Chapter 1 From Classical Mechanics
1.1 Fundamentals of Newtonian Mechanics
Introduction of the Equation of Motion
Newton’s second law, as the 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 $\mathbf{r}$ is the position vector, $m$ is the mass, and $t$ denotes time.
For one-dimensional motion, this is simplified as:
$$
F = m \frac{d^2x }{dt^2 }
$$
Properties of the Equation of Motion
This differential equation is a second-order ordinary differential equation, and a unique solution is determined if the following initial conditions are given:
- Initial position: $x(t_0) = x_0$
- Initial velocity: $\dot{x}(t_0) = v_0$
Physically, this means that if the initial state is given, the subsequent motion is completely determined.
1.2 Derivation of the Law of Energy Conservation
1. Equation of Motion in a Potential Field
The equation of motion of a particle in a conservative force field can be written as a function of position only:
$$
m\frac{d^2x }{dt^2 } = F(x) = -\frac{dV}{dx}
$$
Here $V(x)$ is the potential energy.
2. Transformation of the Equation of Motion
Multiplying both sides by the velocity $dx/dt$:
$$
m\frac{d^2x }{dt^2 }\frac{dx}{dt} = -\frac{dV}{dx} \frac{dx}{dt}
$$
The left-hand 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-hand side is the time derivative of potential energy:
$$
-\frac{dV}{dx}\frac{dx}{dt} = -\frac{dV}{dt}
$$
3. Energy as an Antiderivative
Thus 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 does not depend on time:
$$
E = \frac{1}{2}m(\frac{dx}{dt})^2 + V(x) = \text{constant}
$$
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
The constant is determined by the initial conditions:
- At $t=0$, the initial position $x_0$ and initial velocity $v_0$ give
- $E = \frac{1}{2}mv_0^2 + V(x_0)$
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 following canonical equations are derived from the Hamiltonian:
$$
\frac{dx}{dt} = \frac{\partial H}{\partial p} = \frac{p}{m}
$$
$$
\frac{dp}{dt} = -\frac{\partial H}{\partial x} = -\frac{dV}{dx}
$$
The canonical equations provide a system of differential equations in two variables, $(x,p)$, position and momentum.
While Newtonian mechanics started with equations of motion, in modern formalism, it is often more insightful to start with canonical equations to describe various physical phenomena.
7. Derivation of Energy Conservation
The total time derivative of the Hamiltonian is transformed using the canonical equations above:
$$
\frac{dH}{dt} = \frac{\partial H}{\partial x} \frac{dx}{dt} + \frac{\partial H}{\partial p}\frac{dp}{dt} = 0
$$
This shows that the Hamiltonian does not change with time, representing the law of energy conservation.
Physical Significance
The importance of the Hamiltonian formalism:
- Completely describes the time evolution of the system
- Clearly expresses the symmetry of the canonical variables $(x,p)$
- Provides the foundation for the transition to quantum mechanics
This Hamiltonian formalism plays an essential role as a bridge to quantum mechanics.
Chapter 2 The Quantization Procedure
2.1 Correspondence from Classical to Quantum Mechanics
2.1.1 Basic Idea
In moving from classical to quantum mechanics, physical quantities are expressed not as numbers but as corresponding operators, which simplifies mathematical treatment.
This transformation should be regarded as a mathematical operation only; one must not interpret it as meaning that the essence of physical quantities or matter was actually operators.
2.1.2 Basic Correspondence
The correspondence between main physical quantities and operators is given as:
Position:
$$
x \rightarrow \hat{x} \quad (\text{operation of multiplying the wavefunction by } x)
$$
Momentum:
$$
p \rightarrow \hat{p} = -i\hbar\frac{d}{dx} \quad (\text{operation of differentiating the wavefunction})
$$
This correspondence is based on de Broglie’s relation:
$$
p = \hbar k
$$
Here $k$ is the wave vector, and from its relation with the differential operator $-i\frac{d}{dx}$ acting on the plane wave $e^{ikx}$, the form of the momentum operator is derived.
2.1.3 Derivation of the Momentum Operator
To express the duality of particles and waves, de Broglie proposed the following relation (matter waves):
$$
\lambda = \frac{h}{p}
$$
Here:
- $\lambda$: wavelength of matter wave
- $h$: Planck’s constant
- $p$: momentum of the particle
This equation shows the relationship between the momentum of a particle as a point mass and its wave-like property.
