QM Week 2
2.4
a
< x > = ∫ 0 a Ψ n ∗ x Ψ n d x = ∫ 0 a 2 a x sin 2 ( n π a x ) d x = a 2 − 1 a [ a 2 n π x sin ( 2 n π a x ) + ( a 2 n π ) 2 cos ( 2 n π a x ) ] ∣ 0 a = a 2 \begin{aligned}
<x> &= \int_0^a \Psi_n^* x \Psi_n dx \\
&= \int_0^a \frac{2}{a} x \sin^2\left(\frac{n\pi}{a} x\right) dx \\
&= \frac{a}{2} - \frac{1}{a} \left[\frac{a}{2n\pi} x \sin\left(\frac{2n\pi}{a} x\right) + \left(\frac{a}{2n\pi}\right)^2 \cos\left(\frac{2n\pi}{a} x\right)\right] \Bigg|_0^a\\
&= \frac{a}{2}
\end{aligned} < x > = ∫ 0 a Ψ n ∗ x Ψ n d x = ∫ 0 a a 2 x sin 2 ( a nπ x ) d x = 2 a − a 1 [ 2 nπ a x sin ( a 2 nπ x ) + ( 2 nπ a ) 2 cos ( a 2 nπ x ) ] 0 a = 2 a
b
< x 2 > = ∫ 0 a Ψ n ∗ x 2 Ψ n d x = ∫ 0 a 2 a x 2 sin 2 ( n π a x ) d x = 2 a ∫ 0 a x 2 [ 1 2 − 1 2 cos ( 2 n π a x ) ] d x = a 2 3 − a 2 2 n 2 π 2 \begin{aligned}
<x^2> &= \int_0^a \Psi_n^* x^2 \Psi_n \, dx \\
&= \int_0^a \frac{2}{a} x^2 \sin^2\left(\frac{n\pi}{a} x\right) dx \\
&= \frac{2}{a} \int_0^a x^2 \left[\frac{1}{2} - \frac{1}{2}\cos\left(\frac{2n\pi}{a} x\right)\right] dx \\
&= \frac{a^2}{3} - \frac{a^2}{2n^2\pi^2}
\end{aligned} < x 2 > = ∫ 0 a Ψ n ∗ x 2 Ψ n d x = ∫ 0 a a 2 x 2 sin 2 ( a nπ x ) d x = a 2 ∫ 0 a x 2 [ 2 1 − 2 1 cos ( a 2 nπ x ) ] d x = 3 a 2 − 2 n 2 π 2 a 2
c
< p > = ∫ 0 a − 2 a sin ( n π a x ) i ℏ n π a cos ( n π a x ) d x = 0 <p> = \int_0^a -\frac{2}a\sin(\frac{n\pi}a x)i\hbar \frac{n\pi}a \cos(\frac{n\pi}a x)dx = 0
< p >= ∫ 0 a − a 2 sin ( a nπ x ) i ℏ a nπ cos ( a nπ x ) d x = 0
d
< p 2 > = ℏ 2 ( n π a ) 2 ∫ 0 a Ψ ∗ Ψ d x = ( n ℏ π a ) 2 <p^2> = \hbar^2(\frac{n\pi}a)^2 \int_0^a \Psi^*\Psi dx =(\frac{n\hbar\pi}a)^2
< p 2 >= ℏ 2 ( a nπ ) 2 ∫ 0 a Ψ ∗ Ψ d x = ( a n ℏ π ) 2
e
σ x = a 1 12 − 1 2 n 2 π 2 \sigma_x = a\sqrt{\frac{1}{12} - \frac{1}{2n^2\pi^2}}
σ x = a 12 1 − 2 n 2 π 2 1
f
σ p = n ℏ π a \sigma_p = \frac{n\hbar \pi}a
σ p = a n ℏ π
g
σ x σ p = ℏ 2 n 2 π 2 3 − 2 ≥ ℏ 2 , n = 1 , 2 , … \sigma_x\sigma_p = \frac\hbar 2 \sqrt{\frac{n^2\pi^2}3 - 2}\geq \frac\hbar 2,\quad n = 1,2,\ldots
