One-Dimensional Bound States: Difference between revisions

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When the energy of a particle is less than the potential at both positive and negative infinity, the particle is '''trapped''' in a well and it goes back and forth between the turning points in the potential with its kinetic energy; however, when the energy is larger than the potential at either infinity, the particle is said to be "unbound". For a bound state, the wavefunction must decay at least exponentially at both infinities.
{{Quantum Mechanics A}}
When the energy of a particle moving in one dimension is less than the potential at <math>x\to\pm\infty</math>, the particle is trapped in a potential well and is said to be in a bound state.  However, when the energy is larger than the potential at either infinity, the particle is said to be in a scattering state. The wave function for a bound state must decay at least exponentially at both infinities.


=Basic Properties=  
==Basic Properties==


'''1. Non-degeneracy of the bound states in 1D'''
'''1. The bound states of a one-dimensional potential are non-degenerate.'''


Let's consider a more general property that is the [http://en.wikipedia.org/wiki/Wronskian Wronskian] for 2 solutions of the 1-D Schrodinger equation must equal to a constant.
To prove this, let us suppose that there are two different wave functions, <math>\psi_E^{(1)}</math> and <math>\psi_E^{(2)},</math> for a given bound state with energy <math>E\!</math>.  They must then both satisfy the same Schrödinger equation:
   
Schrodinger equation :
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi</math>
 
is a second-order differential equation. Such equation has 2 linearly independent
<math>\psi_E^{(1)}</math> and <math>\psi_E^{(2)}</math> for each value of <math>E\!</math>:


<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}</math>
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}</math>
Line 17: Line 12:
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}</math>  
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}</math>  


By definition in mathematics, the Wronskian of these functions is:
Multiplying the first equation by <math>\psi_E^{(2)}</math> and the second by <math>\psi_E^{(1)}</math>, then subtracting one equation from the other, we get
<math>W=\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}</math>
 
 
<math>\frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0,</math>
Multiplying equation (2) by <math>\psi_E^{(2)}</math>, equation (3) by <math>\psi_E^{(1)}</math>,
 
then subtracting one equation from the other, we get:
or
 
<math>\rightarrow \frac{d}{dx}\left (\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}\right )=0.</math>
 
We recognize the quatity <math>W(x)=\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}</math> as the [http://en.wikipedia.org/wiki/Wronskian Wronskian] of the two wave functions.  We therefore see that <math>W(x)=C,</math> where <math>C</math> is constant.  For bound states, the wave functions must vanish at infinity, and therefore <math>W=0,</math> or
 
<math>\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0.</math>  
 
Rearranging, we obtain
 
<math>\frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0.</math>
 
We can now integrate this equation to obtain
 
<math>\ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=A,</math>
 
or
 
<math>\psi_E^{(1)}(x)=a\psi_E^{(2)}(x).</math>
 
We see that the two wave functions must be proportional to each other; after normalization, they will become completely identical.  Therefore, we see that the two wave functions correspond to the same state, and thus all bound states in one dimension are non-degenerate.
 
'''2. The wave function for a real potential <math>V(x)</math> can be chosen to be real.'''


<math>\frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0</math>
If we take the complex conjugate of the Schrödinger equation for the system,


<math>\rightarrow \frac{d}{dx}\left(\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}\right)=0</math>
<math> E\psi=-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi,</math>


<math>\rightarrow \frac{dW}{dx}=0</math>
then we obtain
<math> E\psi^{\ast}=-\frac{\hbar^2}{2m}\frac{d^2\psi^{\ast}}{dx^2}+V(x)\psi^{\ast}.</math>


<math>\rightarrow W=C</math>
Since both <math>\psi\!</math> and <math>\psi^{\ast}</math> satisfy the same [[Schrödinger Equation|Schrödinger equation]], their sum <math>\psi+\psi^{\ast}</math> must also be a solution.  This sum is real, and thus we see that the wave function, as asserted, can be chosen to be real.


where <math>C\!</math> is constant.
'''3. The <math>n</math>th excited state has <math>n</math> nodes.'''
So, the [http://en.wikipedia.org/wiki/Wronskian Wronskian] for 2 solutions of the 1-D Schrodinger equation must equal to a constant.
   