The wave number $k$ is related to wavelength $\lambda$ as:
$$
k = \frac{2\pi}{\lambda}
$$
Combining this with the de Broglie relation:
$$
p = \hbar k ,
(\hbar = \frac{h}{2\pi})
$$
Thus, we obtain this relation.
Now, the wavefunction of a free particle is expressed as a plane wave:
$$
\psi(x) = A e^{ikx}
$$
Here $A$ 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 $p$.
Here “corresponds” means correspondence in operation; one must not interpret it as “momentum, an observable reality, 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 the Momentum Operator
Thus, the momentum operator for a general wavefunction is defined as:
$$
\hat{p} = -i\hbar\frac{d}{dx}
$$
This operator satisfies:
- Hermiticity: $\hat{p}^\dagger = \hat{p}$
- Reality of eigenvalues
- Consistency with momentum conservation
Action on a Wavefunction
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:
$$
\langle p \rangle = \int \psi^ * (x)(-i \hbar \frac{d}{dx}) \psi(x) dx
$$
This corresponds to the average momentum observed in experiments.
Thus, the form of the momentum operator is naturally derived through:
- The concept of de Broglie’s matter wave
- Concrete calculations with plane waves
- Extension to general wavefunctions
2.2 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 Schrödinger Equation
Time-dependent Schrödinger equation:
$$
i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi
$$
This is the fundamental equation describing the time evolution of quantum systems.
2.2.3 Eigenvalue Problem
Separating variables in the Schrödinger equation to remove the time-dependent part, we reduce it to an eigenvalue problem.
From the ordinary Schrödinger equation, the time-independent Schrödinger equation is derived.
The time-dependent Schrödinger equation is expressed as:
$$
\begin{equation}
i\hbar\frac{\partial \Psi(r,t)}{\partial t} = \hat{H}\Psi(r,t)
\end{equation}
$$
Here, $\Psi(r,t)$ is the wavefunction, and $\hat{H}$ is the Hamiltonian operator.
Next, consider a stationary state. In a stationary state, the wavefunction $\Psi(r,t)$ does not depend on time and can be expressed by separation of variables as:
\begin{equation}
\Psi(r,t) = \psi(r)e^{-i\frac{E}{\hbar}t}
\end{equation}
Here, $\psi(r)$ is a function only of the spatial coordinate $r$, and $E$ denotes the energy.
Then the left-hand side of the Schrödinger equation transforms as:
$$
\begin{align}
i\hbar\frac{\partial \Psi(r,t)}{\partial t} &= i\hbar\frac{\partial}{\partial t}\left(\psi(r)e^{-i\frac{E}{\hbar}t}\right) \
&= E\psi(r)e^{-i\frac{E}{\hbar}t}
\end{align}
$$
Thus the Schrödinger equation is rewritten as:
$$
E\psi(r)e^{-i\frac{E}{\hbar}t} = \hat{H}\psi(r)e^{-i\frac{E}{\hbar}t}
$$
$$
\left( \hat{H}\psi(r) - E\psi(r) \right) e^{-i\frac{E}{\hbar}t} = 0
$$
For this equation to hold at all times, the term in parentheses must vanish, yielding the following eigenvalue problem:
\begin{equation}
\hat{H}\psi(r) = E\psi(r)
\end{equation}
2.2.4 General Form of the Solution
General solution of the time-dependent wavefunction:
$$
\psi(x,t) = \sum_n c_n \psi_n(x)e^{ -iE_nt/ \hbar}
$$
Here:
- $\psi_n(x)$: energy eigenfunctions
- $E_n$: corresponding eigenvalues
- $c_n$: expansion coefficients determined by initial conditions
Through this quantization procedure:
- Physical quantities are expressed as operators
- States are described as wavefunctions
- Measured values are given as expectation values of operators
This constructs the fundamental framework of quantum mechanics.
Chapter 3 Wavefunctions and Expectation Values of Physical Quantities
1. Fundamental Properties of the Wavefunction
The wavefunction $\psi(x,t)$ is a complex-valued function that completely describes the state of a quantum system.
