Part 2 Bonus Material

No-Nonsense Quantum Mechanics

Exercises

A particle is bound (i.e. $E<0$) in a potential of the form $$V ( x ) = - V _ { 0 } \delta ( x ) \quad V _ { 0 } > 0 \, ,$$ where  $\delta ( x )$ is the delta distribution. There is only one such bound state possible for this potential. Calculate the corresponding energy.
In the two regions $x> 0$ and $x<0$ the Schrödinger equation reads $$\partial _ { x } ^ { 2 } \psi ( x ) - \frac { - 2 m E } { \hbar ^ { 2 } } \psi ( x ) = 0$$ and the two corresponding solutions are $$\psi _ { < } ( x ) = B e ^ { \kappa x }$$ $$\psi_> ( x ) = A e ^ { - \kappa x }$$ where $\kappa \equiv \frac { - 2 m E } { \hbar ^ { 2 } }$. The other possible terms in both regions have to vanish since otherwise the wave function is non-normalizable. The wave function has to be smooth at $x=0$ and therefore we find $A=B$. Next, we integrate the Schrödinger equation in a small interval $[ - \epsilon , \epsilon]$ around $x=0$: $$\int_{-\epsilon}^\epsilon \partial _ { x } ^ { 2 } \psi ( x ) - int_{-\epsilon}^\epsilon \frac { - 2 m E } { \hbar ^ { 2 } } \psi ( x ) = 0$$ $$\therefore \quad - \frac { \hbar ^ { 2 } } { 2 m } \left( \psi _ { > } ^ { \prime } ( 0 ) - \psi _ { < } ^ { \prime } ( 0 ) \right) - V _ { 0 } \psi ( 0 ) = 0 \, .$$ Then we use the explicit solutions from above and find \begin{align} 0 &= - \frac { \hbar ^ { 2 } } { 2 m } \left( \psi _ { > } ^ { \prime } ( 0 ) - \psi _ { < } ^ { \prime } ( 0 ) \right) - V _ { 0 } \psi ( 0 ) \\ &= = A \left( - \frac { \hbar ^ { 2 } } { 2 m } ( - \kappa - \kappa ) - V _ { 0 } \right) \end{align} and we can conclude $$\rightarrow \quad \kappa = \frac { m V _ { 0 } } { \hbar ^ { 2 } }$$ and therefore using the definition of $\kappa$ $$E = - \frac { m V _ { 0 } ^ { 2 } } { 2 \hbar ^ { 2 } }.$$ This is the energy level that we wanted to calculated and, as promised, there is really only one.

A particle is described by the wave function $$\psi ( x ) = \frac { 1 } { \pi ^ { 1 / 4 } d ^ { 1 / 2 } } \exp \left( i k x - \frac { x ^ { 2 } } { 2 d ^ { 2 } } \right)$$ Is this wave function correctly normalized?
Yes: $$| \psi | ^ { 2 } = \int d x \psi ^ { * } ( x ) \psi ( x ) = \int d x | \psi ( x ) | ^ { 2 }$$ $$= \frac { 1 } { \sqrt { \pi } d } \int d x e ^ { - x ^ { 2 } / d ^ { 2 } } = \frac { 1 } { \sqrt { \pi } } \int d x e ^ { - x ^ { 2 } } = 1$$

Show that for 3 operators the following identity is correct $$[ \hat { A } , \hat { B } \hat { C } ] = \hat { B } [ \hat { A } , \hat { C } ] + [ \hat { A } , \hat { B } ] \hat { C } .$$
We calculate: $$\hat { B } [ \hat { A } , \hat { C } ] + [ \hat { A } , \hat { B } ] \hat { C } = \hat { B } ( \hat { A } C - \hat { C } \hat { A } ) + ( \hat { A } \hat { B } - \hat { B } \hat { A } ) \hat { C }$$ $$= \hat { A } \hat { B } \hat { C } - \hat { B } \hat { C } \hat { A } = [ \hat { A } , \hat { B } \hat { C } ]$$

