx}{dt} + \frac{\partial H}{\partial p}\frac{dp}{dt} = 0
$$
This result shows that the Hamiltonian is time-invariant, representing the law of energy conservation.
Physical Significance
Importance of the Hamiltonian formalism:
- Completely describes the system's time evolution
- Clearly expresses the symmetry of canonical variables $(x,p)$
- Provides the foundation for transition to quantum mechanics
This Hamiltonian formalism plays a crucial role as a bridge to quantum mechanics.
Chapter 2: The Quantization Procedure
2.1 Correspondence between Classical and Quantum Mechanics
2.1.1 Basic Concepts
In transitioning from classical to quantum mechanics, expressing physical quantities as operators rather than numerical values leads to a clearer mathematical formulation. This transformation should be viewed as a mathematical operation, and we should avoid the incomprehensible notion that physical quantities or matter are "actually" operators.
2.1.2 Basic Correspondences
The main correspondences between physical quantities and operators are:
Position:
$$
x \rightarrow \hat{x} \text{(multiplication operation on wavefunction)}
$$
Momentum:
$$
p \rightarrow \hat{p} = -i\hbar\frac{d}{dx} \text{(differentiation operation on wavefunction)}
$$
This correspondence is based on de Broglie's relation:
$$
p = \hbar k
$$
where $k$ is the wave vector, and the form of the momentum operator is derived from its relationship with the differential operator $-i\frac{d}{dx}$ acting on plane waves $e^{ikx}$.
2.1.3 Derivation of the Momentum Operator
de Broglie proposed the following relation to express the wave-particle duality (de Broglie's matter waves):
$$
\lambda = \frac{h}{p}
$$
where:
- $\lambda$ is the wavelength of matter waves
- $h$ is Planck's constant
- $p$ is particle momentum
This equation shows the relationship between the particle's momentum and its wave properties.
The wave number $k$ is related to wavelength $\lambda$ by:
$$
k = \frac{2\pi}{\lambda}
$$
Combining this with de Broglie's relation yields:
$$
p = \hbar k ,
(\hbar = \frac{h}{2\pi})
$$
For a free particle, we represent the wavefunction as a plane wave:
$$
\psi(x) = A e^{ikx}
$$
where $A$ is the amplitude.
Applying the differential operator $-i\hbar\frac{d}{dx}$ to this wavefunction:
$$
-i\hbar \frac{d}{dx}(Ae^{ikx}) = -i \hbar(ik)A e^{ikx} = \hbar kA e^{ikx} = pA e^{ikx}
$$
This shows that the operator $-i\hbar\frac{d}{dx}$ corresponds to momentum $p$. This correspondence is operational, and we should not interpret it as "momentum is actually a differential operator."
Extension to General Wavefunctions
Any wavefunction can be expressed as a superposition of plane waves:
$$
\psi(x) = \int A(k)e^{ikx} dk
$$
Definition of Momentum Operator
This leads to the definition of the momentum operator for general wavefunctions:
$$
\hat{p} = -i\hbar\frac{d}{dx}
$$
This operator satisfies:
- Hermiticity: $\hat{p}^\dagger = \hat{p}$
- Reality of eigenvalues
- Consistency with momentum conservation
Action on Wavefunctions
When the momentum operator acts on a wavefunction:
$$
\hat{p}\psi = -i\hbar\frac{d\psi}{dx}
$$
This operator measures the spatial rate of change of the wavefunction.
Interpretation as Observable
The expectation value of momentum is:
$$
\langle p \rangle = \int \psi^ * (x)(-i \hbar \frac{d}{dx}) \psi(x) dx
$$
corresponding to the average momentum observed in experiments.
2.2 The Hamiltonian Operator
2.2.1 Quantization of the Hamiltonian
Classical Hamiltonian:
$$
H = \frac{p^2}{2m} + V(x)
$$
Corresponding quantum Hamiltonian:
$$
\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})
$$
Explicit form:
$$
\hat{H} = -\frac{ \hbar^2 }{2m} \frac{d^2 }{dx^2 } + V(x)
$$
2.2.2 The Schrödinger Equation
Time-dependent Schrödinger equation:
$$
i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi
$$
This is the fundamental equation describing quantum system evolution.
