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One-particle, time-independent Schrodinger equation in spherical coordinates

The Laplacian \nabla^{2} in spherical coordinates can be derived from eq87, eq88 and eq89. Substituting eq78 through eq86 in eq87, eq88 and eq89, adding the resulting three equations and simplifying, we have

\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}+\frac{cos\theta}{r^{2}sin\theta}\frac{\partial}{\partial\theta}+\frac{1}{r^{2}sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\; \; \; \; \; \; \; \; 46

or

\nabla^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r} \right )+\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta}\right )\right ]\; \; \; \; \; \; \; \; 47

It can be easily shown that eq47 is equivalent to eq46 by computing the derivatives in eq47. Hence, eq45 in spherical coordinates is

\small \hat{H}=-\frac{\hbar^{2}}{2m}\left \{ \frac{1}{r^{2}}\frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r}\right ) +\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta} \right )\right ]\right \}+V(r)\; \; \; \; \; \; \; \; 48
For the case of a particle confined to a spherical surface of zero relative potential, \small r is constant and \small V(r)=0. The first term of the Laplacian operating on a wavefunction \small \psi(r,\theta,\phi) will return a result of zero. We can therefore discard it and eq48 becomes:

\hat{H}_{angular}=-\frac{\hbar^{2}}{2mr^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta} \right )\right ]\; \; \; \; \; \; \; \; 49

Substituting eq96 in the above equation gives

\hat{H}_{angular}=\frac{\hat{L}^{2}}{2I}\; \; \; \; \; \; \; \; 50

where I=mr^{2}.

Hence, an eigenvalue of \hat{H}_{angular} is the energy associated with the angular motion of a particle. This implies that the linear (or radial) part of the Hamiltonian is:

\hat{H}_{radial}=-\frac{\hbar^{2}}{2m} \frac{1}{r^{2}}\frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r}\right ) +V(r)=-\frac{\hbar^{2}}{2m} \left ( \frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^{2}}{\partial r^{2}}\right ) +V(r)\; \; \; \; \; \; \; \; 51

Similarly, an eigenvalue of \hat{H}_{radial} is the energy associated with the linear motion of a particle. We can, therefore, rewrite eq48 as:

\hat{H}=\hat{H}_{radial}+\hat{H}_{angular}

 

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One-particle, time-independent Schrodinger equation

The one-particle, time-independent Schrodinger equation is a partial differential equation whose solutions are the one-particle, time-independent wave functions of quantum-mechanical systems.

Even though it is widely regarded as a postulate, we can list the scientific findings that could have inspired Schrodinger to develop the equation.

 

Year Development Formula Scientist
1500s Equations of motion Linear rotational motion equations Many scientists including Galileo and Newton
1746 1-D classical wave equation, whose solutions are wave functions that describe the motion of waves. \frac{\partial^{2}u(x,t)}{\partial x^{2}}=\frac{1}{c^{2}}\frac{\partial^{2}u(x,t)}{\partial t^{2}} Jean d’Alembert
1924 de Broglie’s hypothesis: all matter exhibit wave-like properties p=\frac{h}{\lambda} Louis de Broglie

 

Many scientists at that time were trying to develop models of the atom, in particular, the behaviour of electrons in an atom. It is possible that when de Broglie proposed that all matter have wave—like properties, Schrodinger thought that an equation similar to the classical wave equation could be derived to describe the wave motion of an electron in an atom. Since the classical wave equation is solved by the separation of variables method, we can express u(x,t) as u(x,t)=h(x)g(t)=\psi(x)cos\omega t. Substituting u(x,t) in the classical wave equation, we have:

cos\omega t\frac{\partial^{2}}{\partial x^{2}}\psi(x)=-\frac{1}{c^{2}}\psi(x)\omega^{2}cos\omega t

\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{\omega^{2}}{c^{2}}\psi(x)=0