Show that for an operator $\hat A$ we have $$\frac { d } { d t } e ^ { A t } = A e ^ { A t }.$$
We can check this using the series expansion of the exponential function $$\frac { d } { d t } e ^ { A t } = \frac { d } { d t } \sum _ { n = 0 } ^ { \infty } \frac { A ^ { n } t ^ { n } } { n ! }$$ $$= \sum _ { n = 1 } ^ { \infty } \frac { A ^ { n } t ^ { n - 1 } } { ( n - 1 ) ! }$$ $$= A \sum _ { n = 1 } ^ { \infty } \frac { A ^ { n - 1 } t ^ { n - 1 } } { ( n - 1 ) ! } = A e ^ { A t }$$

Two operators are given explicitly as $$H = \hbar \omega _ { 0 } \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { - 1 } & { 0 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right) , \quad A = a \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { 0 } & { 1 } \\ { 0 } & { 1 } & { 0 } \end{array} \right) ,$$ where $\omega_0$ and $a$ are real constants. Are $H$ and $A$ hermitian? Do they commute?
$H$ and $A$ are hermitian since $H^\dagger = H$ and $A^\dagger = A$. Hermitian conjugation means transposition plus complex conjugtation. Both matrices are real so conjugation has no effect and since both matrices are symmetric the transposition does not have any effect either. Hence both matrices are hermitian. Moreover, both matrices commute as we can check explicitly: $$H A = \hbar \omega _ { 0 } a \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { - 1 } & { 0 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right) \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { 0 } & { 1 } \\ { 0 } & { 1 } & { 0 } \end{array} \right)$$ $$= \hbar \omega _ { 0 } a \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { 0 } & { - 1 } \\ { 0 } & { - 1 } & { 0 } \end{array} \right) = A H$$

The Hamiltonian operator for a system is given by $$H = \hbar \omega _ { 0 } \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { - 1 } & { 0 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right)$$ and the system is in a state described by $$| \Psi \rangle = \frac { 1 } { \sqrt { 2 } } | E _ { 1 } \rangle + \frac { 1 } { 2 } | E _ { 2 } \rangle + \frac { 1 } { 2 } | E _ { 3 } \rangle .$$ What's the energy expectation value?
We calculate $$\langle H \rangle = \langle \Psi ( 0 ) | H | \Psi ( 0 ) \rangle = \sum \langle \Psi ( 0 ) | e _ { i } \rangle H _ { i j } \left\langle e _ { j } | \Psi ( 0 ) \right\rangle ,$$ where $H _ { i j }$ denotes the matrix elements of $H$. Therefore, we need to calculate: $$\langle \Psi ( 0 ) | u _ { i } \rangle = \frac { 1 } { \sqrt { 2 } } \delta _ { 1 i } + \frac { 1 } { 2 } \delta _ { 2 i } + \frac { 1 } { 2 } \delta _ { 3 i }$$ $$\left\langle u _ { j } | \Psi ( 0 ) \right\rangle = \frac { 1 } { \sqrt { 2 } } \delta _ { j 1 } + \frac { 1 } { 2 } \delta _ { j 2 } + \frac { 1 } { 2 } \delta _ { j 3 }.$$ We can now conclude $$\langle H \rangle = \frac { 1 } { 2 } H _ { 11 } + \frac { 1 } { 4 } H _ { 22 } + \frac { 1 } { 4 } H _ { 33 } = 0 .$$

The Hamiltonian operator for a system is given by $$H = \hbar \omega _ { 0 } \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { - 1 } & { 0 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right)$$ and the system is in a state described by $$| \Psi \rangle = \frac { 1 } { \sqrt { 2 } } | E _ { 1 } \rangle + \frac { 1 } { 2 } | E _ { 2 } \rangle + \frac { 1 } { 2 } | E _ { 3 } \rangle .$$ What's the standard deviation of the energy?
The standard deviation is defined as $$\Delta H = \sqrt { \left\langle H ^ { 2 } \right\rangle - \langle H \rangle ^ { 2 } }$$ We already calculated $\langle H \rangle= 0$ in Exercise 6 from which we can immediately conclude $\langle H \rangle ^ { 2 }=0$. And threfore we now only need to calculate $\left\langle H ^ { 2 } \right\rangle$: $$\left\langle H ^ { 2 } \right\rangle = \sum _ { i j k } \langle \Psi ( 0 ) | u _ { i } \rangle H _ { i j } H _ { j k } \left\langle u _ { k } | \Psi ( 0 ) \right\rangle$$ $$= \sum _ { i j k } \left[ \frac { 1 } { \sqrt { 2 } } \delta _ { 1 i } + \frac { 1 } { 2 } \delta _ { 2 i } + \frac { 1 } { 2 } \delta _ { 3 i } \right] H _ { i j } H _ { j k } \left[ \frac { 1 } { \sqrt { 2 } } \delta _ { 1 k } + \frac { 1 } { 2 } \delta _ { 2 k } + \frac { 1 } { 2 } \delta _ { 3 k } \right]$$ $$= \frac { 1 } { 2 } H _ { 11 } ^ { 2 } + \frac { 1 } { 4 } H _ { 22 } ^ { 2 } + \frac { 1 } { 4 } H _ { 33 } ^ { 2 } = \hbar ^ { 2 } \omega ^ { 2 } \, .$$ Therefore $$\Delta H = \hbar \omega .$$

