# Quantum Mechanics Notes

### The Schrodinger Equation

It is not possible to derive the Schrodinger equation from classical physics, however Schrodinger started with the classical equation of wave motion, relating the space and time dependencies of a wave,…
$\nabla^2 \psi = \frac{1}{v^2} \frac{\partial^2\psi}{\partial t^2}$
…with v = velocity of the wave. A general solution of this wave equation, simplified to one dimension, is ,….
$\psi = A e^{i(kx-\omega t)}$
Where $k = \frac{2 \pi}{\lambda}$ is the wave number, and $\omega = 2\pi\nu$ is angular frequency. With this the solution to the spatial dependency becomes,…
$\frac{\partial^2\psi}{\partial x^2} = \frac{\partial}{\partial x} \left (ik\psi\right) = -k^2\psi \cdots$

$\frac{\partial^2\psi}{\partial x^2} = -\frac{4\pi^2}{\lambda^2}\psi$
Substituting the de Broglie relation, $\lambda = \frac{h}{p}$, associating linear momentum and wavelength, into the above,…
$\frac{\partial^2 \psi}{\partial x^2} = -\frac{4\pi^2}{\lambda^2}\psi = -\frac{4\pi^2 p^2}{h^2}\psi = -\frac{p^2}{h^2}\div\frac{1}{4 \pi^2}\psi = -\frac{p^2}{\hbar^2}\psi\cdots$
so that,…
$\frac{\partial^2 \psi}{\partial x^2} = -\frac{p^2}{\hbar^2}\psi\cdots (1)$

Total energy is kinetic plus potential,…
$E = \frac{1}{2}m v^2 + V$ … or in terms of momentum … $E = \frac{p^2}{2m} + V$
Solving for p^2,…
$p^2 = 2m(E-V)$
Substituting this result into (1) above,…
$\frac{\partial^2\psi}{\partial x^2} = -\frac{2m}{\hbar^2} \left (E-V\right )\psi$
Solving this spatial dependency of the wave for $E\psi$,…
$-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} + V\psi = E\psi \cdots (2)$

The temporal dependency of the wave can also be solved for $E\psi$, as long as only the first order derivative is taken with respect to t….
$\psi = A e^{i(kx-\omega t)}$….. again with the general wave solution, so that,…

$\frac{\partial\psi}{\partial t} = -i\omega\psi$ … or … $\frac{\partial\psi}{\partial t} = -i2\pi\nu\psi$ … with $\omega = 2\pi\nu$
Substituting the Planck relation, associating energy and frequency, $E = h\nu$ or $\nu = \frac{E}{h}$, …
$\frac{\partial\psi}{\partial t} = -\frac{i2\pi E}{h}\psi = -\frac{iE}{h}\div\frac{1}{2\pi}\psi = -\frac{iE}{\hbar}\psi$
Now solving this for $E\psi$,…
$\frac{\partial\psi}{\partial t} = -\frac{iE}{\hbar}\psi\cdots$

$i\hbar\frac{\partial\psi}{\partial t} = E\psi \cdots (3)$
Since both (2) and (3) are now expressed in terms of $E\psi$, they can be equated,…
$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} + V\psi$
Or, in three dimensions,…
$\boxed{i\hbar\frac{\partial\psi}{\partial t} = \left [-\frac{\hbar^2}{2m}\nabla^2 + V\right ]\psi}$ …(4) The Schrodinger Equation

This equation can be expressed succinctly using the Hamiltonian operator defined as,…
$\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V$
… implying that the momentum operator is $\hat{p} = -i\hbar\nabla$,… so that…
$i\hbar\dot{\psi} = \hat{H}\psi$

The Schrodinger equation is not relativistic however, as it does not treat the space and time dependencies on an equal footing as is required in the special theory of relativity, since it has a second order partial differential for space and only a first order partial differential for time. Paul Dirac would resolve this issue by using matrices to change the spatial derivative to first order….

### The Dirac Equation

The terms in brackets on the right hand side of the Schrodinger equation (4), represent the Hamiltonian operator. This operator is the quantum total energy operator. With the relativistic mass-energy relation, $E = mc^2$, the total energy of a particle moving at velocity v, is….

$E = \frac{m_0 c^2}{\sqrt{\left (1 - v^2/c^2\right )}}$

The momentum operator, $-i\hbar\nabla$, will be required in quantum mechanics, so using $p = mv$, and $E = mc^2$, and rearranging,…

$E^2 = \frac{m^2_0 c^4}{\left (1 - v^2/c^2\right )}\cdots E^2 \left (1 - v^2/c^2 \right) = m^2_0 c^4 \cdots$

$E^2 - \frac{E^2 v^2}{c^2} = m^2_0 c^4 \cdots E^2 - \frac{m^2_0 c^4 v^2}{c^2} = m^2_0 c^4 \cdots$

$E^2 - p^2 c^2 = m^2_0 c^4 \cdots E^2 = p^2 c^2 + m^2_0 c^4 \cdots$

$E = c\sqrt{p^2 + m^2_0 c^2}$
Substituting the above result into the Schrodinger equation (4) on the right hand side, with the potential term dropped as a freely moving particle is being assumed…
$i\hbar\frac{\partial\psi}{\partial t} = \left [c\sqrt{p^2 + m^2_0 c^2} \right ]\psi$ … (5)

Now, since in quantum mechanics the momentum term will become a differential operator, $-i\hbar\nabla$, the above square root implies a differential equation of infinite order. Paul Dirac needed to find a means of making this equation linear to be compatible with special relativity.