For the bound states, the wave function vanish at infinity, i.e:
<math>\psi_E^{(1)}(\infty)=\psi_E^{(2)}(\infty)=0</math>


From (4), (5) and (6), it follows that <math>W=0\!</math>
See the [[Oscillation Theorem]].
From (4) and (7), we get:
<math>\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0</math>


<math>\rightarrow \frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0</math>
== Parity and the Symmetry of the Wave Functions ==


<math>\rightarrow \frac{d}{dx}\left[\ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})\right]=0</math>
One useful theorem about one-dimensional potentials is that, if the potential <math>V(x)\!</math> is even (i.e. <math>V(x)=V(-x)\!</math>), then the eigenstates can be taken to be even or odd.


<math>\rightarrow \ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=\mbox{const.}</math>
To prove this, we first note that, if <math>\psi(x)</math> satisfies the Schrödinger equation for the potential,


<math>\rightarrow \psi_E^{(1)}=\mbox{const.}\psi_E^{(2)}</math>
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi,</math>


From (8) it follows that <math>\psi_E^{(1)}</math> and <math>\psi_E^{(2)}</math> describe the same state. Therefore, the bound states in 1D are non-degenerate.
then so will <math>\psi(-x)\!</math>.  We may see this by simply taking <math>x\to -x\!</math> in the above equation.  Since both <math>\psi(x)\!</math> and <math>\psi(-x)\!</math> are solutions of the Schrödinger equation, any linear combination of the two must also be a solution. In particular, <math>\psi_e(x)=\psi(x)+\psi(-x)\!</math> and <math>\psi_o(x)=\psi(x)-\psi(-x)\!</math> are solutions.  We may easily see that <math>\psi_e(x)\!</math> is even, while <math>\psi_o(x)\!</math> is odd.


In particular, this implies that all bound states in such a potential ''must'' be either even or odd.  Since <math>\psi(x)\!</math> and <math>\psi(-x)\!</math> both solve the Schrödinger equation, and because the eigenstates must be non-degenerate, we see that <math>\psi(x)\!</math> and <math>\psi(-x)\!</math> must be proportional to one another.  We may easily show that the proportionality constant must be <math>\pm 1\!</math>, corresponding to either an even function or an odd function.


'''2. The wave function for a real potential V(x) can be chosen real.'''
The above discussion is a simple illustration of how the symmetry of the Hamiltonian of a system dictates the transformation properties of the eigenstates, [[Transformations of Operators and Symmetry|as described earlier]].


<math> E\psi(\textbf{r})=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(\textbf{r})}{\partial x^2}+V(\textbf{r})\psi(\textbf{r})</math>
==Infinite Square Well==
Let's consider the motion of a particle in an infinite and symmetric square well: <math>V(x)=+\infty</math> for <math>|x| \ge L/2,</math> and <math>V(x)=0\!</math> otherwise.


so:
A particle subject to this potential is free everywhere except at the two ends <math>(x = \pm L/2),</math> where the infinite potential keeps the particle confined to the well.  Within the well the Schrödinger equation takes the form,
<math> E\psi^*(\textbf{r})=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi^*(\textbf{r})}{\partial x^2}+V(\textbf{r})</math>
:<math>-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi,</math>
<math>\rightarrow\psi^*=C\psi</math>
or equivalently,
<math>\rightarrow\psi=C^*\psi^*</math>
:<math>\frac{d^2\psi}{dx^2}=-k^2\psi,</math>  
<math>\rightarrow\psi^*=|C|^2\psi^*</math>
where     
<math>\rightarrow|C|^2=1</math>
:<math>k=\frac{\sqrt{2mE}}{\hbar}.</math>
Writing the Schrödinger equation in this form, we see that our solution are simply


so:
<math>\psi(x) = A \sin (kx) + B \cos (kx).\!</math>
<math>C=e^{i\theta}\!</math>
                                 
<math>\rightarrow\psi^*=e^{i\theta}\psi</math>
As we showed earlier, the eigenstates of this potential, all of which are bound states, must be either even or odd.  This means that we must either set <math>A=0</math> or <math>B=0.</math> Let us consider each case in turn.


let <math>\theta=0\!</math>
'''Case 1:''' <math>B=0\!</math>.
<math>\rightarrow\psi^*=\psi</math>
So, the wave function is real.