This wavefunction must satisfy the following properties:
- Single-valuedness: uniquely defined at each point in space
- Continuity: the function and its derivatives are continuous
- Boundedness: converges to 0 at infinity
- Orthonormality: different states are orthogonal
2. Mathematical Structure of the Probabilistic 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:
$$
\frac{\partial}{\partial t}|\psi|^2 + \frac{\partial}{\partial x}j = 0
$$
Here $j$ is the probability current density:
$$
j = \frac{\hbar}{2mi}(\psi^* \frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^* }{\partial x})
$$
3. Expectation Values and Their Physical Meaning
The expectation value of an observable $A$ is defined as:
$$
\langle A \rangle = \langle \psi|\hat{A}|\psi \rangle = \int_{-\infty}^{ \infty} \psi^*(x,t)\hat{A}\psi(x,t)dx
$$
Concrete 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 relation is expressed as:
$$
\Delta x \Delta p \geq \frac{\hbar}{2}
$$
This uncertainty arises 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 an eigenstate
Thus, the probabilistic interpretation of the wavefunction and the calculation of expectation values provide the mathematical foundation of 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 $\psi$ as:
$$
i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi
$$
Here 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 an arbitrary observable $\hat{A}$ can be expressed using the expectation value of the commutator of the Hamiltonian and $A$.
The expectation value of $\hat{A}$ is given as:
$$
\langle \hat{A} \rangle = \int \psi^* \hat{A} \psi dx
$$
Its time derivative is:
$$
\frac{d}{dt}\langle \hat{A} \rangle = \int \left(\frac{\partial \psi^* }{\partial t}\hat{A}\psi + \psi^* \hat{A}\frac{\partial \psi}{\partial t}\right)dx
$$
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.
Here the commutator of $A$ and $B$ is defined as:
$$
[A, B] := AB - BA
$$
where:
- $A, B$ are operators
- $AB$ is the product of operator $A$ and $B$
- $BA$ is the product of operator $B$ and $A$
If $[A, B] = 0$, then $A$ and $B$ commute, meaning the order of operators does not matter.
If $[A, B] \neq 0$, then $A$ and $B$ are non-commutative, and operator order is essential.
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 provide the equations of motion for the expectation values of position and momentum.
This represents the canonical equations, but first, let’s confirm how these are derived.
The Hamiltonian operator is given as:
$$
\hat{H} = -\frac{\hbar^2 }{2m} \frac{\partial^2 }{\partial x^2 } + V(x)
$$
Now compute the commutator with the momentum operator $\hat{p}$:
$$
[\hat{H},\hat{p}] = \hat{H}\hat{p} - \hat{p}\hat{H}
$$
First term on the right-hand 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 the right-hand side:
$$
\hat{p}\hat{H} = (-i\hbar\frac{\partial}{\partial x})(-\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 } + V(x))
$$
$$
= (-i\hbar\frac{\partial}{\partial x})(-\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial x^2 }) + (-i\hbar\frac{\partial}{\partial x})V(x)
$$
$$
= i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3 } - i\hbar\frac{\partial V(x)}{\partial x} - i\hbar V(x)\frac{\partial }{\partial x}
$$
In the last term, the product rule is applied when the operator acts on a wavefunction as in $\hat{p}\hat{H}\psi$.
Substituting into the commutator relation
$$
[\hat{H},\hat{p}] = \hat{H} \hat{p} - \hat{p} \hat{H}
$$
gives:
$$
[\hat{H},\hat{p}] = (i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3 } - i\hbar V(x)\frac{\partial}{\partial x}) - (i\frac{\hbar^3 }{2m}\frac{\partial^3 }{\partial x^3} - i\hbar\frac{\partial V(x)}{\partial x} - i\hbar V(x)\frac{\partial}{\partial x})
$$
Now note:
- 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
Thus:
$$
[\hat{H},\hat{p}] = (-i\hbar)\frac{\partial V(x)}{\partial x}
$$
This result shows that contributions from the kinetic energy term cancel out, and only the derivative of the potential energy term remains. This is the key to deriving the quantum version of the canonical equations.
