Quantum Mechanics Note Date: 2024/05/27
Last Updated: 2024-06-05T11:18:36.168Z
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A random variable X X X Ω \Omega Ω R \mathbb{R} R
The probability density function f ( x ) f(x) f ( x ) X X X
The cumulative distribution function F ( x ) F(x) F ( x ) X X X x x x
F ( x ) = P ( X ≤ x ) = ∫ − ∞ x f ( x ) d x . F(x) = P(X \leq x) = \int_{-\infty}^x f(x) dx. F ( x ) = P ( X ≤ x ) = ∫ − ∞ x f ( x ) d x . The expectation of a random variable X X X
< X > = ∫ − ∞ ∞ x f ( x ) d x . <X> = \int_{-\infty}^{\infty} x f(x) dx. < X >= ∫ − ∞ ∞ x f ( x ) d x . The variance of a random variable X X X
Δ X 2 = < ( X − < X > ) 2 > = < X 2 > − < X > 2 . \Delta X^2 = <(X - <X>)^2> = <X^2> - <X>^2. Δ X 2 =< ( X − < X > ) 2 >=< X 2 > − < X > 2 . The standard deviation of a random variable X X X
Δ X = Δ X 2 . \Delta X = \sqrt{\Delta X^2}. Δ X = Δ X 2 . The probability amplitude ψ ( x ) \psi(x) ψ ( x ) X X X x x x ∣ ψ ( x ) ∣ 2 |\psi(x)|^2 ∣ ψ ( x ) ∣ 2
In other words, the probability density function f ( x ) f(x) f ( x )
f ( x ) = ∣ ψ ( x ) ∣ 2 . f(x) = |\psi(x)|^2. f ( x ) = ∣ ψ ( x ) ∣ 2 . The Gaussian integral is given by
∫ − ∞ ∞ e − x 2 σ 2 d x = π σ . \int_{-\infty}^{\infty} e^{-\frac{x^2}{\sigma^2}} dx = \sqrt{\pi}\sigma. ∫ − ∞ ∞ e − σ 2 x 2 d x = π σ . The Gaussian integral with odd powers of x x x
∫ − ∞ ∞ x 2 n + 1 e − x 2 σ 2 d x = 0. \int_{-\infty}^{\infty} x^{2n+1} e^{-\frac{x^2}{\sigma^2}} dx = 0. ∫ − ∞ ∞ x 2 n + 1 e − σ 2 x 2 d x = 0. As this is an odd function, the integral is zero.
The even powers of x x x
∫ − ∞ ∞ x 2 n e − a x 2 d x = ∫ − ∞ ∞ ( − ∂ ∂ a ) n e − a x 2 = ( − ∂ ∂ a ) n ∫ − ∞ ∞ e − a x 2 d x = ( − ∂ ∂ a ) n π a \begin{align}
\int_{-\infty}^{\infty} x^{2n} e^{-ax^2} dx &=
\int_{-\infty}^{\infty} \left(-\frac{\partial}{\partial a}\right)^n e^{-ax^2} \\
&= \left(-\frac{\partial}{\partial a}\right)^n \int_{-\infty}^{\infty} e^{-ax^2} dx \\
&= \left(-\frac{\partial}{\partial a}\right)^n \sqrt{\frac{\pi}{a}}
\end{align} ∫ − ∞ ∞ x 2 n e − a x 2 d x = ∫ − ∞ ∞ ( − ∂ a ∂ ) n e − a x 2 = ( − ∂ a ∂ ) n ∫ − ∞ ∞ e − a x 2 d x = ( − ∂ a ∂ ) n a π We can postulate that the state of a particle is described by a complex value wave function Ψ ( x , t ) \Psi(x, t) Ψ ( x , t ) x x x t t t
The probability density to find the particle at position x x x ∣ Ψ ( x , t ) ∣ 2 |\Psi(x, t)|^2 ∣Ψ ( x , t ) ∣ 2 A A A
P ( A ) = ∫ A ∣ Ψ ( x , t ) ∣ 2 d x . P(A) = \int_A |\Psi(x, t)|^2 dx. P ( A ) = ∫ A ∣Ψ ( x , t ) ∣ 2 d x . The wave function Ψ ( x , t ) \Psi(x, t) Ψ ( x , t )
The wave function Ψ ( x , t ) \Psi(x, t) Ψ ( x , t ) X X X
∫ X ∣ Ψ ( x , t ) ∣ 2 d x = 1. \int_{X} |\Psi(x, t)|^2 dx = 1. ∫ X ∣Ψ ( x , t ) ∣ 2 d x = 1. If the wave function is not normalized, and if A = ∫ X ∣ Ψ ( x , t ) ∣ 2 d x A=\int_{X} |\Psi(x, t)|^2 dx A = ∫ X ∣Ψ ( x , t ) ∣ 2 d x 0 0 0
Ψ normalized ( x , t ) = Ψ ( x , t ) A . \Psi_{\text{normalized}}(x, t) = \frac{\Psi(x, t)}{\sqrt{A}}. Ψ normalized ( x , t ) = A Ψ ( x , t ) . If Ψ 1 ( x , t ) \Psi_1(x, t) Ψ 1 ( x , t ) Ψ 2 ( x , t ) \Psi_2(x, t) Ψ 2 ( x , t )
Ψ ( x , t ) = c 1 Ψ 1 ( x , t ) + c 2 Ψ 2 ( x , t ) , \Psi(x, t) = c_1 \Psi_1(x, t) + c_2 \Psi_2(x, t), Ψ ( x , t ) = c 1 Ψ 1 ( x , t ) + c 2 Ψ 2 ( x , t ) , where c 1 c_1 c 1 c 2 c_2 c 2 Ψ \Psi Ψ
The interference of wave functions is a phenomenon where two wave functions Ψ 1 ( x , t ) \Psi_1(x, t) Ψ 1 ( x , t ) Ψ 2 ( x , t ) \Psi_2(x, t) Ψ 2 ( x , t ) Ψ ( x , t ) \Psi(x, t) Ψ ( x , t )
Let Ψ ( x , t ) = Ψ 1 ( x , t ) + Ψ 2 ( x , t ) \Psi(x, t) = \Psi_1(x, t) + \Psi_2(x, t) Ψ ( x , t ) = Ψ 1 ( x , t ) + Ψ 2 ( x , t )
∣ Ψ ( x , t ) ∣ 2 = ∣ Ψ 1 ( x , t ) ∣ 2 + ∣ Ψ 2 ( x , t ) ∣ 2 + Ψ 1 ( x , t ) ∗ Ψ 2 ( x , t ) + Ψ 1 ( x , t ) Ψ 2 ( x , t ) ∗ |\Psi(x, t)|^2 = |\Psi_1(x, t)|^2 + |\Psi_2(x, t)|^2 + \Psi_1(x, t)^* \Psi_2(x, t) + \Psi_1(x, t) \Psi_2(x, t)^* ∣Ψ ( x , t ) ∣ 2 = ∣ Ψ 1 ( x , t ) ∣ 2 + ∣ Ψ 2 ( x , t ) ∣ 2 + Ψ 1 ( x , t ) ∗ Ψ 2 ( x , t ) + Ψ 1 ( x , t ) Ψ 2 ( x , t ) ∗ And the Ψ 1 ( x , t ) ∗ Ψ 2 ( x , t ) + Ψ 1 ( x , t ) Ψ 2 ( x , t ) ∗ \Psi_1(x, t)^* \Psi_2(x, t) + \Psi_1(x, t) \Psi_2(x, t)^* Ψ 1 ( x , t ) ∗ Ψ 2 ( x , t ) + Ψ 1 ( x , t ) Ψ 2 ( x , t ) ∗
The classical plane wave is given by
Ψ ( x , t ) = A e i k x − i ω t \Psi(x, t) = A e^{ikx - i\omega t} Ψ ( x , t ) = A e ik x − iω t where A A A k k k ω \omega ω
Note: The classical plane wave is not normalizable, as the integral of the probability density is infinite. In this case, we can use Ψ ( x , t ) = A e i k x − i ω t − i a x 2 \Psi(x, t) = A e^{ikx - i\omega t -iax^2} Ψ ( x , t ) = A e ik x − iω t − ia x 2 a a a
The reduced Planck constant is given by
ℏ = h 2 π ≈ 1.0545718 × 1 0 − 34 J s . \hbar = \frac{h}{2\pi} \approx 1.0545718 \times 10^{-34} \text{ J s}. ℏ = 2 π h ≈ 1.0545718 × 1 0 − 34 J s . where h h h
The quantization rule is given by
p ^ = − i ℏ ∇ E ^ = i ℏ ∂ ∂ t \begin{align}
\hat{p} &= -i \hbar \nabla \\
\hat{E} &= i \hbar \frac{\partial}{\partial t}
\end{align} p ^ E ^ = − i ℏ∇ = i ℏ ∂ t ∂ In classical mechanics, the energy of a free particle is given by
E = T + V = p 2 2 m + V ( x ) . E = T + V = \frac{p^2}{2m} + V(x). E = T + V = 2 m p 2 + V ( x ) . By applying the quantization rule, the Schrödinger equation (SE) for a free particle with wave equation ψ \psi ψ
E ^ ψ = [ p ^ 2 2 m + V ] ψ i ℏ ∂ ψ ∂ t = [ − ℏ 2 2 m ∇ 2 + V ] ψ = [ − ℏ 2 2 m Δ + V ] ψ \begin{align}
\hat{E}\psi &= \left[\frac{\hat{p}^2}{2m} + V\right]\psi \\
i \hbar \frac{\partial \psi}{\partial t} &= \left[-\frac{\hbar^2}{2m} \nabla^2 + V\right]\psi = \left[-\frac{\hbar^2}{2m} \Delta + V\right]\psi
\end{align} E ^ ψ i ℏ ∂ t ∂ ψ = [ 2 m p ^ 2 + V ] ψ = [ − 2 m ℏ 2 ∇ 2 + V ] ψ = [ − 2 m ℏ 2 Δ + V ] ψ The Hamiltonian operator H ^ \hat{H} H ^
H ^ = p ^ 2 2 m + V ( x ) \hat{H} = \frac{\hat{p}^2}{2m} + V(x) H ^ = 2 m p ^ 2 + V ( x ) Using the Hamiltonian operator, the Schrödinger equation for a free particle is given by
E ^ ∂ ψ ∂ t = H ^ ψ . \hat{E} \frac{\partial \psi}{\partial t} = \hat{H}\psi. E ^ ∂ t ∂ ψ = H ^ ψ . The Schrödinger equation can be solved using separation of variables.