σ x σ p = 2 ℏ 3 n 2 π 2 − 2 ≥ 2 ℏ , n = 1 , 2 , …
When n = 1 n = 1 n = 1 σ x σ p \sigma_x\sigma_p σ x σ p
2.5
a
Since ω = π 2 ℏ / 2 m a 2 \omega = \pi^2\hbar/2ma^2 ω = π 2 ℏ/2 m a 2
1 = ∫ A 2 ( ψ 1 ∗ ψ 1 + ψ 1 ∗ ψ 2 + ψ 2 ∗ ψ 1 + ψ 2 ∗ ψ 2 ) d x = 2 A 2 A = 1 2 1 = \int A^2 (\psi_1^*\psi_1 + \psi_1^*\psi_2+\psi_2^*\psi_1+\psi_2^*\psi_2)dx = 2A^2\\ A = \frac{1}{\sqrt 2}
1 = ∫ A 2 ( ψ 1 ∗ ψ 1 + ψ 1 ∗ ψ 2 + ψ 2 ∗ ψ 1 + ψ 2 ∗ ψ 2 ) d x = 2 A 2 A = 2 1
b
Ψ ( x , t ) = 1 2 [ ψ 1 e − i ℏ E 1 t / ℏ + ψ 2 e − i E 2 t / ℏ ] = 1 2 [ ψ 1 e − i ω t + ψ 2 e − 4 i ω t ] = 1 a [ sin ( π x a ) e − i ω t + sin ( 2 π x a ) e − 4 i ω t ] \begin{aligned}\Psi(x,t) &= \frac 1{\sqrt{2}}[\psi_1 e^{-i\hbar E_1t/\hbar} + \psi_2 e^{-iE_2t/\hbar}]\\
&=\frac 1{\sqrt{2}}[\psi_1 e^{-i\omega t} + \psi_2 e^{-4i\omega t}]\\
&=\frac{1}{\sqrt a}[\sin(\frac{\pi x }a)e^{-i\omega t} + \sin(\frac{2\pi x }a)e^{-4i\omega t}]\end{aligned} Ψ ( x , t ) = 2 1 [ ψ 1 e − i ℏ E 1 t /ℏ + ψ 2 e − i E 2 t /ℏ ] = 2 1 [ ψ 1 e − iω t + ψ 2 e − 4 iω t ] = a 1 [ sin ( a π x ) e − iω t + sin ( a 2 π x ) e − 4 iω t ]
∣ Ψ ( x , t ) ∣ 2 = 1 a [ sin 2 ( π a x ) + sin 2 ( 2 π a x ) + 2 sin ( π a x ) sin ( 2 π a x ) cos ( 3 ω t ) ] |\Psi(x,t)|^2 = \frac{1}{a}[\sin^2(\frac\pi a x )+ \sin^2(\frac {2\pi}a x ) +2\sin(\frac\pi a x )\sin(\frac {2\pi}a x )\cos(3\omega t)]
∣Ψ ( x , t ) ∣ 2 = a 1 [ sin 2 ( a π x ) + sin 2 ( a 2 π x ) + 2 sin ( a π x ) sin ( a 2 π x ) cos ( 3 ω t )]
c
< x > = ∫ Ψ ∗ x Ψ d x = 1 2 ∫ x [ ψ 1 ∗ ψ 1 + ψ 2 ∗ ψ 2 + 2 ψ 1 ∗ ψ 2 e − 3 i ω t ] d x = 1 2 ∫ ψ 1 ∗ x ψ 1 d x + 1 2 ∫ ψ 2 ∗ x ψ 2 d x + ∫ x ψ 1 ∗ ψ 2 e − 3 i ω t d x = a 4 + a 4 − 16 a 9 π 2 cos ( 3 ω t ) = a 2 − 16 a 9 π 2 cos ( 3 ω t ) \begin{aligned}<x> &= \int \Psi^* x\Psi dx\\
& = \frac{1}2\int x[\psi_1^*\psi_1 + \psi_2^*\psi_2+2\psi_1^*\psi_2 e^{-3i\omega t}]\, dx\\
&= \frac 1 2 \int \psi_1^* x \psi_1 dx + \frac 1 2\int \psi_2^* x \psi_2 dx + \int x\psi_1^*\psi_2 e^{-3i\omega t}\, dx\\