4. Relationship Between Canonical Equations and the Schrödinger Equation
This result corresponds to the classical canonical equation:
$$
\frac{dp}{dt} = -\frac{\partial V}{\partial x}
$$
in expectation value form:
$$
\frac{d}{dt}\langle \hat{p} \rangle = -\left\langle \frac{dV}{dx} \right\rangle
$$
Similarly, for the position operator:
$$
\frac{d}{dt}\langle \hat{x} \rangle = \frac{\langle \hat{p} \rangle}{m}
$$
which corresponds to:
$$
\frac{dx}{dt} = \frac{p}{m} = \frac{\partial H}{\partial p}
$$
Thus, the equations for expectation values obtained from Ehrenfest’s theorem formally coincide with 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 important 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 is a concrete manifestation of the correspondence principle between quantum and classical mechanics.
Moreover, this derivation shows 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 expresses that quantum mechanics is a more general theory encompassing classical mechanics.
5. Correspondence Between the Two
- Canonical equations describe the direct time evolution of position and momentum
- The Schrödinger equation indirectly describes the time evolution of physical quantities through the wavefunction
- Ehrenfest’s theorem provides the correspondence between the two at the level of expectation values
- Classical mechanics emerges as the limiting case of quantum mechanics
Thus, the canonical equations and the Schrödinger equation provide different descriptions of the same physical system, and they correspond through expectation values.
Chapter 5 The Uncertainty Principle and the Essence of Quantum Mechanics
5.1 Measurement and Uncertainty
Heisenberg’s uncertainty relation expresses the impossibility of simultaneously measuring position and momentum with arbitrary precision:
$$
\Delta x \Delta p \geq \frac{\hbar}{2}
$$
This relation is derived from the commutation relation between the position operator $\hat{x}$ and the momentum operator $\hat{p}$:
$$
[\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar
$$
In general, for two observables $A$ and $B$, the following relation holds:
$$
\Delta A \Delta B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|
$$
To derive Heisenberg’s uncertainty relation, we prepare the following.
-
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, starting from the Schwarz inequality, we derive the uncertainty relation.
- Substituting $\hat{f} = \Delta \hat{A}|\psi\rangle$, $\hat{g} = \Delta \hat{B}|\psi\rangle$ 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-hand side is the product of variances:
$(\Delta A)^2(\Delta B)^2$
-
Expanding the expectation value on the right-hand 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 $z$, since $|z|^2 = (\Re z)^2 + (\Im z)^2 \geq (\Im z)^2$:
$$
|\langle\Delta \hat{A}\Delta \hat{B}\rangle|^2 \geq (\Im\langle\Delta \hat{A}\Delta \hat{B}\rangle)^2
$$
- Focusing on the commutator:
$$
\langle[\Delta \hat{A},\Delta \hat{B}]\rangle = \langle\Delta \hat{A}\Delta \hat{B}\rangle - \langle\Delta \hat{B}\Delta \hat{A}\rangle = 2i\Im\langle\Delta \hat{A}\Delta \hat{B}\rangle
$$
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.
In particular, for position and momentum, the commutator is a constant ($i\hbar$), which sets a clear lower bound.
5.2 Superposition of States
From the linearity of the Schrödinger equation, superposition of states is possible:
$$\psi = c_1\psi_1 + c_2\psi_2$$
Here, 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 observable in a superposition state is:
$$
\langle A \rangle = \int \psi^* \hat{A} \psi dx
$$
$$
= |c_1|^2\langle \psi_1|\hat{A}|\psi_1\rangle + |c_2|^2\langle \psi_2|\hat{A}|\psi_2\rangle + (c_1^* c_2 \langle \psi_1|\hat{A}| \psi_2\rangle + c.c.)
$$
As a particularly important example, the wavefunction in the double-slit experiment is:
$$
\psi = \frac{1}{\sqrt{2}}(\psi_1 + \psi_2)
$$
Here, an 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 solution $\psi_n(x)$ represents an energy eigenstate.
Time Evolution
The time evolution of a stationary state takes the form:
$$\psi_n(x,t) = \psi_n(x)e^{-iE_nt/\hbar}$$
The general solution is expressed as a superposition:
$$\psi(x,t) = \sum_n c_n\psi_n(x)e^{-iE_nt/\hbar}$$
Here, the 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:
$$\langle A \rangle = \int \psi_n^*(x)\hat{A}\psi_n(x)dx$$
is independent of time. In particular, the expectation value of energy:
$$\langle E \rangle = E_n$$
remains constant.