2.2.3 The Eigenvalue Problem
The Schrödinger equation can be separated into variables, isolating the spatial dependence from the time dependence, leading to an eigenvalue problem.
From the general Schrödinger equation:
$$
\begin{equation}
i\hbar\frac{\partial \Psi(r,t)}{\partial t} = \hat{H}\Psi(r,t)
\end{equation}
$$
For stationary states, the wavefunction can be separated:
$$
\Psi(r,t) = \psi(r)e^{-i\frac{E}{\hbar}t}
$$
Leading to the time-independent Schrödinger equation:
$$
\hat{H}\psi(r) = E\psi(r)
$$
2.2.4 General Solution Form
The general solution for time-dependent wavefunctions:
$$
\psi(x,t) = \sum_n c_n \psi_n(x)e^{ -iE_nt/ \hbar}
$$
where:
- $\psi_n(x)$ are energy eigenfunctions
- $E_n$ are corresponding eigenvalues
- $c_n$ are expansion coefficients determined by initial conditions
This quantization procedure establishes the fundamental framework of quantum mechanics where:
- Physical quantities are represented as operators
- States are described by wavefunctions
- Measured values are given by operator expectation values
Chapter 3: Wavefunctions and Expectation Values of Physical Quantities
1. Fundamental Properties of Wavefunctions
The wavefunction $\psi(x,t)$ is a complex-valued function that completely describes the state of a quantum system. It must satisfy the following properties:
- Single-valuedness: Uniquely defined at each point in space
- Continuity: Function and its derivatives are continuous
- Boundedness: Converges to zero at infinity
- Orthonormality: Different states are orthogonal
2. Mathematical Structure of Probability Interpretation
The probability density $|\psi(x,t)|^2$ has the following properties:
- Non-negativity:
$$|\psi(x,t)|^2 \geq 0$$ - Normalization condition:
$$\int_{-\infty}^{\infty} |\psi(x,t)|^2 dx = 1$$ - Conservation of probability current:
$$
\frac{\partial}{\partial t}|\psi|^2 + \frac{\partial}{\partial x}j = 0
$$
where $j$ is the probability current density:
$$
j = \frac{\hbar}{2mi}(\psi^* \frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^* }{\partial x})
$$
3. Calculation and Physical Meaning of Expectation Values
The expectation value of an observable $A$ is defined as:
$$
\langle A \rangle = \langle \psi|\hat{A}|\psi \rangle = \int_{-\infty}^{ \infty} \psi^*(x,t)\hat{A}\psi(x,t)dx
$$
Specific examples:
- Expectation value of position:
$$
\langle x \rangle = \int_{-\infty}^{\infty} x|\psi(x,t)|^2dx
$$ - Expectation value of momentum:
$$
\langle p \rangle = \int_{-\infty}^{ \infty} \psi^* (x,t)(-i\hbar \frac{\partial}{\partial x}) \psi(x,t)dx
$$ - Expectation value of energy:
$$
\langle E \rangle = \int_{-\infty}^{ \infty} \psi^* (x,t)(-\frac{\hbar^2 }{2m} \frac{\partial^2 }{\partial x^2 } + V(x))\psi(x,t)dx
$$
4. Quantitative Expression of Uncertainty
The standard deviation of an observable is defined as:
$$
\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}
$$
Using this, the uncertainty principle is expressed as:
$$
\Delta x \Delta p \geq \frac{\hbar}{2}
$$
This uncertainty is derived as a direct mathematical consequence of the probabilistic interpretation of the wavefunction.
5. Time Evolution and Measurement
The time evolution of the wavefunction follows the Schrödinger equation:
$$
i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi
$$
The measurement process is described as a "collapse" of the wavefunction:
- Before measurement: Superposition state
- After measurement: Projection onto eigenstate
Thus, the probabilistic interpretation of the wavefunction and calculation of expectation values provide the mathematical foundation for measurement theory in quantum mechanics.