In this case,


'''3. The nth excited state has n nodes.'''
:<math>\psi(x)=A\sin{kx}.\!</math>


See the ''Oscillation theorem'' below.
We require that the wave function vanish at the ends, so that


= Parity operator and the symmetry of the wavefunctions =
:<math>\sin\left (\frac{kL}{2}\right )=0.\!</math>


The solutions of this equation are


In the above problem, two basic solutions of Schrodinger equation are either odd or even. The general wavefunctions are combinations of odd and even functions. This property originates from the fact that the potential is symmetric or invariant under the inversion <math>x \rightarrow -x</math>, and so does the Hamiltonian. Therefore, the Hamiltonian commutes with the parity operator <math>\hat{P}</math> where <math>\hat{P} \psi(x)=\psi(-x)</math>, or <math> \hat{P}\left |{\mathbf{r}} \right \rangle=\left |{-\mathbf{r}} \right \rangle</math> In this case, the wavefunctions themselves do not need to be odd or even, but they must be some combinations of odd and even functions.
:<math>k=\frac{2n\pi}{L},\!</math>
where <math>n=1,2,3,\ldots\!</math>


One useful law about parity is that, for 1D bound state, if the potential V(x) is inversion symmetric, its wave function is either even or odd.
'''Case 2:''' <math>A=0.\!</math>


We will prove it step by step:
In this case,


1. If potential V(x) is inversion symmetric, <math>\psi (\mathbf{r}))</math> and <math>\psi (-\mathbf{r})</math> are wave functions with the same eigenvalue.
:<math>\psi(x)=B\cos{kx}.\!</math>


Prove:
Again requiring that the wave function vanish at the ends, i.e.
Just change the sign of x, and V(x)=V(-x).


2. If potential V(x) is inversion symmetric, as for every eigenvalue, '''we can find''' a complete set of eigenfunctions, which are either even or odd.
:<math>\cos\left (\frac{kL}{2}\right )=0,\!</math>


Prove:
we find that
If  <math>\psi (\mathbf{r})</math> is a solution of one stationary Schrodinger function, then <math>\psi (-\mathbf{r})</math> is another solution.
Let's formalize
:<math>\text{u(r)}=\psi (\mathbf{r})+\psi (-\mathbf{r})</math> and <math>\text{v(r)=}\psi (\mathbf{r})-\psi (-\mathbf{r})</math>


3. So <math>\text{u(r)}</math> is even and <math>\text{v(r)}</math> is odd. Because they only have one eigenfunction.
:<math>k=\frac{(2n+1)\pi}{L}.\!</math>


4. For 1D bound state, if the potential V('''r''') is inversion symmetric, its wave function is either even or odd.
Combining these two results, we find that the energy eigenvalues are <math> E_n = \frac{\hbar^2}{2m} \left(\frac{\pi n}{L} \right )^2 \!</math>, where <math>n = 1,2,3,\ldots\!</math>  The wave numbers are quantized as a result of the boundary conditions, and thus the energies are quantized as well.  The lowest energy state is the ground state, and it has a non-zero energy, which is due to quantum zero point motion.  The ground state is nodeless, the first excited state has one node, the second excited state has two nodes, and so on.  The wavefunctions are also orthogonal.


[[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/doubledelta]]
==Finite Asymmetric Square Well==


=Infinite square well=
We now consider an asymmetric square well potential <math>V(x),</math> defined as
Let's consider the motion of a particle in an infinite and symmetric square well: <math>V(x)=+\infty</math> for <math>x \ge |L|/2</math>, otherwise  <math>V(x)=0\!</math>


A particle subject to this potential is free everywhere except at the two ends (<math>x = \pm L/2</math>), where the infinite potential keeps the particle confined to the well.  Within the well the Schrodinger equation takes the form:
<math>V(x)=\begin{cases}
:<math>-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi</math>
V_1,\,x<0 \\
or equivalently,
0,\,0<x<a \\
:<math>\frac{d^2\psi}{dx^2}=-k^2\psi</math>  
V_2,\,x>a
where     
\end{cases},</math>
:<math>k=\frac{\sqrt{2mE}}{\hbar}</math>
Writing the Schrodinger equation in this form, we see that our solution are simply