Let ψ ( x , t ) = ϕ ( x ) τ ( t ) \psi(x, t) = \phi(x) \tau(t) ψ ( x , t ) = ϕ ( x ) τ ( t )
E ^ ψ = H ^ ψ ϕ E ^ τ = τ H ^ ϕ E ^ τ τ = H ^ ϕ ϕ = E i ℏ ∂ ∂ t τ = E τ τ = e − i E t / ℏ H ^ ϕ = E ϕ \begin{align}
\hat{E}\psi &= \hat{H}\psi \\
\phi \hat{E} \tau &= \tau \hat{H} \phi \\
\frac{\hat{E} \tau}{\tau} &= \frac{\hat{H} \phi}{\phi} = E \\
i\hbar \frac{\partial}{\partial t} \tau &= E \tau \\
\tau &= e^{-iEt/\hbar} \\
\hat{H} \phi &= E \phi
\end{align} E ^ ψ ϕ E ^ τ τ E ^ τ i ℏ ∂ t ∂ τ τ H ^ ϕ = H ^ ψ = τ H ^ ϕ = ϕ H ^ ϕ = E = E τ = e − i Et /ℏ = Eϕ Assume, ϕ 1 , ϕ 2 , … \phi_1,\phi_2,\ldots ϕ 1 , ϕ 2 , … H ^ ϕ = E ϕ \hat{H} \phi = E \phi H ^ ϕ = Eϕ E 1 , E 2 , … E_1,E_2,\ldots E 1 , E 2 , …
ϕ ( x ) = ∑ n = 1 ∞ c n ϕ n ( x ) e − i E n t / ℏ . \phi(x) = \sum_{n=1}^{\infty} c_n \phi_n(x)e^{-iE_nt/\hbar}. ϕ ( x ) = n = 1 ∑ ∞ c n ϕ n ( x ) e − i E n t /ℏ . By the previous discussion , the task of solving the Schrödinger equation is reduced to solving the equation
H ^ ϕ n = E n ϕ n . \hat{H} \phi_n = E_n \phi_n. H ^ ϕ n = E n ϕ n . This equation is called the time-independent Schrödinger equation (TISE) .
We always call E n E_n E n eigenvalues of H H H
Energy lLevels
Eigenenergies
Energy Eigenvalues
And ϕ n \phi_n ϕ n energy eigenstates of H H H
Stationary States
Energy Eigenfunctions
If two or more energy eigenstates have the same energy eigenvalue, then the energy eigenvalue is said to be degenerate.
If two eigenstates have the same eigenvalue, we call it degeneracy of order 2 or twice degenerate .
Consider the following double slit experiment setup.
We can assume the wave function that go through the slits A A A B B B ψ ( r , t ) = e i k x − i w t r \psi(r, t) = \frac{e^{ikx-iwt}}{r} ψ ( r , t ) = r e ik x − i wt
Then, for any point P P P A A A B B B
Ψ ( x , t ) = ψ A ( r 1 , t ) + ψ B ( r 2 , t ) = e i k r 1 − i w t + e i k r 2 − i w t = e − i w t [ e i k r 1 r 1 + e i k r 2 r 2 ] \begin{align}
\Psi(x, t) &= \psi_A(r_1, t) + \psi_B(r_2, t) \\
&= e^{ikr_1-iwt} + e^{ikr_2-iwt} \\
&= e^{-iwt}\left[\frac{e^{ikr_1}}{r_1} + \frac{e^{ikr_2}}{r_2}\right]
\end{align} Ψ ( x , t ) = ψ A ( r 1 , t ) + ψ B ( r 2 , t ) = e ik r 1 − i wt + e ik r 2 − i wt = e − i wt [ r 1 e ik r 1 + r 2 e ik r 2 ] where r 1 r_1 r 1 r 2 r_2 r 2 A A A B B B P P P
r 1 = L 2 + ( x − d ) 2 r 2 = L 2 + ( x + d ) 2 \begin{align}
r_1 &= \sqrt{L^2 + (x-d)^2} \\
r_2 &= \sqrt{L^2 + (x+d)^2}
\end{align} r 1 r 2 = L 2 + ( x − d ) 2 = L 2 + ( x + d ) 2 Without normalization, the probability density of the wave function is proportional to
∣ Ψ ( x , t ) ∣ 2 = ∣ e i k r 1 r 1 + e i k r 2 r 2 ∣ 2 = 1 r 1 2 + 1 r 2 2 + 1 r 1 r 2 [ e i k ( r 1 − r 2 ) + e − i k ( r 1 − r 2 ) ] ≈ 1 L 2 [ 2 + 2 cos ( k ( r 1 − r 2 ) ) ] = 1 L 2 [ 2 + 2 cos ( k ( L 2 + ( x + d ) 2 − L 2 + ( x − d ) 2 ) ) ] = 1 L 2 [ 2 + 2 cos ( k 4 d x L 2 + ( x + d ) 2 + L 2 + ( x − d ) 2 ) ] ≈ 1 L 2 [ 2 + 2 cos ( k 4 d x 2 L ) ] As L ≫ x , d = 1 L 2 [ 2 + 2 cos ( k 2 d x L ) ] \begin{align}
|\Psi(x, t)|^2 &= \left|\frac{e^{ikr_1}}{r_1} + \frac{e^{ikr_2}}{r_2}\right|^2 \\
&= \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_1r_2}\left[e^{ik(r_1-r_2)} + e^{-ik(r_1-r_2)}\right] \\
&\approx \frac{1}{L^2}\left[2 + 2\cos(k(r_1-r_2))\right] \\
&= \frac{1}{L^2}\left[2 + 2\cos\left(
k
\left(
\sqrt{L^2 + (x+d)^2} - \sqrt{L^2 + (x-d)^2}
\right)
\right) \right] \\
&= \frac{1}{L^2}\left[2 + 2\cos\left(
k
\frac{
4dx
}{
\sqrt{L^2 + (x+d)^2} + \sqrt{L^2 + (x-d)^2}
}
\right) \right] \\
&\approx \frac{1}{L^2}\left[2 + 2\cos\left(
k
\frac{
4dx
}{
2L
}
\right)\right] \quad\text{As } L \gg x,d \\
&= \frac{1}{L^2}\left[2 + 2\cos\left(
k
\frac{
2dx
}{
L
}
\right)\right]
\end{align} ∣Ψ ( x , t ) ∣ 2 = r 1 e ik r 1 + r 2 e ik r 2 2 = r 1 2 1 + r 2 2 1 + r 1 r 2 1 [ e ik ( r 1 − r 2 ) + e − ik ( r 1 − r 2 ) ] ≈ L 2 1 [ 2 + 2 cos ( k ( r 1 − r 2 )) ] = L 2 1 [ 2 + 2 cos ( k ( L 2 + ( x + d ) 2 − L 2 + ( x − d ) 2 ) ) ] = L 2 1 [ 2 + 2 cos ( k L 2 + ( x + d ) 2 + L 2 + ( x − d ) 2 4 d x ) ] ≈ L 2 1 [ 2 + 2 cos ( k 2 L 4 d x ) ] As L ≫ x , d = L 2 1 [ 2 + 2 cos ( k L 2 d x ) ] We can see that the probability density of the wave function is proportional to a transformed cosine function, which explains the interference pattern on the screen.