The Hamiltonian operator for a system is given by $$H = \hbar \omega _ { 0 } \left( \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { - 1 } & { 0 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right)$$ and the system is at $t=0$ in a state described by $$| \Psi(0) \rangle = \frac { 1 } { \sqrt { 2 } } | E _ { 1 } \rangle + \frac { 1 } { 2 } | E _ { 2 } \rangle + \frac { 1 } { 2 } | E _ { 3 } \rangle .$$ Calculate the state at $t \neq 0$.
Using the time-evolution operator we find $$| \Psi ( t ) \rangle = \exp \left( - \frac { i } { \hbar } H t \right) | \Psi ( 0 ) \rangle$$ $$= \left( \begin{array} { c c c } { e ^ { - i \omega _ { 0 } t } } & { 0 } & { 0 } \\ { 0 } & { e ^ { i \omega _ { 0 } t } } & { 0 } \\ { 0 } & { 0 } & { e ^ { i \omega _ { 0 } t } } \end{array} \right) \frac { 1 } { 2 } \left( \begin{array} {c} { \sqrt { 2 } } \\ { 1 } \\ { 1 } \end{array} \right)$$ $$= \frac { 1 } { 2 } \left( \begin{array} {c} { \sqrt { 2 } e ^ { - i \omega _ { 0 } t } } \\ e^{ i \omega_{ 0 } t } \\ e^{ i \omega_{ 0 } t } \end{array} \right)$$

Consider the operator given by the matrix $$B = \left( \begin{array} { l l } { 1 } & { 1 } \\ { 0 } & { 1 } \end{array} \right).$$ Is $B$ hermitian?
No! Since $$B ^ { \dagger } = B ^ { T } = \left( \begin{array} { l l } { 1 } & { 0 } \\ { 1 } & { 1 } \end{array} \right) \neq B$$

A particle is at $t=0$ described by the wave function $$| \Psi ( t = 0 ) \rangle = \frac { 1 } { \sqrt { 2 } } | E _ { 1 } \rangle + \frac { 1 } { \sqrt { 2 } } | E _ { 2 } \rangle .$$ Calculate the time-evolution of $\Psi(t)$ and the corresponding time-dependent probability density.
Using the time-evolution operator, we find $$| \Psi ( t ) \rangle = \frac { 1 } { \sqrt { 2 } } e ^ { - \frac { i E _ { i } t } { \hbar } } | E _ { 1 } \rangle + \frac { 1 } { \sqrt { 2 } } e ^ { - \frac { i E _ { 2 } t } { \hbar } } | E _ { 2 } \rangle .$$ The corresponding probability density is $$\rho ( x , t ) = | \Psi ( x , t ) | ^ { 2 }$$ $$= \frac { 1 } { 2 } \Psi _ { 1 } ^ { 2 } ( x ) + \frac { 1 } { 2 } \Psi _ { 2 } ^ { 2 } ( x ) + \frac { 1 } { 2 } \left[ e ^ { i ( \alpha - \beta ) } + e ^ { i ( \beta - \alpha ) } \right] \Psi _ { 1 } \Psi _ { 2 }$$ $$\rho ( x , t ) = \frac { 1 } { 2 } \Psi _ { 1 } ^ { 2 } ( x ) + \frac { 1 } { 2 } \Psi _ { 2 } ^ { 2 } ( x ) + \cos ( \beta - \alpha ) \Psi _ { 1 } ( x ) \Psi _ { 2 } ( x )$$ (Recall that wave functions are defined under an integral and thus to find the corresponding probability we have to integrate over some spatial region.

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