<math>\psi(x) = A \sin (kx) + B \cos (kx)\!</math>:
where both <math>V_1\!</math> and <math>V_2\!</math> are positive.  We may then split our one-dimensional space into three regions; region I is given by <math>x<0\!</math>, region II by <math>0<x<a\!</math>, and region III by <math>x>a\!</math>.  The Schrödinger equation may be written in each region as
                                 
Now we impose that the solution must vanish at <math>x = \pm L/2</math>:
:<math>-A\sin(kL/2)+B\cos(kL/2)=0\!</math>


:<math>A\sin(kL/2)+B\cos(kL/2)=0\!</math>
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_{I}(x)}{dx^2}+V_1\psi_{I}(x)=E\psi_{I}(x),</math>
Adding the two equations, we get:
:<math>2B\cos(kL/2)=0\!</math>
   
It follows that either <math>B=0\!</math> or <math>\cos(kL/2)=0\!</math>.


Case 1: <math>B=0\!</math>.
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_{II}(x)}{dx^2}=E\psi_{II}(x),</math>
In this case <math>A\ne0\!</math>, otherwise the wavefunction vanishes every where. Furthermore, it is required that:
<math>\sin(kL/2)=0\!</math>


<math>\rightarrow k=2n\pi/L\!</math>
and
where <math>n=1,2,3,...\!</math>


And the wave functions are odd:
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_{III}(x)}{dx^2}+V_2\psi_{III}(x)=E\psi_{III}(x).</math>
<math>\psi(x)=A\sin(2n\pi x/L)\!</math>


Case 2: <math>\cos(kL/2)=0\!</math>   <math>\rightarrow k=(2n+1)\pi/L\!</math>
We will be considering bound states here, so that <math>E<\min(V_1,V_2).\!</math> In this case, the solutions to the above Schrödinger equations are
where <math>n=0,1,2,...\!</math>


In this case <math>A=0\!</math> and <math>B\ne0\!</math>, and the wavefunctions are even:
<math>\psi_{I}(x)=Ae^{k_{1}x},</math>
<math>\psi(x)=B\cos[(2n+1)\pi x/L]\!</math>


The two solutions give the eigenenergies <math> E_n = \frac{\hbar^2}{2m} \left(\frac{\pi n}{L} \right )^2 \!</math>, where <math>n = 1,2,3,...\!</math> These wave numbers are quantized as a result of the boundary conditions, thus making the energy quantized as well.  The lowest energy state is the ground state, and it has a non-zero energy, which is due to quantum zero point motion.  This ground state is also nodeless, the first excited state has one node, the second excited state has two nodes, and so on.  The wavefunctions are also orthogonal.
<math>\psi_{II}(x)=Be^{ik_{2}x}+Ce^{-ik_{2}x},</math>


and


[[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Particle_bouncing_on_the_floor_in_Earth%27s_Gravitational_field]]
<math>\psi_{III}(x)=De^{-k_{3}x},</math>


=Finite asymmetric square well=
where,
<math>k_{1}=\frac{\sqrt{2m(V_{1}-E)}}{\hbar},</math>
<math>k_{2}=\frac{\sqrt{2mE}}{\hbar},</math>
and
<math>k_{3}=\frac{\sqrt{2m(V_{2}-E)}}{\hbar}.</math>


(1)Scattering State:
The wave function and it's first derivative must both be continuous at <math>x=0\!</math> and <math>x=a.\!</math>  The condition at <math>x=0\!</math> yields


If we have the potential <math>V_1</math>for <math>x<0</math>,<math>0</math>for <math>0<x<a</math> and <math>V_2</math> for <math>x>a</math>, then we can write the wave functions for scattering states as:
<math>A=B+C\!</math>


<math>\begin{cases}
and
x<a & \Psi_{1}=A_{1}e^{ik_{1}x}+A_{2}e^{-ik_{2}x}\\
0\leqslant x<a & \Psi_{2}=B_{1}e^{ik_{2}x}+B_{2}e^{-ik_{2}x}\\
x\geqslant a & \Psi_{3}=Ce^{ik_{3}x}\end{cases}</math>