Chapter 4: Relationship between Canonical Equations and the Schrödinger Equation
1. Starting from the Schrödinger Equation
The time-dependent Schrödinger equation describes the time evolution of the wavefunction ψ:
$$
i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi
$$
where the Hamiltonian generally takes the form:
$$
\hat{H} = -\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 } + V(x)
$$
2. Derivation of Ehrenfest's Theorem
The time evolution of the expectation value of any physical quantity $\hat{A}$ can be described using the commutator of the Hamiltonian and A.
The expectation value of physical quantity $\hat{A}$ is expressed as:
$$
\langle \hat{A} \rangle = \int \psi^* \hat{A} \psi dx
$$
Its time derivative becomes:
$$
\frac{d}{dt}\langle \hat{A} \rangle = \int \left(\frac{\partial \psi^* }{\partial t}\hat{A}\psi + \psi^* \hat{A}\frac{\partial \psi}{\partial t}\right)dx
$$
Substituting the Schrödinger equation and its complex conjugate:
$$
\frac{d}{dt}\langle \hat{A} \rangle = \frac{i}{\hbar}\int \psi^* [\hat{H},\hat{A}] \psi dx = \frac{i}{\hbar}\langle [\hat{H},\hat{A}] \rangle
$$
This is Ehrenfest's theorem.
The commutator of A and B is defined as:
$$
[A, B] := AB - BA
$$
where:
- A, B are operators
- AB is the product of operators A and B
- BA is the product of operators B and A
When [A, B] = 0, A and B commute, and the order of operations doesn't matter.
When [A, B] ≠ 0, A and B don't commute, and the order of operations is important.
3. Expectation Values of Position and Momentum
For the position operator $\hat{x}$:
$$
\frac{d}{dt}\langle \hat{x} \rangle = \frac{i}{\hbar}\langle [\hat{H},\hat{x}] \rangle = \frac{\langle \hat{p} \rangle}{m}
$$
For the momentum operator $\hat{p}$:
$$
\frac{d}{dt}\langle \hat{p} \rangle = \frac{i}{\hbar}\langle [\hat{H},\hat{p}] \rangle = -\left\langle \frac{dV}{dx} \right\rangle
$$
These equations give the equations of motion that the expectation values of position and momentum follow. Let's verify the derivation of these equations.
Given the Hamiltonian operator:
$$
\hat{H} = -\frac{\hbar^2 }{2m} \frac{\partial^2 }{\partial x^2 } + V(x)
$$
We calculate its commutator with the momentum operator $\hat{p}$:
$$
[\hat{H},\hat{p}] = \hat{H}\hat{p} - \hat{p}\hat{H}
$$
First term on right side:
$$
\hat{H}\hat{p} = (-\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 } + V(x))(-i\hbar\frac{\partial}{\partial x})
$$
$$
= (-\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 })(-i\hbar\frac{\partial}{\partial x}) + V(x)(-i\hbar\frac{\partial}{\partial x})
$$
$$
= i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3 } - i\hbar V(x)\frac{\partial}{\partial x}
$$
Second term on right side:
$$
\hat{p}\hat{H} = (-i\hbar\frac{\partial}{\partial x})(-\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 } + V(x))
$$
$$
= (-i\hbar\frac{\partial}{\partial x})(-\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 }) + (-i\hbar\frac{\partial}{\partial x})V(x)
$$
$$
= i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3 } - i\hbar\frac{\partial V(x)}{\partial x} - i\hbar V(x)\frac{\partial }{\partial x}
$$
The last term involves the product rule when considering operator action on wavefunctions $\hat{p}\hat{H}\psi$. Substituting these results into the commutator:
$$
[\hat{H},\hat{p}] = (i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3 } - i\hbar V(x)\frac{\partial}{\partial x}) - (i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3} - i\hbar\frac{\partial V(x)}{\partial x} - i\hbar V(x)\frac{\partial}{\partial x})
$$
Note that:
- The terms $i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3 }$ cancel
- The terms $-i\hbar V(x)\frac{\partial}{\partial x}$ cancel
- Only the term $i\hbar\frac{\partial V(x)}{\partial x}$ remains
Therefore:
$$
[\hat{H},\hat{p}] = (-i\hbar)\frac{\partial V(x)}{\partial x}
$$
This result shows that all contributions from the kinetic energy term cancel out, leaving only the derivative of the potential energy term. This is key to deriving the quantum version of the canonical equations.