& = \frac a 4+ \frac a 4 -\frac{16a}{9\pi^2}\cos(3\omega t) \\
& = \frac a 2 - \frac{16a}{9\pi^2}\cos(3\omega t)
\end{aligned} < x > = ∫ Ψ ∗ x Ψ d x = 2 1 ∫ x [ ψ 1 ∗ ψ 1 + ψ 2 ∗ ψ 2 + 2 ψ 1 ∗ ψ 2 e − 3 iω t ] d x = 2 1 ∫ ψ 1 ∗ x ψ 1 d x + 2 1 ∫ ψ 2 ∗ x ψ 2 d x + ∫ x ψ 1 ∗ ψ 2 e − 3 iω t d x = 4 a + 4 a − 9 π 2 16 a cos ( 3 ω t ) = 2 a − 9 π 2 16 a cos ( 3 ω t )
d
< p > = m d < x > d t = 8 ℏ 3 a sin ( 3 ω t ) <p> = m\frac{d<x>}{dt}= \frac{8\hbar}{3a}\sin(3\omega t)
< p >= m d t d < x > = 3 a 8ℏ sin ( 3 ω t )
e
E n = n 2 π 2 ℏ 2 2 m a 2 for: n = 1 , 2 E_n = \frac{n^2\pi^2\hbar^2}{2ma^2} \quad \text{for: } n = 1,2
E n = 2 m a 2 n 2 π 2 ℏ 2 for: n = 1 , 2
Since c 1 = c 2 = 1 2 c_1 = c_2 = \frac{1}{\sqrt 2} c 1 = c 2 = 2 1
< H > = 1 2 ( E 1 + E 2 ) = 5 π 2 ℏ 2 4 m a 2 <H> = \frac{1}2 (E_1+E_2) = \frac{5\pi^2\hbar^2}{4ma^2}
< H >= 2 1 ( E 1 + E 2 ) = 4 m a 2 5 π 2 ℏ 2
2.6
Since e i ϕ ψ 2 e^{i\phi}\psi_2 e i ϕ ψ 2 ψ 1 \psi_1 ψ 1 2.5 (a) remains:
A = 1 2 , Ψ ( x , 0 ) = 1 2 ( ψ 1 + e i ϕ ψ 2 ) A = \frac{1}{\sqrt 2} ,\quad \Psi(x,0) = \frac{1}{\sqrt 2}(\psi_1+e^{i\phi}\psi_2)
A = 2 1 , Ψ ( x , 0 ) = 2 1 ( ψ 1 + e i ϕ ψ 2 )
Ψ ( x , t ) = 1 a [ sin ( π x a ) e − i ω t + sin ( 2 π x a ) e i ( ϕ − 4 ω t ) ] \Psi(x,t) = \frac{1}{\sqrt a}[\sin(\frac{\pi x }a)e^{-i\omega t} + \sin(\frac{2\pi x }a)e^{i(\phi - 4\omega t)}]
Ψ ( x , t ) = a 1 [ sin ( a π x ) e − iω t + sin ( a 2 π x ) e i ( ϕ − 4 ω t ) ]
∣ Ψ ( x , t ) ∣ 2 = 1 a [ sin 2 ( π a x ) + sin 2 ( 2 π a x ) + 2 sin 2 ( π a x ) sin 2 ( 2 π a x ) cos ( 3 ω t − ϕ ) ] |\Psi(x,t)|^2 = \frac{1}{a}[\sin^2(\frac\pi a x )+ \sin^2(\frac {2\pi}a x ) +2\sin^2(\frac\pi a x )\sin^2(\frac {2\pi}a x )\cos(3\omega t-\phi)]
∣Ψ ( x , t ) ∣ 2 = a 1 [ sin 2 ( a π x ) + sin 2 ( a 2 π x ) + 2 sin 2 ( a π x ) sin 2 ( a 2 π x ) cos ( 3 ω t − ϕ )]
< x > = C <x> = C
< x >= C