Further we define:
<math>k_{1}A=ik_{2}(B-C),\!</math>
<math>k_{1}=\frac{\sqrt{2m(E-V_{1})}}{\hbar}</math>
<math>k_{2}=\frac{\sqrt{2mE}}{\hbar}</math>
<math>k_{3}=\frac{\sqrt{2m(E-V_{2})}}{\hbar}</math>


By continuity:
while that at <math>x=a\!</math> gives us
when <math>x=-a</math>


<math>A_{1}e^{-ik_{1}a}+A_{2}e^{ik_{1}a}=B_{1}e^{-ik_{2}a}+B_{2}e^{ik_{2}a}</math>
<math>Be^{ik_{2}a}+Ce^{-ik_{2}a}=De^{-k_{3}a}</math>


<math>ik_{1}A_{1}e^{-ik_{1}a}-ik_{1}A_{2}e^{ik_{1}a}=ik_{2}B_{1}e^{-ik_{2}a}-ik_{2}B_{2}e^{ik_{2}a}</math>
and


when <math>x=a</math>
<math>Be^{ik_{2}a}-Ce^{-ik_{2}a}=\frac{ik_{3}}{k_{2}}De^{-k_{3}a}.</math>


<math>B_{1}e^{ik_{2}a}+B_{2}e^{-ik_{2}a}=Ce^{ik_{3}a}</math>
If we reduce the above set of equations to equations only involving <math>B\!</math> and <math>C\!</math>, we obtain


<math>ik_{2}B_{1}e^{ik_{2}a}-ik_{2}B_{2}e^{-ik_{2}a}=ik_{3}Ce^{ik_{3}a}</math>
<math>(k_1-ik_2)B+(k_1+ik_2)C=0\!</math>


Then we can solve for <math>\frac{C}{A_1}</math> from the above four equations and the transmission rate can be expressed as:
and


<math>T=\frac{k_{3}}{k_{1}}(\frac{C}{A_1})^{2}=</math>
<math>(k_3+ik_2)Be^{ik_2a}+(k_3-ik_2)Ce^{-ik_2a}=0.</math>


To be continued...
In order for this system of homogeneous equations to possess a non-trivial solution in <math>B\!</math> and <math>C,\!</math> the following condition must be met:


(2)Bound State:
<math>(k_1-ik_2)(k_3-ik_2)e^{-ik_2a}-(k_1+ik_2)(k_3+ik_2)e^{ik_2a}=0,</math>


or


If we have the potential <math>V_1</math>for <math>x<0</math>,<math>0</math>for <math>0<x<a</math> and <math>V_2</math> for <math>x>a</math>, then we can write the wave functions for bound states as:
<math>\tan{k_{2}a}=\frac{(k_1+k_3)k_2}{k_2^2-k_1k_3}.</math>


<math>\begin{cases}
If we now substitute in our expressions for <math>k_1,\,k_2,\!</math> and <math>k_3,\!</math> we get
x<0 & \Psi=Ae^{k_{1}x}\\
0<x<a & \Psi=Be^{ik_{2}x}+Ce^{-ik_{2}x}\\
x>a & \Psi=De^{-k_{3}x}\end{cases}</math>


where,
<math>\tan\left (\frac{\sqrt{2mE}}{\hbar}a\right )=\frac{\sqrt{E}(\sqrt{V_{1}-E}+\sqrt{V_{2}-E})}{E-\sqrt{(V_{1}-E)(V_{2}-E)}}.</math>
<math>k_{1}=\sqrt{2m(V_{1}-E)}/\hbar</math>
<math>k_{2}=\sqrt{2mE}/\hbar</math>
<math>k_{3}=\sqrt{2m(V_{2}-E)}/\hbar</math>


By continuity:
The solutions to this equation then give us the bound-state energy spectrum of the system.


when <math>x=0</math>
==Problems==


<math>\begin{cases}
A=B+C\\
k_{1}A=ik_{2}(B-C)\end{cases}</math>


when <math>x=a</math>
'''(1)''' An electron is moving freely inside of a one-dimensional box with walls at <math>x=0\!</math> and <math>x=a.\!</math>  If the electron is initially in the ground state of the box and we suddenly increase the size of the box by moving the right-hand wall instantaneously from <math>x=a\!</math> to <math>x=4a,\!</math> then calculate the probability of finding the electron in


<math>\begin{cases}
'''(a)''' the ground state of the new box, and
Be^{ik_{2}a}+Ce^{-ik_{2}a}=De^{-k_{3}a}\\
Be^{ik_{2}a}-Ce^{-ik_{2}a}=\frac{ik_{3}}{k_{2}}De^{-k_{3}a}\end{cases}</math>


'''(b)''' the first excited state of the new box.