Given a potential V ( x ) : R → C V(x): \mathbb{R}\rightarrow \mathbb{C} V ( x ) : R → C
V ( x ) = V ( − x ) V(x) = V(-x) V ( x ) = V ( − x ) In such cases, we can assume the TISE solution has definite parity , that is:
ϕ ( x ) = ± ϕ ( − x ) \phi(x) = \pm \phi(-x) ϕ ( x ) = ± ϕ ( − x ) Given a real potential V ( x ) V(x) V ( x ) definite parity regarding conjugation , that is:
ϕ ( x ) = ϕ ∗ ( x ) \phi(x) = \phi^*(x) ϕ ( x ) = ϕ ∗ ( x ) In general there are three kinds of states of TISE solutions:
Bound States
Scattering States
Non-physical States
Bound states are states where the energy eigenvalue E n E_n E n
Scattering states are states where the energy eigenvalue E n E_n E n
Scattering states can be approximated by normalizable states by similar method as the classical plane wave .
Non-physical states are states where the energy eigenvalue E n E_n E n
Consider the potential to be real. Define the probability density of a wave function P = Ψ ( x , t ) P = \Psi(x, t) P = Ψ ( x , t ) ∣ Ψ ( x , t ) ∣ 2 |\Psi(x, t)|^2 ∣Ψ ( x , t ) ∣ 2
∂ ∂ t P = ∂ ψ ∂ t ψ ∗ + ψ ∂ ψ ∂ t ∗ = 1 i ℏ [ H ^ ψ ψ ∗ − ψ ( H ^ ψ ) ∗ ] = − ℏ 2 i ℏ 2 m [ ∂ 2 ψ ∂ x 2 ψ ∗ − ψ ∂ 2 ψ ∂ x 2 ∗ ] = − d d x [ ℏ 2 i ℏ 2 m [ ∂ ψ ∂ x ψ ∗ − ∂ ψ ∂ x ∂ ψ ∂ x ∗ − ψ ∂ ψ ∂ x ∗ + ∂ ψ ∂ x ∂ ψ ∂ x ∗ ] ] = − d d x [ ℏ 2 m i [ − ψ ∂ ψ ∂ x ∗ + ψ ∗ ∂ ψ ∂ x ] ] \begin{align}
\frac{\partial}{\partial t} P &=
\frac{\partial \psi}{\partial t} \psi^* + \psi \frac{\partial \psi}{\partial t}^* \\
&= \frac{1}{i\hbar}\left[
\hat{H}\psi\psi^* - \psi(\hat{H}\psi)^*
\right] \\
&= -\frac{\hbar^2}{i\hbar 2m}\left[
\frac{\partial^2 \psi}{\partial x^2}\psi^* - \psi\frac{\partial^2 \psi}{\partial x^2}^*
\right] \\
&= -\frac{d}{dx}\left[\frac{\hbar^2}{i\hbar 2m}\left[
\frac{\partial \psi}{\partial x}\psi^*
- \frac{\partial \psi}{\partial x}\frac{\partial \psi}{\partial x}^*
- \psi\frac{\partial \psi}{\partial x}^*
+ \frac{\partial \psi}{\partial x}\frac{\partial \psi}{\partial x}^*
\right]\right] \\
&= -\frac{d}{dx}\left[\frac{\hbar}{2mi}\left[
-\psi\frac{\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x}
\right]\right]
\end{align} ∂ t ∂ P = ∂ t ∂ ψ ψ ∗ + ψ ∂ t ∂ ψ ∗ = i ℏ 1 [ H ^ ψ ψ ∗ − ψ ( H ^ ψ ) ∗ ] = − i ℏ2 m ℏ 2 [ ∂ x 2 ∂ 2 ψ ψ ∗ − ψ ∂ x 2 ∂ 2 ψ ∗ ] = − d x d [ i ℏ2 m ℏ 2 [ ∂ x ∂ ψ ψ ∗ − ∂ x ∂ ψ ∂ x ∂ ψ ∗ − ψ ∂ x ∂ ψ ∗ + ∂ x ∂ ψ ∂ x ∂ ψ ∗ ] ] = − d x d [ 2 mi ℏ [ − ψ ∂ x ∂ ψ ∗ + ψ ∗ ∂ x ∂ ψ ] ] We thus define the probability current J J J as
J = ℏ 2 m i [ − ψ ∂ ψ ∂ x ∗ + ψ ∗ ∂ ψ ∂ x ] J = \frac{\hbar}{2mi}\left[
-\psi\frac{\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x}
\right] J = 2 mi ℏ [ − ψ ∂ x ∂ ψ ∗ + ψ ∗ ∂ x ∂ ψ ] The above equation become:
∂ P ∂ t = − d J d x \frac{\partial P}{\partial t} = -\frac{dJ}{dx} ∂ t ∂ P = − d x dJ and is called the continuity equation .
We consider a potential barrier of height V 0 V_0 V 0 2 a 2a 2 a
That is, the potential is given by
V ( x ) = { V 0 if ∣ x ∣ < 2 a 0 otherwise V(x) = \begin{cases}
V_0 & \text{if } |x| < 2a \\
0 & \text{otherwise}
\end{cases} V ( x ) = { V 0 0 if ∣ x ∣ < 2 a otherwise Then, the TISE can be divided into three regions:
Region I: x < − a x < -a x < − a
d 2 ψ d x 2 = − 2 m ℏ 2 E ψ \frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}E\psi d x 2 d 2 ψ = − ℏ 2 2 m E ψ
Region II: − a < x < a -a < x < a − a < x < a
d 2 ψ d x 2 = − 2 m ℏ 2 ( E − V 0 ) ψ \frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}(E-V_0)\psi d x 2 d 2 ψ = − ℏ 2 2 m ( E − V 0 ) ψ
Region III: x > a x > a x > a
d 2 ψ d x 2 = − 2 m ℏ 2 E ψ \frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}E\psi d x 2 d 2 ψ = − ℏ 2 2 m E ψ For the sake of simplicity,
let k = 2 m E ℏ 2 k = \sqrt{\frac{2mE}{\hbar^2}} k = ℏ 2 2 m E
In region I:
ψ ( x ) = A e i k x + B e − i k x \psi(x) = A e^{ikx} + B e^{-ikx} ψ ( x ) = A e ik x + B e − ik x In region III:
ψ ( x ) = F e i k x \psi(x) = F e^{ikx} ψ ( x ) = F e ik x
The wave A e i k x A e^{ikx} A e ik x B e − i k x B e^{-ikx} B e − ik x F e i k x F e^{ikx} F e ik x
As there is no probability for the particle to be in region II, the probability current must be same at the point x = a x = a x = a x = − a x = -a x = − a
In region I:
J = ℏ 2 m i [ ψ − ∂ ψ ∂ x ∗ + ψ ∗ ∂ ψ ∂ x ] = ℏ k m ( ∣ A ∣ 2 − ∣ B ∣ 2 ) J = \frac{\hbar}{2mi}\left[
\psi\frac{-\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x}
\right] = \frac{\hbar k}{m}(|A|^2 - |B|^2) J = 2 mi ℏ [ ψ ∂ x − ∂ ψ ∗ + ψ ∗ ∂ x ∂ ψ ] = m ℏ k ( ∣ A ∣ 2 − ∣ B ∣ 2 ) In region III:
J = ℏ 2 m i [ ψ − ∂ ψ ∂ x ∗ + ψ ∗ ∂ ψ ∂ x ] = ℏ k m ∣ F ∣ 2 J = \frac{\hbar}{2mi}\left[
\psi\frac{-\partial \psi}{\partial x}^* + \psi^*\frac{\partial \psi}{\partial x}
\right] = \frac{\hbar k}{m}|F|^2 J = 2 mi ℏ [ ψ ∂ x − ∂ ψ ∗ + ψ ∗ ∂ x ∂ ψ ] = m ℏ k ∣ F ∣ 2 Thus, we have
∣ A ∣ 2 − ∣ B ∣ 2 = ∣ F ∣ 2 |A|^2 - |B|^2 = |F|^2 ∣ A ∣ 2 − ∣ B ∣ 2 = ∣ F ∣ 2 We define the reflection probability R R R and transmission probability T T T as
R = ∣ B ∣ 2 ∣ A ∣ 2 T = ∣ F ∣ 2 ∣ A ∣ 2 \begin{align}
R &= \frac{|B|^2}{|A|^2} \\
T &= \frac{|F|^2}{|A|^2}
\end{align} R T = ∣ A ∣ 2 ∣ B ∣ 2 = ∣ A ∣ 2 ∣ F ∣ 2 Then, we have
R + T = 1 R + T = 1 R + T = 1 We consider a barrier that is infinitely high and thin at origin.