4. Relationship between Canonical Equations and the Schrödinger Equation
This result appears as the expectation value version of the classical canonical equation:
$$
\frac{dp}{dt} = -\frac{\partial V}{\partial x}
$$
becoming:
$$
\frac{d}{dt}\langle \hat{p} \rangle = -\left\langle \frac{dV}{dx} \right\rangle
$$
Similarly, for the position operator:
$$
\frac{d}{dt}\langle \hat{x} \rangle = \frac{\langle \hat{p} \rangle}{m}
$$
corresponds to:
$$
\frac{dx}{dt} = \frac{p}{m} = \frac{\partial H}{\partial p}
$$
Thus, the equations of expectation values derived from Ehrenfest's theorem formally match the classical canonical equations. However, in quantum mechanics, these equations are for expectation values.
| Classical Canonical Equations | Corresponding Quantum Mechanical Expression |
|---|---|
| $$\frac{dx}{dt} = \frac{\partial H}{\partial p}= \frac{p}{m} $$ $$\frac{dp}{dt} = -\frac{\partial H}{\partial x}= -\frac{\partial V}{\partial x} $$ |
$$\frac{d}{dt}\langle \hat{x} \rangle = \frac{\langle \hat{p} \rangle}{m}$$ $$\frac{d}{dt}\langle \hat{p} \rangle = -\left\langle \frac{dV}{dx} \right\rangle$$ |
This correspondence is crucial for understanding the transition from quantum mechanics to classical mechanics. When the wave packet is sufficiently localized, the time evolution of expectation values approaches classical trajectories. This represents a concrete manifestation of the correspondence principle between quantum and classical mechanics.
Furthermore, this derivation demonstrates that the Schrödinger equation is the fundamental equation describing the time evolution of quantum systems, and the canonical equations are derived as a special case (time evolution of expectation values). This shows that quantum mechanics is a more general theory that encompasses classical mechanics.
5. Correspondence Between the Two
- Canonical equations directly describe the time evolution of position and momentum
- The Schrödinger equation indirectly describes the time evolution of physical quantities through the wavefunction
- Through Ehrenfest's theorem, the two correspond at the level of expectation values
- Classical mechanics emerges as a limit of quantum mechanics
Thus, canonical equations and the Schrödinger equation provide different descriptions of the same physical system and maintain a correspondence through expectation values.
Chapter 5: The Uncertainty Principle and the Essence of Quantum Mechanics
5.1 Measurement and Uncertainty
Heisenberg's uncertainty principle expresses the uncertainty in simultaneous measurements of position and momentum:
$$
\Delta x \Delta p \geq \frac{\hbar}{2}
$$
This relation is derived from the commutation relation between the position operator $\hat{x}$ and momentum operator $\hat{p}$:
$$
[\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar
$$
In general, for two physical quantities $A$ and $B$, the following relation holds:
$$
\Delta A \Delta B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|
$$
For the derivation of Heisenberg's uncertainty relation, we prepare:
- Schwarz inequality:
$$
\langle \hat{f}^\dagger \hat{f}\rangle \langle \hat{g}^\dagger \hat{g}\rangle \geq |\langle \hat{f}^\dagger \hat{g}\rangle|^2