For nontrivial solution of <math>B_1</math> and <math>B_2</math>, we must have:
[[Phy5645/One dimensional problem|Solution]]


<math>\frac{ik_{3}}{k_{2}}=\frac{(k_{1}+ik_{2})e^{ik_{2}a}-(-k_{1}+ik_{2})e^{-ik_{2}a}}{(k_{1}+ik_{2})e^{ik_{2}a}+(-k_{1}+ik_{2})e^{-ik_{2}a}}</math>


<math>tgk_{2}a=\frac{k_{2}(k_{1}+k_{2})}{k_{2}^{2}-k_{1}k_{2}}</math>
'''(2)''' Consider a particle of mass <math>m\!</math> bouncing vertically from a smooth floor in Earth's gravitational field.  The potential is given by


Substitute <math>k_1</math><math>k_2</math>and<math>k_3</math>, we get:
<math>V(z)=\begin{cases}
mgz, \; z>0 \\
\infty,\; z\leq 0
\end{cases},
</math>


<math>tg\frac{\sqrt{2mE}}{\hbar}a=\frac{\sqrt{E}(\sqrt{V_{1}-E}+\sqrt{V_{2}-E})}{E-\sqrt{(V_{1}-E)(V_{2}-E)}}</math>
where <math>g\!</math> is the acceleration due to gravity at Earth's surface.  Find the energy levels and corresponding wave functions of the particle.


This is the quantization condition of the energy, which gives us energy spectrum.
[[Phy5645/Particle bouncing on the floor in Earth's Gravitational field|Solution]]

Latest revision as of 00:00, 1 September 2013

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

When the energy of a particle moving in one dimension is less than the potential at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\to\pm\infty} , the particle is trapped in a potential well and is said to be in a bound state. However, when the energy is larger than the potential at either infinity, the particle is said to be in a scattering state. The wave function for a bound state must decay at least exponentially at both infinities.

Basic Properties

1. The bound states of a one-dimensional potential are non-degenerate.

To prove this, let us suppose that there are two different wave functions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)},} for a given bound state with energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\!} . They must then both satisfy the same Schrödinger equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}}

Multiplying the first equation by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)}} and the second by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} , then subtracting one equation from the other, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \frac{d}{dx}\left (\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}\right )=0.}

We recognize the quatity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x)=\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}} as the Wronskian of the two wave functions. We therefore see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(x)=C,} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} is constant. For bound states, the wave functions must vanish at infinity, and therefore Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W=0,} or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0.}

Rearranging, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0.}

We can now integrate this equation to obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=A,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}(x)=a\psi_E^{(2)}(x).}

We see that the two wave functions must be proportional to each other; after normalization, they will become completely identical. Therefore, we see that the two wave functions correspond to the same state, and thus all bound states in one dimension are non-degenerate.

2. The wave function for a real potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)} can be chosen to be real.

If we take the complex conjugate of the Schrödinger equation for the system,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi=-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi,}

then we obtain Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi^{\ast}=-\frac{\hbar^2}{2m}\frac{d^2\psi^{\ast}}{dx^2}+V(x)\psi^{\ast}.}

Since both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi^{\ast}} satisfy the same Schrödinger equation, their sum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi+\psi^{\ast}} must also be a solution. This sum is real, and thus we see that the wave function, as asserted, can be chosen to be real.

3. The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th excited state has Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} nodes.

See the Oscillation Theorem.

Parity and the Symmetry of the Wave Functions

One useful theorem about one-dimensional potentials is that, if the potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)\!} is even (i.e. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=V(-x)\!} ), then the eigenstates can be taken to be even or odd.