Dividing out $\psi$, squaring both sides of (5), and using the quantum operator form of momentum, $\hat{p} = -i\hbar\nabla$,…

$-\frac{\hbar^2}{c^2}\frac{\partial^2}{\partial t^2} = \left(\hbar^2\frac{\partial^2}{\partial x^2} + \hbar^2\frac{\partial^2}{\partial y^2} + \hbar^2\frac{\partial^2}{\partial z^2} \right ) + m^2_0 c^2\cdots$

$\left (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right ) = \frac{-m^2_0 c^2}{\hbar^2}$

Taking the square root,…

$\sqrt{\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{c^2}\frac{\partial^2}{\partial t^2}} = \frac{im_0 c}{\hbar}$

Now Dirac would use coefficients, A,B,C,D to factor out the square root such that,…

$\sqrt{\left (A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y} + C\frac{\partial}{\partial z} + D\frac{\partial}{\partial t}\right )\left (A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y} + C\frac{\partial}{\partial z} + D\frac{\partial}{\partial t}\right )} =$
$A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y} + C\frac{\partial}{\partial z} + D\frac{\partial}{\partial t}$

Which requires that the coefficients, A,B,C,D serve to cancel out the cross terms produced by the product in the square root,…

$\left (A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y} + C\frac{\partial}{\partial z} + D\frac{\partial}{\partial t}\right ) \left (A\frac{\partial}{\partial x} + B\frac{\partial}{\partial y} + C\frac{\partial}{\partial z} + D\frac{\partial}{\partial t}\right ) =$
$AA\frac{\partial}{\partial x}\frac{\partial}{\partial x} + AB\frac{\partial}{\partial x}\frac{\partial}{\partial y} + BA\frac{\partial}{\partial y}\frac{\partial}{\partial x} + AC\frac{\partial}{\partial x}\frac{\partial}{\partial z} + CA\frac{\partial}{\partial z}\frac{\partial}{\partial x} + BB\frac{\partial}{\partial y}\frac{\partial}{\partial y} + BC\frac{\partial}{\partial y}\frac{\partial}{\partial z} + CB\frac{\partial}{\partial z}\frac{\partial}{\partial y} + CC\frac{\partial}{\partial z}\frac{\partial}{\partial z} + DD\frac{\partial}{\partial t}\frac{\partial}{\partial t} + DA\frac{\partial}{\partial t}\frac{\partial}{\partial x} + AD\frac{\partial}{\partial x}\frac{\partial}{\partial t} + DB\frac{\partial}{\partial t}\frac{\partial}{\partial y} + BD\frac{\partial}{\partial y}\frac{\partial}{\partial t} + DC\frac{\partial}{\partial t}\frac{\partial}{\partial z} + CD\frac{\partial}{\partial z}\frac{\partial}{\partial t}$

This requires that the coefficients, A,B,C,D have the property that $AB = -BA$, etc and that $A^2 = B^2 = C^2 = D^2 = 1$, which will effectively remove all the cross terms.

These conditions imply that the coefficients are each 4 X 4 matrices, which in turn implies that the wavefucntion itself, $\psi$ becomes a column vector of four elements,… in other words, that there are now four wavefucntion solutions. Two of these solutions are interpreted as positive energy solutions corresponding to the spin-up and spin-down configuration of the electron. The other two solutions were to be negative energy solutions corresponding to the spin-up and spin-down configuration of the positron. But no one knew what a positron was at the time,… Dirac initially thinking it was the proton. Remarkably, Dirac’s relativistic equation for the electron predicted the existence of a subatomic particle,… in fact the existence of anti-particles generally.

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### Momentum As An Operator

The probability of finding a particle between a and b at time t, is,…
$\int\limits_a^b \left |\psi(x,t)\right |^2 dx$
Since the particle must be somewhere the statistical interpretation of quantum mechanics requires the normalization condition,…
$\int\limits_\infty^\infty \left |\psi(x,t)\right |^2 dx = 1$
In general the average value of some function f() is (called the ‘expectation value’ in QM),…
$\textless f(j) \textgreater = \sum\limits_{j=0}^\infty f(j)p(j)$ ,….. where p(j) is the probability of f(j).
So, for example, the expectation value for position, x, would be,…
$\textless x \textgreater = \int\limits_{-\infty}^\infty x \left |\psi(x,t)\right |^2 dx \text{ ....or .... } = \int\limits_{-\infty}^\infty x \psi^* \psi dx$
Observables in quantum mechanics are represented by Hermitian operators, self-adjoint operators or matricies that are equal to their own transpose-conjugate. (i.e. position, momentum, Hamiltonian, etc). This is required so that eiginvalues are Real numbers, and to facilitate the association of these operators with the orthonormal basis underlying Hilbert space. On account of this, the x-operator above, can be moved over to operate on the second factor of the inner product…
$\textless x \psi^* , \psi \textgreater = \textless \psi^*, x \psi \textgreater$
$\textless x \textgreater = \int\limits_{-\infty}^\infty \psi^* x \psi \text{ }dx$