To state this formally, we consider the potential to be the limiting behaviour of
the Dirac delta function potential.
δ L ( x ) = { 1 L if − L 2 < x < L 2 0 otherwise \delta_L(x) = \begin{cases}
\frac{1}{L} & \text{if } -\frac{L}{2} < x < \frac{L}{2} \\
0 & \text{otherwise}
\end{cases} δ L ( x ) = { L 1 0 if − 2 L < x < 2 L otherwise and
δ = lim L → 0 δ L ( x ) \delta = \lim_{L \to 0} \delta_L(x) δ = L → 0 lim δ L ( x ) Also, for all function g ( x ) g(x) g ( x )
lim ϵ → 0 ∫ − ϵ ϵ g ( x ) δ ( x ) d x = g ( 0 ) \lim_{\epsilon \to 0} \int_{-\epsilon}^{\epsilon} g(x) \delta(x) dx = g(0) ϵ → 0 lim ∫ − ϵ ϵ g ( x ) δ ( x ) d x = g ( 0 ) Then, the potential is given by
V ( x ) = α δ ( x ) V(x) = \alpha \delta(x) V ( x ) = α δ ( x ) We again, consider the simplified model in the previous section .
In region I:
ψ ( x ) = A e i k x + B e − i k x \psi(x) = A e^{ikx} + B e^{-ikx} ψ ( x ) = A e ik x + B e − ik x In region III:
ψ ( x ) = F e i k x \psi(x) = F e^{ikx} ψ ( x ) = F e ik x As the region II is infinity thin,
the right hand side of the region I and the left hand side of the region III must be the same.
Thus, we have
A + B = F A + B = F A + B = F Also, by integrating the TISE over the region [ − ϵ , ϵ ] [-\epsilon,\epsilon] [ − ϵ , ϵ ]
∫ − ϵ ϵ d 2 ψ d x 2 d x = − 2 m ℏ 2 ∫ − ϵ ϵ ( E − α δ ( x ) ) ψ d x [ d ψ d x ] − ϵ ϵ = − 2 m ℏ 2 [ E ψ ( ϵ ) − E ψ ( − ϵ ) − α ∫ − ϵ ϵ ψ δ d x ] \begin{align}
\int_{-\epsilon}^{\epsilon} \frac{d^2 \psi}{dx^2} dx &= -\frac{2m}{\hbar^2} \int_{-\epsilon}^{\epsilon} (E-\alpha\delta(x))\psi dx \\
\left[\frac{d \psi}{dx}\right]_{-\epsilon}^{\epsilon} &= -\frac{2m}{\hbar^2}
\left[E\psi(\epsilon)-E\psi(-\epsilon) - \alpha\int_{-\epsilon}^{\epsilon} \psi\delta dx\right] \\
\end{align} ∫ − ϵ ϵ d x 2 d 2 ψ d x [ d x d ψ ] − ϵ ϵ = − ℏ 2 2 m ∫ − ϵ ϵ ( E − α δ ( x )) ψ d x = − ℏ 2 2 m [ E ψ ( ϵ ) − E ψ ( − ϵ ) − α ∫ − ϵ ϵ ψ δ d x ] Taking the limit ϵ → 0 \epsilon \to 0 ϵ → 0
[ d ψ d x ] 0 + − [ d ψ d x ] 0 − = 2 m ℏ 2 α ψ ( 0 ) \begin{align}
\left[\frac{d \psi}{dx}\right]_{0^+} - \left[\frac{d \psi}{dx}\right]_{0^-} &= \frac{2m}{\hbar^2} \alpha \psi(0)
\end{align} [ d x d ψ ] 0 + − [ d x d ψ ] 0 − = ℏ 2 2 m α ψ ( 0 ) Substituting the solution of the TISE in region I and region III, we get
i k ( A − B ) − i k F = 2 m ℏ 2 α F ik(A-B) - ikF = \frac{2m}{\hbar^2} \alpha F ik ( A − B ) − ik F = ℏ 2 2 m α F Solving the above equation, we can get the reflection and transmission probability.
Define the Hilbert space H \mathbf{H} H as the space of all possible wave functions Ψ ( x , t ) \Psi(x, t) Ψ ( x , t )
H = { Ψ ( x , t ) ∣ ∫ − ∞ ∞ ∣ Ψ ( x , t ) ∣ 2 d x < ∞ } \mathbf{H} = \left\{
\Psi(x, t) \mid \int_{-\infty}^{\infty} |\Psi(x, t)|^2 dx < \infty
\right\} H = { Ψ ( x , t ) ∣ ∫ − ∞ ∞ ∣Ψ ( x , t ) ∣ 2 d x < ∞ } The Hilbert space is a complex vector space .
We define the inner product of two wave functions Ψ 1 ( x , t ) \Psi_1(x, t) Ψ 1 ( x , t ) Ψ 2 ( x , t ) \Psi_2(x, t) Ψ 2 ( x , t )
< Ψ 1 , Ψ 2 > = ∫ − ∞ ∞ Ψ 1 ∗ ( x , t ) Ψ 2 ( x , t ) d x \left<\Psi_1, \Psi_2\right> = \int_{-\infty}^{\infty} \Psi_1^*(x, t) \Psi_2(x, t) dx ⟨ Ψ 1 , Ψ 2 ⟩ = ∫ − ∞ ∞ Ψ 1 ∗ ( x , t ) Ψ 2 ( x , t ) d x
Linearity in Second Argument : For all Ψ 1 , Ψ 2 ∈ H \Psi_1, \Psi_2 \in \mathbf{H} Ψ 1 , Ψ 2 ∈ H a , b ∈ C a,b \in \mathbb{C} a , b ∈ C < a Ψ 1 + b Ψ 2 , Ψ 3 > = a < Ψ 1 , Ψ 3 > + b < Ψ 2 , Ψ 3 > \left<a\Psi_1 + b\Psi_2, \Psi_3\right> = a\left<\Psi_1, \Psi_3\right> + b\left<\Psi_2, \Psi_3\right> ⟨ a Ψ 1 + b Ψ 2 , Ψ 3 ⟩ = a ⟨ Ψ 1 , Ψ 3 ⟩ + b ⟨ Ψ 2 , Ψ 3 ⟩ Anti-linearity in First Argument : For all Ψ 1 , Ψ 2 ∈ H \Psi_1, \Psi_2 \in \mathbf{H} Ψ 1 , Ψ 2 ∈ H a , b ∈ C a,b \in \mathbb{C} a , b ∈ C < Ψ 1 , a Ψ 2 + b Ψ 3 > = a ∗ < Ψ 1 , Ψ 2 > + b ∗ < Ψ 1 , Ψ 3 > \left<\Psi_1, a\Psi_2 + b\Psi_3\right> = a^*\left<\Psi_1, \Psi_2\right> + b^*\left<\Psi_1, \Psi_3\right> ⟨ Ψ 1 , a Ψ 2 + b Ψ 3 ⟩ = a ∗ ⟨ Ψ 1 , Ψ 2 ⟩ + b ∗ ⟨ Ψ 1 , Ψ 3 ⟩ Positive Definite : For all Ψ ∈ H \Psi \in \mathbf{H} Ψ ∈ H < Ψ , Ψ > ≥ 0 \left<\Psi, \Psi\right> \geq 0 ⟨ Ψ , Ψ ⟩ ≥ 0 < Ψ , Ψ > = 0 \left<\Psi, \Psi\right> = 0 ⟨ Ψ , Ψ ⟩ = 0 Ψ = 0 \Psi = 0 Ψ = 0 Conjugate Symmetric (Skew Symmetric) : For all Ψ 1 , Ψ 2 ∈ H \Psi_1, \Psi_2 \in \mathbf{H} Ψ 1 , Ψ 2 ∈ H < Ψ 1 , Ψ 2 > = < Ψ 2 , Ψ 1 > ∗ \left<\Psi_1, \Psi_2\right> = \left<\Psi_2, \Psi_1\right>^* ⟨ Ψ 1 , Ψ 2 ⟩ = ⟨ Ψ 2 , Ψ 1 ⟩ ∗
The norm of a wave function Ψ ( x , t ) \Psi(x, t) Ψ ( x , t )
∣ ∣ Ψ ∣ ∣ = < Ψ , Ψ > ||\Psi|| = \sqrt{\left<\Psi, \Psi\right>} ∣∣Ψ∣∣ = ⟨ Ψ , Ψ ⟩ Two wave functions Ψ 1 ( x , t ) \Psi_1(x, t) Ψ 1 ( x , t ) Ψ 2 ( x , t ) \Psi_2(x, t) Ψ 2 ( x , t ) orthogonal if
< Ψ 1 , Ψ 2 > = 0 \left<\Psi_1, \Psi_2\right> = 0 ⟨ Ψ 1 , Ψ 2 ⟩ = 0 The angle between two wave functions Ψ 1 ( x , t ) \Psi_1(x, t) Ψ 1 ( x , t ) Ψ 2 ( x , t ) \Psi_2(x, t) Ψ 2 ( x , t )
cos θ = < Ψ 1 , Ψ 2 > ∣ ∣ Ψ 1 ∣ ∣ ∣ ∣ Ψ 2 ∣ ∣ \cos\theta = \frac{\left<\Psi_1, \Psi_2\right>}{||\Psi_1|| ||\Psi_2||} cos θ = ∣∣ Ψ 1 ∣∣∣∣ Ψ 2 ∣∣ ⟨ Ψ 1 , Ψ 2 ⟩ A set of wave functions { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … } \{\Psi_1(x, t), \Psi_2(x, t), \ldots\} { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … } orthonormal basis if
The set is orthogonal.