$$ - Define operator deviations:
$$\Delta \hat{A} = \hat{A} - \langle \hat{A}\rangle$$
$$\Delta \hat{B} = \hat{B} - \langle \hat{B}\rangle$$
Now, let's derive the uncertainty relation starting from the Schwarz inequality:
-
First, substitute $\hat{f} = \Delta \hat{A}|\psi\rangle$, $\hat{g} = \Delta \hat{B}|\psi\rangle$ into the inequality:
$$
\langle\psi|(\Delta \hat{A})^2|\psi\rangle \langle\psi|(\Delta \hat{B})^2|\psi\rangle \geq |\langle\psi|\Delta \hat{A}\Delta \hat{B}|\psi\rangle|^2
$$
The left side is the product of standard deviations:
$(\Delta A)^2(\Delta B)^2$ -
Expand the expectation value on the right side:
$$
\langle\psi|\Delta \hat{A}\Delta \hat{B}|\psi\rangle = \langle\psi|(\hat{A} - \langle \hat{A}\rangle)(\hat{B} - \langle \hat{B}\rangle)|\psi\rangle
$$ -
For a complex number $z$, using $|z|^2 = (\Re z)^2 + (\Im z)^2 \geq (\Im z)^2$:
$$
|\langle\Delta \hat{A}\Delta \hat{B}\rangle|^2 \geq (\Im\langle\Delta \hat{A}\Delta \hat{B}\rangle)^2
$$ -
Looking at the commutation relation:
$$
\langle[\Delta \hat{A},\Delta \hat{B}]\rangle = \langle\Delta \hat{A}\Delta \hat{B}\rangle - \langle\Delta \hat{B}\Delta \hat{A}\rangle = 2i\Im\langle\Delta \hat{A}\Delta \hat{B}\rangle
$$
Therefore:
$$
\Im\langle\Delta \hat{A}\Delta \hat{B}\rangle = \frac{1}{2i}\langle[\Delta \hat{A},\Delta \hat{B}]\rangle
$$ -
Returning to the original inequality:
$$
(\Delta A)^2(\Delta B)^2 \geq |\langle\Delta \hat{A}\Delta \hat{B}\rangle|^2 \geq (\Im\langle\Delta \hat{A}\Delta \hat{B}\rangle)^2 = \frac{1}{4}|\langle[\Delta \hat{A},\Delta \hat{B}]\rangle|^2
$$ -
Finally, using $[\Delta \hat{A},\Delta \hat{B}] = [\hat{A},\hat{B}]$:
$$
\Delta A \Delta B \geq \frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|
$$ -
For position and momentum, since $[\hat{x},\hat{p}] = i\hbar$:
$$
\Delta x \Delta p \geq \frac{\hbar}{2}
$$
This is the general derivation of the uncertainty principle. Particularly for position and momentum, the lower bound is clearly determined because the commutation relation is a constant ($i\hbar$).
5.2 Superposition of States
From the linearity of the Schrödinger equation, superposition of states is possible:
$$\psi = c_1\psi_1 + c_2\psi_2$$
where the complex coefficients $c_1$, $c_2$ must satisfy the normalization condition:
$$ |c_1|^2 + |c_2|^2 = 1 $$
The expectation value of a physical quantity in a superposition state is:
$$
\langle A \rangle = \int \psi^* \hat{A} \psi dx
$$
$$
= |c_1|^2\langle \psi_1|\hat{A}|\psi_1\rangle + |c_2|^2\langle \psi_2|\hat{A}|\psi_2\rangle + (c_1^* c_2 \langle \psi_1|\hat{A}| \psi_2\rangle + c.c.)
$$
A particularly important example is the wavefunction in the double-slit experiment:
$$
\psi = \frac{1}{\sqrt{2}}(\psi_1 + \psi_2)
$$
where the interference term appears:
$$
|\psi|^2 = \frac{1}{2}(|\psi_1|^2 + |\psi_2|^2 + 2\Re(\psi_1^* \psi_2))
$$
This interference term represents the essence of quantum superposition.
Chapter 6: Solutions of the Schrödinger Equation and Physical Interpretation
6.1 Stationary States
Time-Independent Schrödinger Equation
When the Hamiltonian does not explicitly depend on time, we consider the following eigenvalue equation:
$$\hat{H}\psi_n(x) = E_n\psi_n(x)$$
The solutions $\psi_n(x)$ represent energy eigenstates.