Substitute \omega=2\pi v and c=v\lambda in the above equation

\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{4\pi^{2}}{\lambda^{2}}\psi(x)=0\; \; \; \; \; \; \; \; 43

The total energy of an electron is E=KE+PE=\frac{p^{2}}{2m}+V(x) or p=\sqrt{2m[E-V(x)]}, where p  is the momentum of the electron. Substituting p in de Broglie’s relation, we have \sqrt{2m[E-V(x)]}=\frac{h}{\lambda}, which we then substitute in eq43 to give:

\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{2m}{\hbar^{2}}[E-V(x)]\psi(x)=0

where \hbar=\frac{h}{2\pi}, or

\hat{H}\psi=E\psi\; \; \; \; \; \; \; \; 44

where \hat{H}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V.

\hat{H} is called the Hamiltonian operator and Eq44 is the one-dimensional, one-particle time-independent Schrodinger equation. As it is an eigenvalue equation, \psi (known as a wave function) is an eigenfunction and E is the corresponding eigenvalue. For a three-dimensional, one-particle system, the Hamiltonian is:

\hat{H}=-\frac{\hbar^{2}}{2m}\nabla^{2} +V(x,y,z)\; \; \; \; \; \; \; \; 45

where \nabla^{2}=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} and is called the Laplacian.

 

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Postulates of quantum mechanics

The postulates of quantum mechanics are fundamental mathematical statements that cannot be proven. Nevertheless, they are statements that everyone agrees with.

Examples of other postulates of science and mathematics are Newton’s 2nd law, F=ma, and the Euclidean statement that a line is defined by two points respectively.

Generally, the postulates of quantum mechanics are expressed as the following 7 statements:

1) The state of a physical system at a particular time t is represented by a vector in a Hilbert space.

We call such a vector, a wavefunction \psi(x,t), which is a function of time and space . A wavefunction contains all assessable physical information about a system in a particular state. For example, the energy of the stationary state of a system is obtained by solving the time-independent Schrodinger equation \hat{H}\psi=E\psi.

2) Every measurable physical property of a system is described by a Hermitian operator \hat{O} that acts on a wavefunction representing the state of the system.

In other words, to every observable quantity in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics. The most well-known Hermitian operator in quantum mechanics is the Hamiltonian \hat{H}, which is the energy operator. The wavefunctions that \hat{H} acts on are called eigenfunctions. Eigenfunctions of a Hermitian operator in quantum mechanics are further postulated to form a complete set. Other Hermitian operators frequently encountered in quantum mechanics are the momentum operator \hat{p} and the position operator \hat{x}.

3) The result of the measurement of a physical property of a system must be one of the eigenvalues of an operator \hat{O}.

The state of a system is expressed as a wavefunction, which can be a single basis wavefunction or a linear combination of a complete set of basis wavefunctions. Since basis wavefunctions of a Hermitian operator form a complete set, all wavefunctions can be written as a linear combination of basis wavefunctions.

 

Question

What about a system described by a stationary state?

Answer

The wavefunction can be expressed, though trivially, as \psi=\sum_{n=0}^{\infty}c_n\phi_n, where c_i=1 and c_{n\not\equiv i}=0 (we have assumed that the spectrum is discrete, i.e. the eigenvalues are separated from one another).

 

It is generally accepted by scientists that the values of the coefficients c_n are unknown prior to a measurement. Upon measurement, the result obtained is an eigenvalue associated with one of the eigenfunctions, and hence the phrase ‘the initial wavefunction collapses to one of the eigenfunctions’. This implies that the measurement alters the initial wavefunction such that a 2nd measurement, if made quickly, yields that same result (this obviously refers to wavefunctions describing non-stationary states, as wavefunctions of stationary states are independent of time). If we prepare a large number of identical systems and simultaneously measure them, the values of c_n are found; with \sum_{n=0}^{\infty}\vert c_n\vert^{2}=1, and the expectation value of the measurements being \langle\psi\vert\hat{O}\vert\psi\rangle.