The set is normalized.
The set spans the Hilbert space. That is, for all Ψ ( x , t ) ∈ H \Psi(x, t) \in \mathbf{H} Ψ ( x , t ) ∈ H { c 1 , c 2 , … } \{c_1, c_2, \ldots\} { c 1 , c 2 , … }
Ψ ( x , t ) = ∑ n = 1 ∞ c n Ψ n ( x , t ) \Psi(x, t) = \sum_{n=1}^{\infty} c_n \Psi_n(x, t) Ψ ( x , t ) = n = 1 ∑ ∞ c n Ψ n ( x , t ) An operator is a function that maps a wave function to another wave function.
The Hermitian conjugate of an operator A ^ \hat{A} A ^ A ^ † \hat{A}^\dagger A ^ †
< A ^ Ψ 1 , Ψ 2 > = < Ψ 1 , A ^ † Ψ 2 > \left<\hat{A}\Psi_1, \Psi_2\right> = \left<\Psi_1, \hat{A}^\dagger\Psi_2\right> ⟨ A ^ Ψ 1 , Ψ 2 ⟩ = ⟨ Ψ 1 , A ^ † Ψ 2 ⟩ An Hermitian operator is an operator that satisfies
< A ^ Ψ 1 , Ψ 2 > = < Ψ 1 , A ^ Ψ 2 > \left<\hat{A}\Psi_1, \Psi_2\right> = \left<\Psi_1, \hat{A}\Psi_2\right> ⟨ A ^ Ψ 1 , Ψ 2 ⟩ = ⟨ Ψ 1 , A ^ Ψ 2 ⟩ for all Ψ 1 , Ψ 2 ∈ H \Psi_1, \Psi_2 \in \mathbf{H} Ψ 1 , Ψ 2 ∈ H
An equivalent definition is that the operator is equal to its Hermitian conjugate.
The spectral theorem states that for all Hermitian operators A ^ \hat{A} A ^ { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … } \{\Psi_1(x, t), \Psi_2(x, t), \ldots\} { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … }
A ^ Ψ n ( x , t ) = a n Ψ n ( x , t ) \hat{A}\Psi_n(x, t) = a_n\Psi_n(x, t) A ^ Ψ n ( x , t ) = a n Ψ n ( x , t ) where a n a_n a n A ^ \hat{A} A ^
The Hamiltonian operator H ^ \hat{H} H ^
Proof:
< H ^ Ψ 1 , Ψ 2 > = ∫ − ∞ ∞ Ψ 1 ∗ ( x , t ) H ^ Ψ 2 ( x , t ) d x = ∫ − ∞ ∞ Ψ 1 ∗ ( x , t ) [ − ℏ 2 2 m ∂ 2 Ψ 2 ∂ x 2 + V ( x ) Ψ 2 ( x , t ) ] d x By Integration By Part = ∫ − ∞ ∞ [ − ℏ 2 2 m ∂ 2 Ψ 1 ∗ ∂ x 2 + V ( x ) Ψ 1 ∗ ( x , t ) ] Ψ 2 ( x , t ) d x = < Ψ 1 , H ^ Ψ 2 > \begin{align}
\left<\hat{H}\Psi_1, \Psi_2\right> &= \int_{-\infty}^{\infty} \Psi_1^*(x, t) \hat{H}\Psi_2(x, t) dx \\
&= \int_{-\infty}^{\infty} \Psi_1^*(x, t) \left[-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi_2}{\partial x^2} + V(x)\Psi_2(x, t)\right] dx \quad \text{By Integration By Part} \\
&= \int_{-\infty}^{\infty} \left[-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi_1^*}{\partial x^2} + V(x)\Psi_1^*(x, t)\right] \Psi_2(x, t) dx \\
&= \left<\Psi_1, \hat{H}\Psi_2\right>
\end{align} ⟨ H ^ Ψ 1 , Ψ 2 ⟩ = ∫ − ∞ ∞ Ψ 1 ∗ ( x , t ) H ^ Ψ 2 ( x , t ) d x = ∫ − ∞ ∞ Ψ 1 ∗ ( x , t ) [ − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ 2 + V ( x ) Ψ 2 ( x , t ) ] d x By Integration By Part = ∫ − ∞ ∞ [ − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ 1 ∗ + V ( x ) Ψ 1 ∗ ( x , t ) ] Ψ 2 ( x , t ) d x = ⟨ Ψ 1 , H ^ Ψ 2 ⟩ An operator A ^ \hat{A} A ^ positive definite if
give any non-zero wave function Ψ ( x , t ) \Psi(x, t) Ψ ( x , t )
< A ^ Ψ , Ψ > > 0 \left<\hat{A}\Psi, \Psi\right> > 0 ⟨ A ^ Ψ , Ψ ⟩ > 0 An operator A ^ \hat{A} A ^ positive semi-definite if
give any wave function Ψ ( x , t ) \Psi(x, t) Ψ ( x , t )
< A ^ Ψ , Ψ > ≥ 0 \left<\hat{A}\Psi, \Psi\right> \geq 0 ⟨ A ^ Ψ , Ψ ⟩ ≥ 0 Any operator given by A ^ = B ^ † B ^ \hat{A} = \hat{B}^\dagger \hat{B} A ^ = B ^ † B ^
Given a orthonormal basis { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … } \{\Psi_1(x, t), \Psi_2(x, t), \ldots\} { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … } Ψ ( x , t ) \Psi(x, t) Ψ ( x , t ) Ψ n ( x , t ) \Psi_n(x, t) Ψ n ( x , t )
P n = ∣ < Ψ n , Ψ > ∣ 2 P_n = \left|\left<\Psi_n, \Psi\right>\right|^2 P n = ∣ ⟨ Ψ n , Ψ ⟩ ∣ 2 After the measurement, the state of the wave function will collapse to the state that is measured.