6.2 Probability Current and Conservation Law
Probability Current Density
In three dimensions, the probability current density is given by:
$$
\mathbf{j} = \frac{\hbar}{2mi}(\psi^* \nabla\psi - \psi\nabla\psi^* )
$$
This represents the density of particle flow.
Continuity Equation
The probability density $\rho = |\psi|^2$ and the probability current density $\mathbf{j}$ satisfy the continuity equation:
$$
\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{j} = 0
$$
This follows 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:
$$
-i\hbar\frac{\partial\psi^* }{\partial t} = -\frac{\hbar^2 }{2m} \nabla^2 \psi^* + V\psi^*
$$ -
From these, the continuity equation is derived.
Physical Significance
-
Local conservation of probability:
The time variation of probability in any closed region arises only through inflow and outflow across the boundary. -
Conservation of total probability:
$$
\frac{d}{dt}\int |\psi|^2 d^3 x = 0
$$ -
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 the wavefunction and underpin 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 passing through a non-uniform magnetic field separates into two spots. This is also a principle behind quantum computing.
Overview of the Stern–Gerlach Experiment
The Stern–Gerlach experiment was conducted in 1922 by German physicists Otto Stern and Walther Gerlach.
It is known as the first experimental demonstration of the relationship between electron spin and magnetic moment.
The experiment proceeds as follows:
- Emit an electron beam from an electron gun
- Pass the electron beam through a magnetic field $\mathbf{B}$
- Inside the magnetic field, the beam splits into two
The apparatus consists of:
- Electron gun: accelerates electrons to form a beam
- Magnet assembly: generates a non-uniform magnetic field through which the beam passes
- Detector screen: detects the separated electron beams
When the electron beam passes through the non-uniform magnetic field, two beams are observed depending on the electron spin:
- Electrons with spin $+1/2$ are deflected upward
- Electrons with spin $-1/2$ are deflected downward
This result shows that electrons possess an internal degree of freedom called spin, and that the magnetic moment arising from spin interacts with the magnetic field to cause beam separation.
The Stern–Gerlach experiment was a groundbreaking demonstration of the quantum mechanical concept of spin and greatly influenced the development of quantum mechanics.
Electron Spin States
In the Heisenberg picture, the electron spin states are represented by the following two state vectors:
Up-spin state: $\left|\uparrow\right> = \begin{pmatrix} 1 \ 0 \end{pmatrix}$
Down-spin state: $\left|\downarrow\right> = \begin{pmatrix} 0 \ 1 \end{pmatrix}$
Pauli Spin Matrices
The Pauli spin matrices represent spin operators of the electron 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 $\hat{\sigma}_z$ are the Pauli spin operators.
(1) State Immediately After Emission
Electrons emitted from the electron gun are in eigenstates of the Hamiltonian $\hat{H_0}$.
If the state vector is $|\psi\rangle$, then:
$$ \hat{H_0}| \psi \rangle = E| \psi \rangle $$
$$\hat{H_0} = \frac{\hat{\mathbf{p}}^2 }{2m_e} $$
Here $E$ is the electron’s energy, $\hat{\mathbf{p}}$ is the momentum operator, and $m_e$ is the electron mass.
The state vector is a superposition of up-spin and down-spin states:
$$|\psi\rangle = a|\uparrow\rangle + b|\downarrow\rangle$$
Here, $a$ and $b$ are complex coefficients satisfying $|a|^2 + |b|^2 = 1$.
(2) State Just After Passing Through the Non-Uniform Magnetic Field
When an electron passes through a non-uniform magnetic field $\mathbf{B}= \begin{pmatrix} 0 \ 0 \ B_z \end{pmatrix}$, the Hamiltonian changes to $\hat{H} = \hat{H_0} - \hat{\mathbf{\mu}} \cdot \mathbf{B}$.
Here $\hat{\mathbf{\mu}} = -\mu_B \hat{\mathbf{\sigma}}$ is the spin magnetic moment, and $\mu_B$ is the Bohr magneton.
In the non-uniform magnetic field, the electron’s energy is expressed as the interaction energy between the magnetic moment and the magnetic field.
This energy difference causes the separation observed in the Stern–Gerlach experiment.