To prove this, we first note that, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)} satisfies the Schrödinger equation for the potential,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi,}

then so will Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} . We may see this by simply taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\to -x\!} in the above equation. Since both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} are solutions of the Schrödinger equation, any linear combination of the two must also be a solution. In particular, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_e(x)=\psi(x)+\psi(-x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_o(x)=\psi(x)-\psi(-x)\!} are solutions. We may easily see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_e(x)\!} is even, while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_o(x)\!} is odd.

In particular, this implies that all bound states in such a potential must be either even or odd. Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} both solve the Schrödinger equation, and because the eigenstates must be non-degenerate, we see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(-x)\!} must be proportional to one another. We may easily show that the proportionality constant must be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm 1\!} , corresponding to either an even function or an odd function.

The above discussion is a simple illustration of how the symmetry of the Hamiltonian of a system dictates the transformation properties of the eigenstates, as described earlier.

Infinite Square Well

Let's consider the motion of a particle in an infinite and symmetric square well: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=+\infty} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x| \ge L/2,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=0\!} otherwise.

A particle subject to this potential is free everywhere except at the two ends Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x = \pm L/2),} where the infinite potential keeps the particle confined to the well. Within the well the Schrödinger equation takes the form,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi,}

or equivalently,

where

Writing the Schrödinger equation in this form, we see that our solution are simply

As we showed earlier, the eigenstates of this potential, all of which are bound states, must be either even or odd. This means that we must either set or Let us consider each case in turn.

Case 1: .

In this case,

We require that the wave function vanish at the ends, so that

The solutions of this equation are

where

Case 2:

In this case,

Again requiring that the wave function vanish at the ends, i.e.

we find that

Combining these two results, we find that the energy eigenvalues are , where The wave numbers are quantized as a result of the boundary conditions, and thus the energies are quantized as well. The lowest energy state is the ground state, and it has a non-zero energy, which is due to quantum zero point motion. The ground state is nodeless, the first excited state has one node, the second excited state has two nodes, and so on. The wavefunctions are also orthogonal.

Finite Asymmetric Square Well

We now consider an asymmetric square well potential defined as

where both and are positive. We may then split our one-dimensional space into three regions; region I is given by , region II by , and region III by . The Schrödinger equation may be written in each region as

and

We will be considering bound states here, so that In this case, the solutions to the above Schrödinger equations are

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{III}(x)=De^{-k_{3}x},}

where, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{1}=\frac{\sqrt{2m(V_{1}-E)}}{\hbar},} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{2}=\frac{\sqrt{2mE}}{\hbar},} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{3}=\frac{\sqrt{2m(V_{2}-E)}}{\hbar}.}

The wave function and it's first derivative must both be continuous at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a.\!} The condition at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0\!} yields

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=B+C\!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{1}A=ik_{2}(B-C),\!}

while that at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a\!} gives us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Be^{ik_{2}a}+Ce^{-ik_{2}a}=De^{-k_{3}a}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Be^{ik_{2}a}-Ce^{-ik_{2}a}=\frac{ik_{3}}{k_{2}}De^{-k_{3}a}.}

If we reduce the above set of equations to equations only involving Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} , we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (k_1-ik_2)B+(k_1+ik_2)C=0\!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (k_3+ik_2)Be^{ik_2a}+(k_3-ik_2)Ce^{-ik_2a}=0.}

In order for this system of homogeneous equations to possess a non-trivial solution in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C,\!} the following condition must be met:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (k_1-ik_2)(k_3-ik_2)e^{-ik_2a}-(k_1+ik_2)(k_3+ik_2)e^{ik_2a}=0,}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan{k_{2}a}=\frac{(k_1+k_3)k_2}{k_2^2-k_1k_3}.}

If we now substitute in our expressions for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1,\,k_2,\!} and we get

The solutions to this equation then give us the bound-state energy spectrum of the system.

Problems

(1) An electron is moving freely inside of a one-dimensional box with walls at and If the electron is initially in the ground state of the box and we suddenly increase the size of the box by moving the right-hand wall instantaneously from to then calculate the probability of finding the electron in

(a) the ground state of the new box, and

(b) the first excited state of the new box.

Solution


(2) Consider a particle of mass bouncing vertically from a smooth floor in Earth's gravitational field. The potential is given by

where is the acceleration due to gravity at Earth's surface. Find the energy levels and corresponding wave functions of the particle.

Solution

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