Given an observable A ^ \hat{A} A ^ A ^ \hat{A} A ^ { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … } \{\Psi_1(x, t), \Psi_2(x, t), \ldots\} { Ψ 1 ( x , t ) , Ψ 2 ( x , t ) , … }
The expectation value of the observable A ^ \hat{A} A ^ ψ \psi ψ
< A ^ > = ∑ n = 1 ∞ a n P n = < ψ ∣ A ^ ψ > \left<\hat{A}\right> = \sum_{n=1}^{\infty} a_n P_n = \left<\psi|\hat{A}\psi\right> ⟨ A ^ ⟩ = n = 1 ∑ ∞ a n P n = ⟨ ψ ∣ A ^ ψ ⟩ In general, two operators A ^ \hat{A} A ^ B ^ \hat{B} B ^
We define the commutator (Lie Bracket) of two operators A ^ \hat{A} A ^ B ^ \hat{B} B ^
[ A ^ , B ^ ] = A ^ B ^ − B ^ A ^ [\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} [ A ^ , B ^ ] = A ^ B ^ − B ^ A ^
Linearity : [ a A ^ + b B ^ , C ^ ] = a [ A ^ , C ^ ] + b [ B ^ , C ^ ] [a\hat{A} + b\hat{B}, \hat{C}] = a[\hat{A}, \hat{C}] + b[\hat{B}, \hat{C}] [ a A ^ + b B ^ , C ^ ] = a [ A ^ , C ^ ] + b [ B ^ , C ^ ] Anti-linearity : [ A ^ , a B ^ + b C ^ ] = a [ A ^ , B ^ ] + b [ A ^ , C ^ ] [\hat{A}, a\hat{B} + b\hat{C}] = a[\hat{A}, \hat{B}] + b[\hat{A}, \hat{C}] [ A ^ , a B ^ + b C ^ ] = a [ A ^ , B ^ ] + b [ A ^ , C ^ ] Linearity in Second Argument : [ C ^ , a A ^ + b B ^ ] = a [ C ^ , A ^ ] + b [ C ^ , B ^ ] [\hat{C}, a\hat{A} + b\hat{B}] = a[\hat{C},\hat{A}] + b[\hat{C},\hat{B}] [ C ^ , a A ^ + b B ^ ] = a [ C ^ , A ^ ] + b [ C ^ , B ^ ] Distributivity : [ A ^ , B ^ C ^ ] = [ A ^ , B ^ ] C ^ + B ^ [ A ^ , C ^ ] [\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}] [ A ^ , B ^ C ^ ] = [ A ^ , B ^ ] C ^ + B ^ [ A ^ , C ^ ]
The commutator of position operator x ^ \hat{x} x ^ p ^ \hat{p} p ^
[ x ^ , p ^ ] = x ^ p ^ − p ^ x ^ = x ( − i ℏ ∂ ∂ x ) − ( − i ℏ ∂ ∂ x ) x = x ( − i ℏ ∂ ∂ x ) − ( − i ℏ ) − x ( − i ℏ ∂ ∂ x ) = i ℏ \begin{align}
[\hat{x},\hat{p}] &= \hat{x}\hat{p} - \hat{p}\hat{x} \\
&= x\left(-i\hbar\frac{\partial}{\partial x}\right) - \left(-i\hbar\frac{\partial}{\partial x}\right)x \\
&= x\left(-i\hbar\frac{\partial}{\partial x}\right) - \left(-i\hbar\right) - x\left(-i\hbar\frac{\partial}{\partial x}\right) \\
&= i\hbar
\end{align} [ x ^ , p ^ ] = x ^ p ^ − p ^ x ^ = x ( − i ℏ ∂ x ∂ ) − ( − i ℏ ∂ x ∂ ) x = x ( − i ℏ ∂ x ∂ ) − ( − i ℏ ) − x ( − i ℏ ∂ x ∂ ) = i ℏ Two observables A ^ \hat{A} A ^ B ^ \hat{B} B ^ compatible if they commute.
The Robertson Inequality states that for any two observables A ^ \hat{A} A ^ B ^ \hat{B} B ^
Δ A ^ Δ B ^ ≥ 1 2 ∣ < [ A ^ , B ^ ] > ∣ \Delta \hat{A} \Delta \hat{B} \ge \frac{1}{2} \left|\left<[\hat{A}, \hat{B}]\right>\right| Δ A ^ Δ B ^ ≥ 2 1 ⟨ [ A ^ , B ^ ] ⟩ Proof:
Without loss of generality, we can assume < A ^ > = 0 \left<\hat{A}\right> = 0 ⟨ A ^ ⟩ = 0 < B ^ > = 0 \left<\hat{B}\right> = 0 ⟨ B ^ ⟩ = 0 A ^ \hat{A} A ^ A ^ − < A ^ > \hat{A} - \left<\hat{A}\right> A ^ − ⟨ A ^ ⟩ B ^ \hat{B} B ^ B ^ − < B ^ > \hat{B} - \left<\hat{B}\right> B ^ − ⟨ B ^ ⟩
Then, we have
Δ A ^ Δ B ^ = < A ^ 2 > < B ^ 2 > = < A ^ ψ ∣ A ^ ψ > < B ^ ψ ∣ B ^ ψ > ≥ ∣ < A ^ ψ ∣ B ^ ψ > ∣ Cauchy-Schwartz = ∣ < ψ ∣ A ^ B ^ ψ > ∣ \begin{align}
\Delta \hat{A} \Delta \hat{B} &= \sqrt{\left<\hat{A}^2\right>\left<\hat{B}^2\right>} \\
&= \sqrt{\left<\hat{A}\psi|\hat{A}\psi\right>\left<\hat{B}\psi|\hat{B}\psi\right>} \\
&\ge \left|\left<\hat{A}\psi|\hat{B}\psi\right>\right| \quad \text{Cauchy-Schwartz}\\
&= \left|\left<\psi|\hat{A}\hat{B}\psi\right>\right| \\
\end{align} Δ A ^ Δ B ^ = ⟨ A ^ 2 ⟩ ⟨ B ^ 2 ⟩ = ⟨ A ^ ψ ∣ A ^ ψ ⟩ ⟨ B ^ ψ ∣ B ^ ψ ⟩ ≥ ⟨ A ^ ψ ∣ B ^ ψ ⟩ Cauchy-Schwartz = ⟨ ψ ∣ A ^ B ^ ψ ⟩ By similar argument, we have
Δ A ^ Δ B ^ ≥ ∣ < ψ ∣ B ^ A ^ ψ > ∣ \Delta \hat{A} \Delta \hat{B} \ge \left|\left<\psi|\hat{B}\hat{A}\psi\right>\right| Δ A ^ Δ B ^ ≥ ⟨ ψ ∣ B ^ A ^ ψ ⟩ Thus,
Δ A ^ Δ B ^ = 1 2 ( Δ A ^ Δ B ^ + Δ A ^ Δ B ^ ) ≥ 1 2 ( ∣ < ψ ∣ A ^ B ^ ψ > ∣ + ∣ < ψ ∣ B ^ A ^ ψ > ∣ ) ≥ 1 2 ∣ < ψ ∣ A ^ B ^ ψ > − < ψ ∣ B ^ A ^ ψ > ∣ = 1 2 ∣ < ψ ∣ [ A ^ , B ^ ] ψ > ∣ \begin{align}
\Delta \hat{A} \Delta \hat{B} &= \frac{1}{2}\left(\Delta \hat{A} \Delta \hat{B} + \Delta \hat{A} \Delta \hat{B}\right) \\
&\ge \frac{1}{2}\left(
\left|\left<\psi|\hat{A}\hat{B}\psi\right>\right| + \left|\left<\psi|\hat{B}\hat{A}\psi\right>\right|
\right)\\
&\ge \frac{1}{2}\left|
\left<\psi|\hat{A}\hat{B}\psi\right> - \left<\psi|\hat{B}\hat{A}\psi\right>
\right| \\
&= \frac{1}{2}\left|
\left<\psi|[\hat{A}, \hat{B}]\psi\right>
\right| \\
\end{align} Δ A ^ Δ B ^ = 2 1 ( Δ A ^ Δ B ^ + Δ A ^ Δ B ^ ) ≥ 2 1 ( ⟨ ψ ∣ A ^ B ^ ψ ⟩ + ⟨ ψ ∣ B ^ A ^ ψ ⟩ ) ≥ 2 1 ⟨ ψ ∣ A ^ B ^ ψ ⟩ − ⟨ ψ ∣ B ^ A ^ ψ ⟩ = 2 1 ⟨ ψ ∣ [ A ^ , B ^ ] ψ ⟩ Given the position operator x ^ \hat{x} x ^ p ^ \hat{p} p ^