Specifically:
- The up-spin state $|\uparrow\rangle$ is deflected upward
- The down-spin state $|\downarrow\rangle$ is deflected downward
Each spin state corresponds to an eigenstate of the Pauli spin matrix:
$$\hat{\sigma}_z|\uparrow\rangle = +1|\uparrow \rangle$$
$$\hat{\sigma}_z|\downarrow\rangle = -1|\downarrow \rangle$$
Thus, the Hamiltonian eigenvalues in the magnetic field are:
$$\hat{H}|\uparrow\rangle = (E + \mu_B B_z)| \uparrow\rangle $$
$$\hat{H}|\downarrow \rangle = (E - \mu_B B_z)|\downarrow\rangle$$
This means that the eigenvalue of the Hamiltonian for the up-spin state is $(E + \mu_B B_z)$.
The time evolution operator in the magnetic field is:
$$U(t) = e^{-i\frac{\hat{H}}{\hbar}t}$$
The state after passing through the magnetic field is:
$$|\psi'\rangle = U(t)|\psi\rangle$$
Explicitly:
$$|\psi'\rangle = e^{-i\frac{E+\mu_B B_z}{\hbar}t}a|\uparrow\rangle + e^{-i\frac{E-\mu_B B_z}{\hbar}t}b|\downarrow\rangle$$
Thus, passing through the magnetic field attaches phase factors $e^{-i\frac{(E\pm\mu_B B)t}{\hbar}}$ to each spin component of the initial state.
(3) Measurement Results by Observation
On the detector screen, two spots corresponding to up-spin and down-spin states appear.
The detection probabilities are given by projections of $|\psi'\rangle$:
- Probability of detecting up-spin: $P_\uparrow = |\langle\uparrow|\psi'\rangle|^2 = |a|^2$
- Probability of detecting down-spin: $P_\downarrow = |\langle\downarrow|\psi'\rangle|^2 = |b|^2$
These probabilities match the experimental observation of beam separation.
By describing the interaction with the magnetic field using the Hamiltonian and expressing electron states with Pauli matrices, the Stern–Gerlach results are explained quantum mechanically.
Case of an Electron Passing Through Another Magnetic Field
Note that the absolute values remove time-dependent terms. Thus, whether or not the electron passes through the magnetic field, the probability distribution of outcomes remains unchanged.
Now consider the case where the electron passes through a non-uniform magnetic field $\mathbf{B}= \begin{pmatrix} B_x \ 0 \ 0 \end{pmatrix}$.
Let the initial state be $|\psi\rangle = a|\uparrow\rangle + b|\downarrow\rangle$.
The state vector after passing through the magnetic field is:
$$|\psi'\rangle = U(t)|\psi\rangle$$
with Hamiltonian $\hat{H} = \hat{H}_0 - \hat{\mathbf{\mu}} \cdot \mathbf{B}$, where $\hat{\mathbf{\mu}} = -\mu_B \hat{\mathbf{\sigma}}$.
For $\mathbf{B} = \begin{pmatrix} B_x \ 0 \ 0 \end{pmatrix}$, the Hamiltonian becomes:
$$\hat{H} = \hat{H_0} - \mu_B B_x \hat{\sigma_x}$$
The corresponding time evolution operator is:
$$U(t) = e^{-i\frac{\hat{H}t}{\hbar}} = e^{ -i\frac{\hat{H}_0 t}{\hbar}} e^{ i\frac{\mu_B B_x t}{\hbar} \hat{\sigma}_x} $$
Acting on the initial state:
$$|\psi'\rangle = e^{ -i\frac{E}{\hbar}t} e^{i\frac{\mu_B B_x t}{\hbar}\hat{\sigma}_x}(a|\uparrow\rangle + b|\downarrow\rangle)$$
Considering $\hat{\sigma}_x|\uparrow\rangle = |\downarrow\rangle$, $\hat{\sigma}_x|\downarrow\rangle = |\uparrow\rangle$:
$$
|\psi'\rangle = a e^{i\frac{-E +\mu_B B_x }{\hbar}t}|\downarrow\rangle + b e^{-i\frac{E+ \mu_B B_x }{\hbar}t}|\uparrow\rangle
$$
Finally, the detection probabilities for up-spin and down-spin states are:
$$P_\uparrow = |b|^2$$
$$P_\downarrow = |a|^2$$
Thus, passing through a non-uniform magnetic field oriented along the x-axis results in the swapping of detection probabilities between up-spin and down-spin components of the initial state.
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