4) The probability of obtaining an eigenvalue E_i upon measuring a system is given by the square of the inner product of the normalised \psi with the orthonormal eigenfunction \phi_i.

In other words,

\vert\langle\phi_i\vert\psi\rangle \vert^{2}=\vert\langle\phi_i\vert\sum_{n=0}^{\infty}c_n\phi_n\rangle \vert^{2}=\vert\langle c_i\phi_i\vert\phi_i\rangle \vert^{2}=\vert c_i\vert^{2}

If the spectrum is continuous, \psi=\int_{-\infty}^{\infty}c_n\phi_n\, dn and the probability of obtaining an eigenvalue in the range dn is \vert\langle\phi_n\vert\psi\rangle \vert^{2}\, dn.

5) The state of a system immediately after a measurement yielding the eigenvalue E_i is described by the normalised eigenfunction \phi_i.

We have explained in the postulate 3 that this is commonly known as the collapse of the wavefunction \psi to one of the eigenfunctions \phi_i. It is also known as the projection of \psi onto \phi_i, i.e., \hat{P}_i\vert\psi\rangle; or if \psi is not normalised:

\frac{\hat{P}_i\vert\psi\rangle}{\sqrt{\langle\psi\vert\hat{P}_i\vert\psi\rangle}}

 

Question

Show that \sqrt{\langle\psi\vert\hat{P}_i\vert\psi\rangle} is the normalisation constant.

Answer

To normalise a wavefunction,

NN\int c_{i}^{*}\phi_{i}^{*}c_i\phi_i\, d\tau=1\; \; \; \Rightarrow \; \; \; N^{2}=\frac{1}{\vert c_i\vert^{2}\langle\phi_i\vert\phi_i\rangle}\; \; \; \Rightarrow \; \; \; N=\frac{1}{ c_i\sqrt{\langle\phi_i\vert\phi_i\rangle}}

So,

\frac{1}{ \sqrt{\langle\psi\vert\hat{P}_i\vert\psi\rangle}}=\frac{1}{ \sqrt{\langle\psi\vert c_i\phi_i\rangle}}=\frac{1}{ \sqrt{\langle c_i\phi_i\vert\psi\rangle^{*}}}=\frac{1}{ \sqrt{(c_{i}^{*}c_i\langle \phi_i\vert\phi_i\rangle)^{*}}}

=\frac{1}{ \sqrt{c_{i}^{*}c_i\langle \phi_i\vert\phi_i\rangle}}=\frac{1}{ c_i\sqrt{\langle \phi_i\vert\phi_i\rangle}}

 

6) The time evolution of the wavefunction \psi(t) is governed the time-dependent Schrodinger equation i\hbar\frac{d}{dt}\vert\psi(t)\rangle=\hat{H}(t)\vert\psi(t)\rangle.

7) The total wavefunction of a system of identical particles must be either symmetric (for bosons) or antisymmetric (for fermions) under exchange of any two particles. This postulate is often referred to as the symmetrisation postulate.

 

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Linear combination of wavefunctions

A linear combination of wavefunctions is the weighted sum of a complete set of basis wavefunctions \phi_n.

It is commonly used as a technique in quantum chemistry to approximate a multi-electron wavefunction \psi, where

\psi=\sum_{n=0}^{\infty}c_n\phi_n

Because the Hamiltonian operator \hat{H}=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx}+V(x) is a linear operator, a linear combination of \phi_n is a solution to \hat{H} if \{\phi_n\} are solutions to \hat{H}. However, it is only meaningful to solve the eigenvalue equation \hat{H}\psi=E\psi, not just finding solutions to \hat{H}. The problem is \psi (e.g. \psi=c_1\phi_1+c_2\phi_2) is not an eigenfunction of \hat{H}:

\hat{H}(c_1\phi_1+c_2\phi_2)=c_1\hat{H}\phi_1+c_2\hat{H}\phi_2=c_1E_1\phi_1+c_2E_2\phi_2\neq E\psi\; \; \; \; \; \; \; \; 42

The terms on the RHS of the 2nd equality of the above equation cannot be expressed as E\psi, which implies that a state described by \psi does not have a well-defined eigenvalue. Nevertheless, we can find the average value (or expectation value) of \psi. This means that for wavefunctions that are not eigenfunctions of \hat{H}, we can use approximations like \langle H\rangle to find E. The exception to eq42 is if all the basis functions describe a degenerate eigenstate, where E_1=E_2=E:

\hat{H}(c_1\phi_1+c_2\phi_2)=c_1E\phi_1+c_2E\phi_2= E(c_1\phi_1+c_2\phi_2)

In general, a linear combination of wavefunctions (or a linear combination of atomic orbitals, LCAO) is a method of finding an easy or solvable solution to an otherwise complicated or unsolvable one. LCAO is often used in:

    1. Finding the energy of a system using the Hartree-Fock-Roothan method.
    2. Eliminating complex wave functions to find the formula of hydrogenic p/d/f-orbitals that describe their corresponding p/d/f degenerate states.
    3. Constructing wavefunctions that satisfy Pauli’s exclusion principle.
    4. Explaining the relationship between symmetry and degeneracy in group theory.
    5. Constructing symmetry-adapted linear combinations (SALCs) of orbitals, which are building blocks of a molecular orbital (MO) of a molecule.
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Eigenfunctions and eigenvalue equations

An eigenfunction in a vector space of function is a non-zero function, f, that when acted upon by an operator \hat{O}, returns the same function but multiplied by a constant c, which is known as an eigenvalue.

\hat{O}f=cf

The above equation is called an eigenvalue equation. The time-independent Schrodinger equation

\hat{H}\psi=E\psi\; \; \; \; \; \; \; \; 40

where \hat{H} is a Hermitian operator called the Hamiltonian, is an example of an eigenvalue equation. We say that the eigenfunction \psi of a stationary state is a solution to the Schrodinger equation. For a one-dimensional, one-particle system in a stationary state, \hat{H}=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x), and

\left [-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+V(x)\right ]\psi=E\psi\; \; \; \; \; \; \; \; 41

where V(x)=\frac{Ze^{2}}{4\pi\varepsilon _0x}.

 

Question

Show that an eigenfunction representing a stationary state satisfies eq40.

Answer

Stationary states of \hat{H} are determinate states, meaning a measurement of the energy of a particle in that state results in the same value of E. Hence, the variance \sigma^{2} of the observable of \hat{H} for a stationary state would be zero, i.e. \sigma^{2}=\frac{\sum_{i=1}^{N}(x_i-\mu)^{2}}{N}=0. Since \frac{\sum_{i=1}^{N}(x_i-\mu)^{2}}{N} is also the average value of (x_i-\mu)^{2},

\sigma^{2}=\langle(H-\langle H\rangle)^{2}\rangle=\langle(H-E)^{2}\rangle=0

where H is the observable of \hat{H} and \langle H\rangle=\langle\psi\vert\hat{H}\vert\psi\rangle denotes the average value of \hat{H}, which is E since \psi is a stationary state of \hat{H}.

Next, we assume that the operator for the observable (H-E)^{2} is (\hat{H}-E)^{2}. Our assumption is only valid if (\hat{H}-E)^{2} is Hermitian, because for any observable in quantum mechanics, there is a corresponding linear, Hermitian operator. This assumption can be verified as follows:

\langle(H-E)^{2}\rangle=\langle\psi\vert(\hat{H}-E)^{2}\vert\psi\rangle=\int \psi^{*}(\hat{H}-E) ^{2}\psi d\tau

