QM Week 3
2.36
Consider x ∈ [ − a , a ] x\in [-a,a] x ∈ [ − a , a ] 2.22 ~ 2.31 :2.26 : ψ ( − a ) = ψ ( a ) = 0 \psi(-a) = \psi(a) = 0 ψ ( − a ) = ψ ( a ) = 0
A sin ( − a k ) + B cos ( − a k ) = A sin ( a k ) + B cos ( a k ) = 0 A\sin(-ak) + B\cos(-ak) = A\sin(ak) + B\cos(ak) = 0
A sin ( − ak ) + B cos ( − ak ) = A sin ( ak ) + B cos ( ak ) = 0
⟹ { A sin ( a k ) = 0 B cos ( a k ) = 0 \implies \begin{cases}A\sin(ak) = 0\\ B\cos(ak) = 0 \end{cases}
⟹ { A sin ( ak ) = 0 B cos ( ak ) = 0
Case 1: if B = 0 , A ≠ 0 , sin ( a k ) = 0 B = 0,\,A\neq 0,\,\sin (ak) = 0 B = 0 , A = 0 , sin ( ak ) = 0
k n = n π a , n = 1 , 2 , … k_n = \frac{n\pi}a,\quad n = 1,2,\ldots
k n = a nπ , n = 1 , 2 , …
E n = n 2 π 2 ℏ 2 2 m a 2 E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}
E n = 2 m a 2 n 2 π 2 ℏ 2
∫ − a a ∣ A ∣ 2 sin 2 ( k x ) d x = ∣ A ∣ 2 a = 1 ⟹ A = 1 a \int_{-a}^a |A|^2 \sin^2(kx)dx = |A|^2 a = 1 \implies A = \sqrt{\frac 1 a}
∫ − a a ∣ A ∣ 2 sin 2 ( k x ) d x = ∣ A ∣ 2 a = 1 ⟹ A = a 1
ψ n ( x ) = 1 a sin ( n π a x ) \psi_n (x)= \sqrt\frac{1}{a} \sin(\frac{n\pi }a x)
ψ n ( x ) = a 1 sin ( a nπ x )
Case 2: if A = 0 , B ≠ 0 , cos ( a k ) = 0 A=0,\,B\neq 0,\, \cos(ak) = 0 A = 0 , B = 0 , cos ( ak ) = 0
k n = ( 2 n − 1 ) π 2 a , n = 1 , 2 , … k_n = \frac{(2n-1)\pi}{2a},\quad n = 1, 2,\ldots
k n = 2 a ( 2 n − 1 ) π , n = 1 , 2 , …
E n = ( 2 n − 1 ) 2 ℏ 2 π 2 8 m a 2 E_n = \frac{(2n-1)^2 \hbar^2\pi^2}{8ma^2}
E n = 8 m a 2 ( 2 n − 1 ) 2 ℏ 2 π 2
∫ − a a ∣ A ∣ 2 cos 2 ( k x ) d x = ∣ A ∣ 2 a = 1 ⟹ A = 1 a \int_{-a}^a |A|^2 \cos^2(kx)dx = |A|^2 a =1 \implies A = \sqrt\frac{1}a
∫ − a a ∣ A ∣ 2 cos 2 ( k x ) d x = ∣ A ∣ 2 a = 1 ⟹ A = a 1
ψ n ( x ) = 1 a cos ( ( 2 n − 1 ) π 2 a x ) \psi_n(x) = \sqrt\frac 1 a \cos(\frac{(2n-1)\pi}{2a}x)
ψ n ( x ) = a 1 cos ( 2 a ( 2 n − 1 ) π x )
2.38
a
2.18 :
Ψ n ( x , t ) = 2 a sin ( n π a x ) exp [ − i ( n 2 π 2 ℏ 2 m a 2 ) t ] \Psi_n (x,t) = \sqrt\frac 2 a \sin(\frac{n\pi}a x )\exp \left[-i\left(\frac{n^2\pi^2 \hbar}{2ma^2}\right)t\right ]
Ψ n ( x , t ) = a 2 sin ( a nπ x ) exp [ − i ( 2 m a 2 n 2 π 2 ℏ ) t ]
Ψ n ( x , 0 ) = 2 a sin ( n π a x ) \Psi_n (x, 0) = \sqrt\frac 2 a \sin(\frac{n\pi}a x )
Ψ n ( x , 0 ) = a 2 sin ( a nπ x )
Ψ n ( x , T ) = 2 a sin ( n π a x ) exp [ − i ( n 2 π 2 ℏ 2 m a 2 ) 4 m a 2 π ℏ ] = 2 a sin ( n π a x ) exp ( − 2 n 2 π i ) = Ψ n ( x , 0 ) \begin{aligned}
\Psi_n (x,T) &= \sqrt\frac 2 a \sin(\frac{n\pi}a x )\exp \left[-i\left(\frac{n^2\pi^2 \hbar}{2ma^2}\right)\frac{4ma^2}{\pi \hbar}\right ]\\
&=\sqrt\frac 2 a \sin(\frac{n\pi}a x ) \exp (-2n^2 \pi i)\\
&=\Psi_n (x,0)
\end{aligned}
Ψ n ( x , T ) = a 2 sin ( a nπ x ) exp [ − i ( 2 m a 2 n 2 π 2 ℏ ) π ℏ 4 m a 2 ] = a 2 sin ( a nπ x ) exp ( − 2 n 2 πi ) = Ψ n ( x , 0 )
Ψ ( x , T ) = ∑ c n Ψ n ( x , T ) = ∑ c n Ψ n ( x , 0 ) = Ψ ( x , 0 ) \Psi(x,T) = \sum c_n \Psi_n (x,T) = \sum c_n \Psi_n(x,0) = \Psi(x,0)