Time Evolution
The time evolution of stationary states takes the form:
$$\psi_n(x,t) = \psi_n(x)e^{-iE_nt/\hbar}$$
The general solution is a superposition of these states:
$$\psi(x,t) = \sum_n c_n\psi_n(x)e^{-iE_nt/\hbar}$$
where coefficients $c_n$ are determined by initial conditions and satisfy the normalization condition:
$$\sum_n |c_n|^2 = 1$$
Physical Properties
The expectation value of an observable in a stationary state:
$$\langle A \rangle = \int \psi_n^*(x)\hat{A}\psi_n(x)dx$$
does not depend on time. In particular, the expectation value of energy:
$$\langle E \rangle = E_n$$
remains constant.
6.2 Probability Current and Conservation Laws
Probability Current Density
The probability current density in three dimensions:
$$
\mathbf{j} = \frac{\hbar}{2mi}(\psi^* \nabla\psi - \psi\nabla\psi^* )
$$
represents the density of particle flow.
Continuity Equation
The probability density $\rho = |\psi|^2$ and probability current density $\mathbf{j}$ satisfy the following continuity equation:
$$
\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{j} = 0
$$
This is derived directly from the Schrödinger equation:
-
Schrödinger equation:
$$
i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2 }{2m}\nabla^2 \psi + V\psi
$$ -
Its complex conjugate:
$$
-i\hbar\frac{\partial\psi^* }{\partial t} = -\frac{\hbar^2 }{2m} \nabla^2 \psi^* + V\psi^*
$$ -
These lead to the above continuity equation.
Physical Significance
- Local probability conservation:
- The time change of probability in any closed region is only due to flow through its boundaries
-
Total probability conservation:
$$
\frac{d}{dt}\int |\psi|^2 d^3 x = 0
$$ -
Time invariance of normalization condition:
$$
\int |\psi(x,t)|^2 d^3 x = 1 \quad \text{for all } t
$$
These relations ensure the mathematical consistency of the probabilistic interpretation of wavefunctions and support the fundamental framework of quantum mechanics.
Chapter 7: The Stern-Gerlach Experiment and Pauli Spin Matrices
In the Stern-Gerlach experiment, it was observed that an electron beam splits into two spots when passing through a non-uniform magnetic field. This is also the operating principle of quantum computers.
Overview of the Stern-Gerlach Experiment
The Stern-Gerlach experiment was conducted in 1922 by Otto Stern and Walter Gerlach in Germany. This experiment is known for first experimentally demonstrating the relationship between electron spin and magnetic moment.
The specific content of the experiment is as follows:
- Emit an electron beam from an electron gun
- Pass the electron beam through a magnetic field $\mathbf{B}$
- The electron beam splits into two in the magnetic field
The experimental apparatus consists of:
- Electron gun: Accelerates electrons to form an electron beam
- Magnetic field generator: Generates a non-uniform magnetic field for the electron beam to pass through
- Detection plate: Captures and detects the separated electron beams
When the electron beam was passed through a non-uniform magnetic field, it was observed to separate into two beams depending on the electron spin:
- Electrons with spin +1/2 are deflected upward in the magnetic field
- Electrons with spin -1/2 are deflected downward in the magnetic field
This result indicated that electrons possess an internal degree of freedom called spin, and the separation of the electron beam occurred due to the interaction between the magnetic moment arising from this spin and the magnetic field.
The Stern-Gerlach experiment is evaluated as a groundbreaking achievement that experimentally demonstrated the quantum mechanical concept of electron spin, and had a significant impact on the development of quantum mechanics.
Electron Spin States
In the Heisenberg picture, electron spin states are represented by two state vectors:
Spin-up state: $\left|\uparrow\right> = \begin{pmatrix} 1 \ 0 \end{pmatrix}$
Spin-down state: $\left|\downarrow\right> = \begin{pmatrix} 0 \ 1 \end{pmatrix}$
Pauli Spin Matrices
The Pauli spin matrices represent electron spin operators and are defined as:
$$
\hat{ \mathbf{ \sigma }} = \begin{pmatrix}
\hat{\sigma_x} & \hat{\sigma_y} & \hat{\sigma_z}
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}, \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
$$
Here, $\hat{\sigma}_x$, $\hat{\sigma}_y$, and $\hat{\sigma}_z$ represent the Pauli spin operators.