Δ x ^ Δ p ^ ≥ ℏ 2 \Delta \hat{x} \Delta \hat{p} \ge \frac{\hbar}{2} Δ x ^ Δ p ^ ≥ 2 ℏ A quantum harmonic oscillator is a system where the potential energy is given by
V ( x ) = 1 2 m ω 2 x 2 V(x) = \frac{1}{2}m\omega^2x^2 V ( x ) = 2 1 m ω 2 x 2 Given the Hamiltonian operator H ^ \hat{H} H ^ annihilation operator a ^ \hat{a} a ^ and creation operator a ^ † \hat{a}^\dagger a ^ † as
a ^ = u x ^ + v p ^ i a ^ † = u x ^ − v p ^ i \begin{align}
\hat{a} &= u\hat{x} + v\hat{p} i \\
\hat{a}^\dagger &= u\hat{x} - v\hat{p} i
\end{align} a ^ a ^ † = u x ^ + v p ^ i = u x ^ − v p ^ i Thus, we have
a ^ a ^ † = u 2 x ^ 2 − v 2 p ^ 2 − u v x ^ p ^ i + u v p ^ x ^ i = u 2 x ^ 2 − v 2 p ^ 2 + u v ℏ \begin{align}
\hat{a}\hat{a}^\dagger &= u^2 \hat{x}^2 - v^2 \hat{p}^2
- uv \hat{x}\hat{p} i + uv \hat{p}\hat{x} i \\
&= u^2 \hat{x}^2 - v^2 \hat{p}^2 + uv\hbar \\
\end{align} a ^ a ^ † = u 2 x ^ 2 − v 2 p ^ 2 − uv x ^ p ^ i + uv p ^ x ^ i = u 2 x ^ 2 − v 2 p ^ 2 + uv ℏ If we take:
u 2 = 1 2 ℏ m ω v 2 = 1 2 m ω ℏ \begin{align}
u^2 &= \frac{1}{2\hbar}m\omega \\
v^2 &= \frac{1}{2 m\omega\hbar} \\
\end{align} u 2 v 2 = 2ℏ 1 mω = 2 mω ℏ 1 Then, we have
a ^ a ^ † = 1 2 ℏ m ω x ^ 2 + 1 2 m ω ℏ p ^ 2 + 1 2 = 1 ℏ ω H ^ + 1 2 \begin{align}
\hat{a}\hat{a}^\dagger &= \frac{1}{2\hbar}m\omega \hat{x}^2 + \frac{1}{2 m\omega\hbar} \hat{p}^2 + \frac{1}{2} \\
&= \frac{1}{\hbar\omega}\hat{H} + \frac{1}{2}
\end{align} a ^ a ^ † = 2ℏ 1 mω x ^ 2 + 2 mω ℏ 1 p ^ 2 + 2 1 = ℏ ω 1 H ^ + 2 1 Given the annihilation operator a ^ \hat{a} a ^ a ^ † \hat{a}^\dagger a ^ †
[ a ^ , a ^ † ] = 2 u v h = 1 [\hat{a}, \hat{a}^\dagger] = 2uvh = 1 [ a ^ , a ^ † ] = 2 uv h = 1 Given the annihilation operator a ^ \hat{a} a ^ a ^ † \hat{a}^\dagger a ^ † number operator N ^ \hat{N} N ^ as
N ^ = a ^ † a ^ \hat{N} = \hat{a}^\dagger\hat{a} N ^ = a ^ † a ^ Thus, we have
H ^ = ℏ ω ( N ^ + 1 2 ) \begin{align}
\hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right)
\end{align} H ^ = ℏ ω ( N ^ + 2 1 ) As the number operator N ^ \hat{N} N ^ { Ψ 0 ( x ) , Ψ 1 ( x ) , … } \{\Psi_0(x), \Psi_1(x), \ldots\} { Ψ 0 ( x ) , Ψ 1 ( x ) , … }
N ^ Ψ n ( x ) = E n Ψ n ( x ) \hat{N}\Psi_n(x) = E_n\Psi_n(x) N ^ Ψ n ( x ) = E n Ψ n ( x ) We next prove that E n E_n E n
Given Ψ n \Psi_n Ψ n N ^ \hat{N} N ^ E n E_n E n
N ^ a ^ Ψ n = a ^ † a ^ 2 Ψ n = ( a ^ a ^ † − 1 ) a ^ Ψ n = a ^ N ^ − a ^ Ψ n = ( E n − 1 ) a ^ Ψ n \begin{align}
\hat{N}\hat{a}\Psi_n &= \hat{a}^\dagger\hat{a}^2\Psi_n \\
&= (\hat{a}\hat{a}^\dagger - 1)\hat{a}\Psi_n \\
&= \hat{a}\hat{N} - \hat{a}\Psi_n \\
&= (E_n-1)\hat{a}\Psi_n \\
\end{align} N ^ a ^ Ψ n = a ^ † a ^ 2 Ψ n = ( a ^ a ^ † − 1 ) a ^ Ψ n = a ^ N ^ − a ^ Ψ n = ( E n − 1 ) a ^ Ψ n Thus, if a ^ Ψ n \hat{a}\Psi_n a ^ Ψ n a ^ Ψ n \hat{a}\Psi_n a ^ Ψ n N ^ \hat{N} N ^ E n − 1 E_n-1 E n − 1
We next prove that a ^ Ψ n = 0 \hat{a}\Psi_n = 0 a ^ Ψ n = 0 E n = 0 E_n = 0 E n = 0
If E n = 0 E_n = 0 E n = 0
< a ^ Ψ n ∣ a ^ Ψ n > = < Ψ n ∣ a ^ † a ^ Ψ n > = < Ψ n ∣ 0 > = 0 \begin{align}
\left<\hat{a}\Psi_n|\hat{a}\Psi_n\right> &= \left<\Psi_n|\hat{a}^\dagger\hat{a}\Psi_n\right> \\
&= \left<\Psi_n|0\right> \\
&= 0
\end{align} ⟨ a ^ Ψ n ∣ a ^ Ψ n ⟩ = ⟨ Ψ n ∣ a ^ † a ^ Ψ n ⟩ = ⟨ Ψ n ∣0 ⟩ = 0 Thus, a ^ Ψ n = 0 \hat{a}\Psi_n = 0 a ^ Ψ n = 0
If a ^ Ψ n = 0 \hat{a}\Psi_n = 0 a ^ Ψ n = 0
< a ^ Ψ n ∣ a ^ Ψ n > = < Ψ n ∣ a ^ † a ^ Ψ n > = < Ψ n ∣ N ^ Ψ n > = E n < Ψ n ∣ Ψ n > = 0 \begin{align}
\left<\hat{a}\Psi_n|\hat{a}\Psi_n\right> &= \left<\Psi_n|\hat{a}^\dagger\hat{a}\Psi_n\right> \\
&= \left<\Psi_n|\hat{N}\Psi_n\right> \\
&= E_n\left<\Psi_n|\Psi_n\right> \\
&= 0
\end{align} ⟨ a ^ Ψ n ∣ a ^ Ψ n ⟩ = ⟨ Ψ n ∣ a ^ † a ^ Ψ n ⟩ = ⟨ Ψ n ∣ N ^ Ψ n ⟩ = E n ⟨ Ψ n ∣ Ψ n ⟩ = 0 Thus, E n = 0 E_n = 0 E n = 0
By proceeding the previous argument, we conclude that a ^ k Ψ n \hat{a}^k\Psi_n a ^ k Ψ n N ^ \hat{N} N ^ E n − k E_n-k E n − k
If E n En E n k k k E n − k E_n - k E n − k k k k E n − k < 0 E_n - k < 0 E n − k < 0 N ^ = a ^ † a ^ \hat{N} = \hat{a}^\dagger\hat{a} N ^ = a ^ † a ^
Thus, E n E_n E n
By discussion in the previous section ,
we can find an normalised eigenstate Ψ 0 ( x ) \Psi_0(x) Ψ 0 ( x ) N ^ \hat{N} N ^ E 0 = 0 E_0 = 0 E 0 = 0
In this section, we prove that it is unique, up to a complex constant.
By previous discussion, we see that a ^ Ψ 0 = 0 \hat{a}\Psi_0 = 0 a ^ Ψ 0 = 0
u x ^ Ψ 0 + v p ^ i Ψ 0 = 0 u x ^ Ψ 0 + v ℏ d d x Ψ 0 = 0 \begin{align}
u\hat{x}\Psi_0 + v\hat{p} i \Psi_0 &= 0 \\
u\hat{x}\Psi_0 + v\hbar \frac{d}{dx}\Psi_0 &= 0 \\
\end{align} u x ^ Ψ 0 + v p ^ i Ψ 0 u x ^ Ψ 0 + v ℏ d x d Ψ 0 = 0 = 0 As the above equation is a first oder homogeneous ODE, and the solution is unique up to a complex constant.