Ψ ( x , T ) = ∑ c n Ψ n ( x , T ) = ∑ c n Ψ n ( x , 0 ) = Ψ ( x , 0 )
b
E = 1 2 m v 2 , 2 a = t v , ∴ t = a 2 m E E = \frac 1 2 mv^2 ,2a = tv,\;\therefore t = a \sqrt \frac{2m}{E} E = 2 1 m v 2 , 2 a = t v , ∴ t = a E 2 m
c
t = T t = T t = T
E = π 2 ℏ 2 8 m a 2 E = \frac{\pi^2\hbar^2}{8ma^2}
E = 8 m a 2 π 2 ℏ 2
2.40
a
ξ = m ω ℏ x \xi = \sqrt{\frac{m\omega}\hbar}x ξ = ℏ mω x 2.86 :
ψ n ( x ) = ( m ω π ℏ ) 1 / 4 1 2 n n ! H n ( ξ ) e − ξ 2 / 2 \psi_n (x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^n n!}}H_n(\xi) e^{-\xi^2/2}
ψ n ( x ) = ( π ℏ mω ) 1/4 2 n n ! 1 H n ( ξ ) e − ξ 2 /2
Denote ( m ω π ℏ ) 1 / 4 \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} ( π ℏ mω ) 1/4 α \alpha α
ψ 0 = α e − ξ 2 / 2 ψ 1 = α 2 ξ e − ξ 2 / 2 ψ 2 = α ( 2 ξ 2 − 1 2 ) e − ξ 2 / 2 \psi_0 = \alpha e^{-\xi^2/2}\qquad \psi_1 = \alpha\sqrt 2\xi e^{-\xi^2/2} \qquad \psi_2 = \alpha (\sqrt 2 \xi^2 - \frac{1}{\sqrt 2})e^{-\xi^2/2}
ψ 0 = α e − ξ 2 /2 ψ 1 = α 2 ξ e − ξ 2 /2 ψ 2 = α ( 2 ξ 2 − 2 1 ) e − ξ 2 /2
ψ ( x , 0 ) = A exp ( − m ω 2 ℏ x 2 ) − 4 A m ω ℏ x exp ( − m ω 2 ℏ x 2 ) + 4 A m ω 2 ℏ x 2 exp ( − m ω 2 ℏ x 2 ) = A e − ξ 2 / 2 − 4 A ξ e − ξ 2 / 2 + 4 A ξ 2 e − ξ 2 / 2 = A α ( 3 ψ 0 − 2 2 ψ 1 + 2 2 ψ 2 ) \begin{aligned}
\psi(x,0) &= A \exp(-\frac{m\omega}{2\hbar}x^2) -4A\sqrt{\frac{m\omega}\hbar}x \exp(-\frac{m\omega}{2\hbar}x^2)+ 4A\frac{m\omega}{2\hbar}x^2\exp(-\frac{m\omega}{2\hbar}x^2)\\
& = A e^{- \xi^2/2 } - 4A \xi e^{- \xi^2/2 } + 4A \xi^2 e^{- \xi^2/2 }\\
& = \frac{A}\alpha (3 \psi_0 -2\sqrt 2 \psi_1 + 2\sqrt 2 \psi_2 )
\end{aligned}
ψ ( x , 0 ) = A exp ( − 2ℏ mω x 2 ) − 4 A ℏ mω x exp ( − 2ℏ mω x 2 ) + 4 A 2ℏ mω x 2 exp ( − 2ℏ mω x 2 ) = A e − ξ 2 /2 − 4 A ξ e − ξ 2 /2 + 4 A ξ 2 e − ξ 2 /2 = α A ( 3 ψ 0 − 2 2 ψ 1 + 2 2 ψ 2 )
1 = ∑ ∣ c n ∣ 2 = A 2 α 2 ( 9 + 8 + 8 ) ∴ A = ( m ω π ℏ ) 1 / 4 5 ; c 1 = 3 5 , c 2 = − 2 2 5 , c 3 = 2 2 5 1 = \sum |c_n|^2 = \frac{A^2}{\alpha^2}(9+8+8)\\\therefore A = \frac{\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}}5;\; c_1 = \frac 3 5,\, c_2 = -\frac{2\sqrt 2}{5},\, c_3 = \frac{2\sqrt 2}{5}
1 = ∑ ∣ c n ∣ 2 = α 2 A 2 ( 9 + 8 + 8 ) ∴ A = 5 ( π ℏ mω ) 1/4 ; c 1 = 5 3 , c 2 = − 5 2 2 , c 3 = 5 2 2
b
The possible energies are: E 0 = 1 2 ℏ ω , E 1 = 3 2 ℏ ω , E 2 = 5 2 ℏ ω E_0=\frac 1 2 \hbar\omega, E_1 = \frac 3 2 \hbar \omega , E_2 = \frac 5 2 \hbar \omega E 0 = 2 1 ℏ ω , E 1 = 2 3 ℏ ω , E 2 = 2 5 ℏ ω
⟨ H ⟩ = ∑ ∣ c n ∣ 2 E n = 73 50 ℏ ω \langle H \rangle = \sum |c_n|^2 E_n = \frac{73}{50}\hbar \omega
⟨ H ⟩ = ∑ ∣ c n ∣ 2 E n = 50 73 ℏ ω
c
T = π ω T = \frac \pi\omega T = ω π
2.41
Since it is impossible for the partical to appear at x < 0 x< 0 x < 0
E n = ( 2 n − 1 2 ) ℏ ω , n = 1 , 2 , … E_n = (2n-\frac 1 2) \hbar \omega,\quad n = 1,2,\ldots
E n = ( 2 n − 2 1 ) ℏ ω , n = 1 , 2 , …