①State Immediately After Electron Emission
After being emitted from the electron gun, the electron is in an eigenstate of the Hamiltonian $\hat{H_0}$. If we denote this state vector as $|\psi\rangle$:
$$ \hat{H_0}| \psi \rangle = E| \psi \rangle $$
$$\hat{H_0} = \frac{\hat{\mathbf{p}}^2 }{2m_e} $$
where $E$ is the electron's energy, $\hat{\mathbf{p}}$ is the electron momentum operator, and $m_e$ is the electron mass. The state vector is a superposition of spin-up and spin-down states:
$$|\psi\rangle = a|\uparrow\rangle + b|\downarrow\rangle$$
where $a$ and $b$ are complex coefficients satisfying $|a|^2 + |b|^2 = 1$.
②State Immediately After Passing Through Non-uniform Magnetic Field
When the electron passes through a non-uniform magnetic field $\mathbf{B}= \begin{pmatrix} 0 \ 0 \ B_z \end{pmatrix}$, the Hamiltonian changes to $\hat{H} = \hat{H_0} - \hat{\mathbf{\mu}} \cdot \mathbf{B}$, where $\hat{\mathbf{\mu}} = -\mu_B \hat{\mathbf{\sigma}}$ is the spin magnetic moment and $\mu_B$ is the Bohr magneton.
The electron's kinetic energy in the non-uniform magnetic field is expressed as the interaction energy between the above magnetic moment and the magnetic field. The difference in this interaction energy causes the separation of electron beams observed in the Stern-Gerlach experiment.
Specifically, since the spin-up state |↑⟩ and spin-down state |↓⟩ have different interaction energies, they are deflected in different directions. Noting that each spin state has the following relationship as eigenstates of the Pauli spin matrix:
$$\hat{\sigma}_z|\uparrow\rangle = +1|\uparrow \rangle$$
$$\hat{\sigma}_z|\downarrow\rangle = -1|\downarrow \rangle$$
The eigenstates of the Hamiltonian $\hat{H}$ in the magnetic field can be calculated for spin-up and spin-down states:
$$\hat{H}|\uparrow\rangle = E|\uparrow \rangle + \mu_B B_z \hat{\sigma_z} |\uparrow \rangle = (E + \mu_B B_z)| \uparrow\rangle $$
$$\hat{H}|\downarrow \rangle = E|\downarrow \rangle + \mu_B B_z \hat{\sigma_z} |\downarrow \rangle = (E - \mu_B B_z)|\downarrow\rangle$$
In other words, the eigenvalue of the Hamiltonian for the spin-up state $|\uparrow\rangle$ becomes $(E + \mu_B B_z)$, where $E$ is the energy of the free electron.
Let's explain how the initial state $|\psi\rangle$ changes as it passes through the magnetic field.
Passing through the magnetic field means that the Hamiltonian changes to $\hat{H} = \hat{H_0} - \hat{\mathbf{\mu}} \cdot \mathbf{B}$, where $\hat{\mathbf{\mu}} = -\mu_B \hat{\mathbf{\sigma}}$ is the spin magnetic moment.
The eigenstates of this Hamiltonian $\hat{H}$ in the magnetic field separate into spin-up and spin-down states $|\uparrow\rangle$, $|\downarrow\rangle$:
$$\hat{H}|\uparrow\rangle = (E + \mu_B B_z)|\uparrow\rangle$$
$$\hat{H}|\downarrow\rangle = (E - \mu_B B_z)|\downarrow\rangle$$
The time evolution operator for this Hamiltonian in the magnetic field is:
$$U(t) = e^{-i\frac{\hat{H}}{\hbar}t}$$
The state $|\psi'\rangle$ after the initial state $|\psi\rangle$ passes through the magnetic field can be written as:
$$|\psi'\rangle = U(t)|\psi\rangle$$
Calculating explicitly, the state vector after passing through the magnetic field becomes:
$$|\psi'\rangle = e^{-i\frac{E+\mu_B B_z}{\hbar}t}a|\uparrow\rangle + e^{-i\frac{E-\mu_B B_z}{\hbar}t}b|\downarrow\rangle$$
In other words, passing through the magnetic field causes the initial state $|\psi\rangle$ to change to state $|\psi'\rangle$ where each spin component has acquired a phase factor $e^{-i\frac{(E\pm\mu_B B)t}{\hbar}}$.