Given the ground state Ψ 0 ( x ) \Psi_0(x) Ψ 0 ( x ) Ψ n ( x ) \Psi_n(x) Ψ n ( x ) N ^ \hat{N} N ^ E n = n E_n = n E n = n
N ^ a ^ † Ψ n = ( E n + 1 ) a ^ † Ψ n \hat{N}\hat{a}^\dagger\Psi_n = (E_n+1) \hat{a}^\dagger\Psi_n N ^ a ^ † Ψ n = ( E n + 1 ) a ^ † Ψ n and
< a ^ † Ψ n ∣ a ^ † Ψ n > = < Ψ n ∣ a ^ a ^ † Ψ n > = < Ψ n ∣ [ a ^ , a ^ † ] + a ^ † a ^ Ψ n > = < Ψ n ∣ Ψ n > + < Ψ n ∣ N ^ Ψ n > = E n + 1 \begin{align}
\left<\hat{a}^\dagger\Psi_n|\hat{a}^\dagger\Psi_n\right> &=
\left<\Psi_n|\hat{a}\hat{a}^\dagger\Psi_n\right> \\
&= \left<\Psi_n|[\hat{a},\hat{a}^\dagger]+\hat{a}^\dagger\hat{a}\Psi_n\right> \\
&= \left<\Psi_n|\Psi_n\right> + \left<\Psi_n|\hat{N}\Psi_n\right> \\
&= E_n + 1
\end{align} ⟨ a ^ † Ψ n ∣ a ^ † Ψ n ⟩ = ⟨ Ψ n ∣ a ^ a ^ † Ψ n ⟩ = ⟨ Ψ n ∣ [ a ^ , a ^ † ] + a ^ † a ^ Ψ n ⟩ = ⟨ Ψ n ∣ Ψ n ⟩ + ⟨ Ψ n ∣ N ^ Ψ n ⟩ = E n + 1 Thus, we could define the recursive relation:
Ψ n + 1 = a ^ † Ψ n E n + 1 \Psi_{n+1} = \frac{\hat{a}^\dagger\Psi_n}{\sqrt{E_n+1}} Ψ n + 1 = E n + 1 a ^ † Ψ n Repeating the formula, we can get all the eigenstates of the number operator N ^ \hat{N} N ^ Ψ 0 ( x ) \Psi_0(x) Ψ 0 ( x )
Ψ n ( x ) = ( a ^ † ) n Ψ 0 ( x ) n ! \Psi_n(x) = \frac{(\hat{a}^\dagger)^n\Psi_0(x)}{\sqrt{n!}} Ψ n ( x ) = n ! ( a ^ † ) n Ψ 0 ( x ) As the creation operator a ^ † \hat{a}^\dagger a ^ † a ^ \hat{a} a ^ x ^ \hat{x} x ^ p ^ \hat{p} p ^ x ^ \hat{x} x ^ p ^ \hat{p} p ^ a ^ \hat{a} a ^ a ^ † \hat{a}^\dagger a ^ †
x ^ = 1 2 m ℏ ω ( a ^ + a ^ † ) p ^ = i 2 m ℏ ω ( a ^ − a ^ † ) \begin{align}
\hat{x} &= \frac{1}{\sqrt{2m\hbar\omega}}(\hat{a} + \hat{a}^\dagger) \\
\hat{p} &= \frac{i}{\sqrt{2m\hbar\omega}}(\hat{a} - \hat{a}^\dagger)
\end{align} x ^ p ^ = 2 m ℏ ω 1 ( a ^ + a ^ † ) = 2 m ℏ ω i ( a ^ − a ^ † ) Thus, we have
< Ψ n ∣ x ^ Ψ n > = 1 2 m ℏ ω < Ψ n ∣ ( a ^ + a ^ † ) Ψ n > = 0 \begin{align}
\left<\Psi_n|\hat{x}\Psi_n\right> &= \frac{1}{\sqrt{2m\hbar\omega}}\left<\Psi_n|(\hat{a} + \hat{a}^\dagger)\Psi_n\right> = 0 \\
\end{align} ⟨ Ψ n ∣ x ^ Ψ n ⟩ = 2 m ℏ ω 1 ⟨ Ψ n ∣ ( a ^ + a ^ † ) Ψ n ⟩ = 0 and
< Ψ n ∣ p ^ Ψ n > = i 2 m ℏ ω < Ψ n ∣ ( a ^ − a ^ † ) Ψ n > = 0 \begin{align}
\left<\Psi_n|\hat{p}\Psi_n\right> &= \frac{i}{\sqrt{2m\hbar\omega}}\left<\Psi_n|(\hat{a} - \hat{a}^\dagger)\Psi_n\right> = 0 \\
\end{align} ⟨ Ψ n ∣ p ^ Ψ n ⟩ = 2 m ℏ ω i ⟨ Ψ n ∣ ( a ^ − a ^ † ) Ψ n ⟩ = 0 Given an operator A ^ \hat{A} A ^
d d t < A > = d d t < Ψ ∣ A ^ Ψ > = d d t ∫ − ∞ ∞ Ψ ∗ A ^ Ψ d x = ∫ − ∞ ∞ ( ∂ Ψ ∗ ∂ t A ^ Ψ + Ψ ∗ ∂ A ^ ∂ t Ψ + Ψ ∗ A ^ ∂ Ψ ∂ t ) d x = < Ψ ∣ ∂ A ^ ∂ t Ψ > + ∫ − ∞ ∞ ( ∂ Ψ ∗ ∂ t A ^ Ψ + Ψ ∗ A ^ ∂ Ψ ∂ t ) d x = < Ψ ∣ ∂ A ^ ∂ t Ψ > + < ∂ Ψ ∂ t ∣ A ^ Ψ > + < Ψ ∣ A ^ ∂ Ψ ∂ t > = < Ψ ∣ ∂ A ^ ∂ t Ψ > + < ( 1 i ℏ H ^ Ψ ) ∣ A ^ Ψ > + < Ψ ∣ A ^ ( 1 i ℏ H ^ Ψ ) > = < Ψ ∣ ∂ A ^ ∂ t Ψ > + 1 i ℏ [ − < Ψ ∣ H ^ A ^ Ψ > + < Ψ ∣ A ^ H ^ Ψ > ] = < Ψ ∣ ∂ A ^ ∂ t Ψ > + 1 i ℏ < [ A ^ , H ^ ] > \begin{align}
\frac{d}{dt}\left<A\right> &= \frac{d}{dt}\left<\Psi|\hat{A}\Psi\right> \\
&= \frac{d}{dt} \int_{-\infty}^{\infty} \Psi^*\hat{A}\Psi dx \\
&= \int_{-\infty}^{\infty} \left(\frac{\partial \Psi^*}{\partial t}\hat{A}\Psi + \Psi^*\frac{\partial \hat{A}}{\partial t}\Psi + \Psi^*\hat{A}\frac{\partial \Psi}{\partial t}\right) dx \\
&= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right>
+\int_{-\infty}^{\infty}
\left(\frac{\partial \Psi^*}{\partial t}\hat{A}\Psi + \Psi^*\hat{A}\frac{\partial \Psi}{\partial t}\right) dx \\
&= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right>
+ \left<\frac{\partial \Psi}{\partial t}|\hat{A}\Psi\right>
+ \left<\Psi|\hat{A}\frac{\partial \Psi}{\partial t}\right> \\
&= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right>
+ \left<\left(\frac{1}{i\hbar}\hat{H}\Psi\right)|\hat{A}\Psi\right>
+ \left<\Psi|\hat{A}\left(\frac{1}{i\hbar}\hat{H}\Psi\right)\right> \\
&= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right>
+ \frac{1}{i\hbar}\left[
- \left<\Psi|\hat{H}\hat{A}\Psi\right>
+ \left<\Psi|\hat{A}\hat{H}\Psi\right>
\right] \\
&= \left<\Psi|\frac{\partial \hat{A}}{\partial t}\Psi\right>
+ \frac{1}{i\hbar}
\left<[\hat{A},\hat{H}]\right>
\end{align} d t d ⟨ A ⟩ = d t d ⟨ Ψ∣ A ^ Ψ ⟩ = d t d ∫ − ∞ ∞ Ψ ∗ A ^ Ψ d x = ∫ − ∞ ∞ ( ∂ t ∂ Ψ ∗ A ^ Ψ + Ψ ∗ ∂ t ∂ A ^ Ψ + Ψ ∗ A ^ ∂ t ∂ Ψ ) d x = ⟨ Ψ∣ ∂ t ∂ A ^ Ψ ⟩ + ∫ − ∞ ∞ ( ∂ t ∂ Ψ ∗ A ^ Ψ + Ψ ∗ A ^ ∂ t ∂ Ψ ) d x = ⟨ Ψ∣ ∂ t ∂ A ^ Ψ ⟩ + ⟨ ∂ t ∂ Ψ ∣ A ^ Ψ ⟩ + ⟨ Ψ∣ A ^ ∂ t ∂ Ψ ⟩ = ⟨ Ψ∣ ∂ t ∂ A ^ Ψ ⟩ + ⟨ ( i ℏ 1 H ^ Ψ ) ∣ A ^ Ψ ⟩ + ⟨ Ψ∣ A ^ ( i ℏ 1 H ^ Ψ ) ⟩ = ⟨ Ψ∣ ∂ t ∂ A ^ Ψ ⟩ + i ℏ 1 [ − ⟨ Ψ∣ H ^ A ^ Ψ ⟩ + ⟨ Ψ∣ A ^ H ^ Ψ ⟩ ] = ⟨ Ψ∣ ∂ t ∂ A ^ Ψ ⟩ + i ℏ 1 ⟨ [ A ^ , H ^ ] ⟩ 