When ϕ = π / 2 \phi = \pi/2 ϕ = π /2 Ψ ( x , 0 ) = A [ ψ 1 − i ψ 2 ] \Psi(x,0) = A[\psi_1 - i\psi_2] Ψ ( x , 0 ) = A [ ψ 1 − i ψ 2 ] < x > ∣ t = 0 = a / 2 <x>|_{t=0} = a/2 < x > ∣ t = 0 = a /2 ϕ = π \phi = \pi ϕ = π Ψ ( x , 0 ) = A [ ψ 1 − ψ 2 ] \Psi(x,0) = A[\psi_1 - \psi_2] Ψ ( x , 0 ) = A [ ψ 1 − ψ 2 ] < x > ∣ t = 0 = a 2 + 16 a 9 π 2 <x>|_{t=0} = \frac a 2 + \frac{16a}{9\pi^2} < x > ∣ t = 0 = 2 a + 9 π 2 16 a
2.7
a
1 = ∫ 0 a ∣ Ψ ∣ 2 d x = 2 ∫ 0 a 2 ( A x ) 2 d x ⟹ A = 12 a 3 1 = \int_0^a |\Psi|^2 dx = 2\int_0^{\frac a 2} (Ax)^2 dx \implies A = \sqrt{\frac{12}{a^3}}
1 = ∫ 0 a ∣Ψ ∣ 2 d x = 2 ∫ 0 2 a ( A x ) 2 d x ⟹ A = a 3 12
b
∑ c n 2 a sin ( n π a x ) \sum c_n \sqrt\frac{2}{a}\sin(\frac{n\pi}a x) ∑ c n a 2 sin ( a nπ x ) Ψ ( x , 0 ) \Psi(x,0) Ψ ( x , 0 )
c n = ∫ 0 a ψ n ∗ Ψ ( x , 0 ) d x = ∫ 0 a / 2 2 a sin ( n π a x ) A x d x + ∫ a / 2 a 2 a sin ( n π a x ) A ( a − x ) d x let: y = x − a , x = y − a = ∫ 0 a / 2 2 a sin ( n π a x ) A x d x − ∫ 0 a / 2 2 a sin ( n π a ( a − y ) ) A y d y = { A 2 a ∫ 0 a / 2 2 x sin ( n π a x ) d x n is odd 0 n is even = { 4 6 n 2 π 2 sin ( n π 2 ) n is odd 0 n is even \begin{aligned}c_n &= \int_0^a \psi_n^* \Psi(x,0) dx\\
&=\int_0^{a/2} \sqrt\frac{2}{a}\sin(\frac{n\pi}{a}x)Ax \,dx + \int_{a/2}^a \sqrt\frac{2}{a}\sin(\frac{n\pi}{a}x)A(a-x) dx\\
&\quad \text{let: } y=x-a,\,x = y-a\\
&=\int_0^{a/2} \sqrt\frac{2}{a}\sin(\frac{n\pi}{a}x)Ax \,dx - \int_0^{a/2} \sqrt\frac{2}{a}\sin(\frac{n\pi}{a}(a-y))Ay\, dy\\
&=\begin{cases}A\sqrt\frac{2}{a}\int_0^{a/2} 2x\sin(\frac{n\pi}{a}x)\,dx\quad & n \text{ is odd}\\
0 & n \text{ is even}\end{cases} \\
&= \begin{cases}
\frac{4\sqrt 6}{n^2\pi^2}\sin(\frac{n\pi}2) &n \text{ is odd}\\
0 & n \text{ is even}\end{cases}
\end{aligned} c n = ∫ 0 a ψ n ∗ Ψ ( x , 0 ) d x = ∫ 0 a /2 a 2 sin ( a nπ x ) A x d x + ∫ a /2 a a 2 sin ( a nπ x ) A ( a − x ) d x let: y = x − a , x = y − a = ∫ 0 a /2 a 2 sin ( a nπ x ) A x d x − ∫ 0 a /2 a 2 sin ( a nπ ( a − y )) A y d y = { A a 2 ∫ 0 a /2 2 x sin ( a nπ x ) d x 0 n is odd n is even = { n 2 π 2 4 6 sin ( 2 nπ ) 0 n is odd n is even