Now, this position expectation value changes as the wavefunction $\psi$ evolves in time. The expectation value of this rate of change, the “velocity”, would be,…
$\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty x \frac{\partial}{\partial t} (\psi^* \psi) \text{ } dx$ , ….. or via Leibniz rule ,….

$\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty x (\psi^* \frac{\partial\psi} {\partial t} + \frac{\partial\psi^*} {\partial t}\psi) \text{ } dx$ ………………..(1)

The Schrödinger equation shows what the rate of change of $\psi$ with respect to time must be,…
$\frac{\partial\psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2} - \frac{i}{\hbar} V \psi$ … and its complex conjugate,… $\frac{\partial\psi^*}{\partial t} = -\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{i}{\hbar} V \psi^*$

Substituting these into equation (1) above, allows the potential terms, $\pm \frac{i}{\hbar} V\psi^*\psi$, to cancel out, and results in,….
$\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty x \frac{i\hbar}{2m}\left (\psi^*\frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi^*}{\partial x^2}\psi \right ) \text{ } dx$
Factoring out a partial derivative operator and moving the constants outside the integral,…
$\frac{d\textless x \textgreater}{dt} = \frac{i\hbar}{2m} \int\limits_{-\infty}^\infty x \frac{\partial}{\partial x} \left (\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi \right ) \text{ } dx$ ………………..(2)

This equation can be simplified using ‘integration by parts’ which works as follows;
———————————————————-
Starting with a derivative of a product and expanding via Leibniz rule, ….
$\frac{d(fg)}{dx} = f \frac{dg}{dx} + \frac{df}{dx} g$ …. or,…. $\frac{d(fg)}{dx} - \frac{df}{dx} g = f \frac{dg}{dx}$
Integrating each side,….
$fg\bigg|_{-\infty}^\infty - \int\frac{df}{dx} g = \int f \frac{dg}{dx}$
The wavefunction will be zero at the boundary or at infinity, so that term can be dropped.
$\int f \frac{dg}{dx} = - \int\frac{df}{dx} g$
The result is equivalent to moving the f into the derivative and the g out, and then negating.
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Performing an ‘integration by parts’ on equation (2) above will move the $x$ into the $\frac{\partial}{\partial x}$, and the $\left (\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi \right )$ out, plus negating results in,…
$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^\infty \frac{\partial x}{\partial x} \left (\psi^*\frac{\partial\psi}{\partial x} - \frac{\partial\psi^*}{\partial x}\psi \right ) \text{ } dx$

Performing a integration by parts again, on the second term in parentheses, will move the $\psi$ into the partial derivative, and the $\psi^*$ out, and negating,….
$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^\infty \left (\psi^*\frac{\partial\psi}{\partial x} + \psi^*\frac{\partial\psi}{\partial x} \right ) \text{ } dx$ ,….

$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{2m} \int\limits_{-\infty}^\infty 2 \psi^*\frac{\partial\psi}{\partial x} \text{ } dx$ ,….

$\frac{d\textless x \textgreater}{dt} = -\frac{i\hbar}{m} \int\limits_{-\infty}^\infty \psi^*\frac{\partial\psi}{\partial x} \text{ } dx$ ,….

$\frac{d\textless x \textgreater}{dt} = \frac{1}{m} \int\limits_{-\infty}^\infty \psi^*\frac{\hbar}{i}\frac{\partial}{\partial x}\psi\text{ } dx$

Now, momentum is mass times velocity, so multiplying both sides by m, will result in the ‘expectation value’ of momentum,….
$m\frac{d\textless x \textgreater}{dt} = \int\limits_{-\infty}^\infty \psi^*\frac{\hbar}{i}\frac{\partial}{\partial x}\psi\text{ } dx$
This is similar in form to the expectation value equation for position above,….
$\textless x \textgreater = \int\limits_{-\infty}^\infty \psi^* x \psi \text{ }dx$
Where the position operator, x, was placed between $\psi^*$ and $\psi$. Therefore, momentum in quantum mechanics is associated with an operator,…
$p = -i\hbar\frac{\partial}{\partial x}$

The first term on the right side of the Schrödinger equation is the momentum form of kinetic energy….

$i\hbar\frac{\partial\psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} - V \psi$

kinetic energy $= \frac{1}{2}mv^2 = \frac{1}{2}\frac{m^2v^2}{m} = \frac{p^2}{2m} = \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$