In this calculation process, we obtain the final state $|\psi'\rangle$ by deriving the time evolution operator $U(t)$ using the eigenstates and eigenvalues of the Hamiltonian, and applying it to the initial state $|\psi\rangle$.
③Measurement Results from Observation
When observing the electron position on the detection plate, two spots corresponding to spin-up and spin-down states are obtained.
Noting that the inner product with the spin-up state $|\uparrow\rangle$ is the complex number $\langle\uparrow|\psi'\rangle = a$, the observation probabilities are given by the projection of the state vector $|\psi'\rangle$:
- Probability of detecting spin-up: $P_\uparrow = |\langle\uparrow|\psi'\rangle|^2 = |a|^2$
- Probability of detecting spin-down: $P_\downarrow = |\langle\downarrow|\psi'\rangle|^2 = |b|^2$
These probabilities match the observed separation pattern of the electron beam. We can explain the Stern-Gerlach experiment results quantum mechanically by describing the interaction with the magnetic field through changes in the Hamiltonian and representing the electron state using Pauli spin matrices.
Case When Electron Passes Through a Different Magnetic Field
Note that the time-dependent terms vanish when taking the absolute value. In the end, whether the electron passes through the magnetic field or not, the probability distribution remains unchanged.
Let's now consider the case where an electron passes through a non-uniform magnetic field $\mathbf{B}= \begin{pmatrix} B_x \ 0 \ 0 \end{pmatrix}$.
Let the initial state be $|\psi\rangle = a|\uparrow\rangle + b|\downarrow\rangle$.
When this state passes through the non-uniform magnetic field $\mathbf{B} = \begin{pmatrix} B_x \ 0 \ 0 \end{pmatrix}$, the state vector $|\psi'\rangle$ can be expressed using the Hamiltonian's time evolution operator $U(t)$ as:
$$|\psi'\rangle = U(t)|\psi\rangle$$
Here, the Hamiltonian becomes $\hat{H} = \hat{H}_0 - \hat{\mathbf{\mu}} \cdot \mathbf{B}$. The spin magnetic moment is expressed as $\hat{\mathbf{\mu}} = -\mu_B \hat{\mathbf{\sigma}}$.
For the magnetic field $\mathbf{B} = \begin{pmatrix} B_x \ 0 \ 0 \end{pmatrix}$, the Hamiltonian can be written as:
$$\hat{H} = \hat{H_0} - \mu_B B_x \hat{\sigma_x}$$
Here, $\hat{\sigma}_x$ is the x-component of the Pauli spin operator.
The time evolution operator $U(t)$ for this $\hat{H}$ is:
$$U(t) = e^{-i\frac{\hat{H}t}{\hbar}} = e^{ -i\frac{\hat{H}_0 t}{\hbar}} e^{ i\frac{\mu_B B_x t}{\hbar} \hat{\sigma}_x} $$
Applying this to the initial state $|\psi\rangle$:
$$|\psi'\rangle = e^{ -i\frac{E}{\hbar}t} e^{i\frac{\mu_B B_x t}{\hbar}\hat{\sigma}_x}(a|\uparrow\rangle + b|\downarrow\rangle)$$
Considering that $\hat{\sigma}_x |\uparrow\rangle = |\downarrow \rangle$, $\hat{\sigma}_x |\downarrow\rangle = |\uparrow\rangle$:
$$
|\psi'\rangle = a e^{i\frac{-E +\mu_B B_x }{\hbar}t}|\downarrow\rangle + b e^{-i\frac{E+ \mu_B B_x }{\hbar}t}|\uparrow\rangle
$$
Finally, the detection probabilities for spin-up state $|\uparrow\rangle$ and spin-down state $|\downarrow\rangle$ become:
$$P_\uparrow = |b|^2$$
$$P_\downarrow = |a|^2$$
This shows that when passing through a non-uniform magnetic field in the x direction, the up and down spin components of the initial state are detected in reversed form.
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