c
c 1 = 4 6 π 2 , P 1 = ∣ c 1 ∣ 2 = 96 π 4 c_1 =\frac{4\sqrt 6}{\pi^2}, \quad P_1 = |c_1|^2 = \frac{96}{\pi^4}
c 1 = π 2 4 6 , P 1 = ∣ c 1 ∣ 2 = π 4 96
d
P n = ∣ c n ∣ 2 = { 96 n 4 π 4 n is odd 0 n is even P_n = |c_n|^2 = \begin{cases}\frac{96}{n^4\pi^4} &n \text{ is odd}\\0 & n \text{ is even}\end{cases}
P n = ∣ c n ∣ 2 = { n 4 π 4 96 0 n is odd n is even
< H > = ∑ n P n E n = ∑ n = 1 96 π 4 1 n 4 n 2 π 2 ℏ 2 2 m a 2 for n = 1 , 3 , 5 , … = 48 ℏ 2 π 2 m a 2 ∑ n = 1 1 n 2 for n = 1 , 3 , 5 , … = 6 ℏ 2 m a 2 \begin{aligned}<H> &= \sum_n P_n E_n\\
&= \sum_{n=1}\frac{96}{\pi^4}\frac{1}{n^4} \frac{n^2\pi^2\hbar^2}{2ma^2} \text{\; for } n = 1,3,5,\ldots\\
&=\frac{48 \hbar^2}{\pi^2 ma^2} \sum_{n=1}\frac{1}{n^2}\text{\; for } n = 1,3,5,\ldots \\
&=\frac{6\hbar^2}{ma^2}
\end{aligned} < H > = n ∑ P n E n = n = 1 ∑ π 4 96 n 4 1 2 m a 2 n 2 π 2 ℏ 2 for n = 1 , 3 , 5 , … = π 2 m a 2 48 ℏ 2 n = 1 ∑ n 2 1 for n = 1 , 3 , 5 , … = m a 2 6 ℏ 2
2.8
E 1 = π 2 ℏ 2 2 m a 2 E_1 = \frac{\pi^2\hbar^2}{2ma^2}
E 1 = 2 m a 2 π 2 ℏ 2
c 1 = ∫ 0 a Ψ ∗ ψ 1 d x = ∫ 0 a / 2 2 a A sin ( π a x ) d x = A 2 a 2 \begin{aligned}c_1 &= \int_0^a \Psi^* \psi_1 dx \\
&= \int_0^{a/2}\sqrt{ \frac{2}{a}}A\sin(\frac{\pi}{a}x) dx\\
&= \frac{A\sqrt{ 2a}}{2}
\end{aligned} c 1 = ∫ 0 a Ψ ∗ ψ 1 d x = ∫ 0 a /2 a 2 A sin ( a π x ) d x = 2 A 2 a
∫ ∣ Ψ ∣ 2 d x = 1 ⟹ A = 2 a \int |\Psi|^2 dx =1\implies A = \sqrt{\frac{2}{a}}
∫ ∣Ψ ∣ 2 d x = 1 ⟹ A = a 2
∴ c 1 = 2 π , P 1 = 4 π 2 \therefore c_1 = \frac 2 \pi,\quad P_1 = \frac 4 {\pi^2}
∴ c 1 = π 2 , P 1 = π 2 4
2.9
< H > = ∫ 0 a Ψ ∗ H ^ Ψ d x = ∫ 0 a A 2 x ( a − x ) ℏ 2 m d x = 5 ℏ 2 m a 2 \begin{aligned}
<H> &= \int_0^a \Psi^* \hat H \Psi dx\\
&=\int_0^a A^2 x(a-x) \frac{\hbar^2}{m} dx\\
&=\frac{5\hbar^2}{ma^2}\end{aligned}
< H > = ∫ 0 a Ψ ∗ H ^ Ψ d x = ∫ 0 a A 2 x ( a − x ) m ℏ 2 d x = m a 2 5 ℏ 2
Which is the same as the result in Example 2.21