\small =\int \psi^{*}(\hat{H}-E)[(\hat{H}-E)\psi]\, d\tau=\int \psi^{*}\hat{H}[(\hat{H}-E)\psi]\, d\tau-\int \psi^{*}E[(\hat{H}-E)\psi]\, d\tau

Using the property of the Hermitian operator \hat{H} and the condition that E is real (E=E^{*}),

\langle(H-E)^{2}\rangle=\int \psi\{\hat{H}[(\hat{H}-E)\psi]\} ^{*}d\tau-\int \psi\{E[(\hat{H}-E)\psi]\} ^{*}d\tau

=\int \psi\{(\hat{H}-E) [(\hat{H}-E)\psi]\}^{*} d\tau=\int \psi[(\hat{H}-E)]^{2}\psi]^{*}\, d\tau

Since , is a Hermitian operator. So,

\langle(H-E)^{2}\rangle=\langle\psi\vert(\hat{H}-E)^{2}\vert\psi\rangle=\langle\psi\vert(\hat{H}-E)[(\hat{H}-E)\psi]\rangle =0

Because (\hat{H}-E)\psi produces another ket, the property of a Hermitian operator gives

\langle\psi\vert(\hat{H}-E)[(\hat{H}-E)\psi]\rangle =\langle[\hat{H}-E)\psi]\vert[(\hat{H}-E)\psi]\rangle ^{*}=0

From the positive semi-definiteness property of the inner product space, where \langle f\vert f\rangle \geq 0, with \langle f\vert f\rangle = 0 if f=0,

\hat{H}\psi=E\psi

 

Examples of functions that are eigenfunctions of \hat{H} are \psi=sinx and \psi=e^{-kx}, and examples of functions that are not eigenfunctions of \hat{H} are \psi=sin(ax)+sin(bx) and \psi=e^{-k_1x}+e^{-k_2x}. We can easily verify whether a function is an eigenfunction or not one by substituting it into eq41.

 

Question

Why is an eigenfunction defined as a non-zero function?

Answer

Since \hat{H}0=E0=0, an eigenfunction that is zero would have an infinite number of eigenvalues (any value of E multiplied by zero is zero), which contradicts the Born interpretation that an acceptable eigenfunction representing a stationary state in quantum mechanics must be single-valued.

 

 

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Hermitian operator

An operator with domain D_{\hat{O}} is Hermitian if

\langle\boldsymbol{\mathit{m}}\vert\hat{O}\vert\boldsymbol{\mathit{n}}\rangle=\langle\boldsymbol{\mathit{n}}\vert\hat{O}\vert\boldsymbol{\mathit{m}}\rangle^{*}\; \; \; \; \; \; \; \; 35

for all \boldsymbol{\mathit{m}},\boldsymbol{\mathit{n}}\in D_{\hat{O}}.

Eq35 can also be expressed as \langle f_m\vert\hat{O}\vert f_n\rangle=\langle\hat{O}f_m\vert f_n\rangle because

\langle f_m\vert\hat{O}\vert f_n\rangle=\int f_{m}^{*}\hat{O}f_n\: d\tau =\int f_n(\hat{O}f_{m})^{*}\: d\tau=\langle\hat{O}f_m\vert f_n\rangle\; \; \; \; \; \; \; \; 36

where we have used the Hermitian property of the operator in the 2nd equality.

Another way of defining a Hermitian operator is by noting that the average value of a physical property \langle O\rangle must be a real number, i.e. \langle O\rangle=\langle O\rangle^{*} or

\int \psi^{*}\hat{O}\psi\, d\tau =\left ( \int \psi^{*}\hat{O}\psi\: d\tau\right )^{*}=\int \psi(\hat{O}\psi)^{*}d\tau=\int(\hat{O}\psi)^{*}\psi\, d\tau= \langle\hat{O}\psi\vert\psi\rangle\; \; \; \; \; \; \; \; 37

 

Question

Show that eq37 is equivalent to eq36.

Answer

Substitute \psi=f_m+af_n where a is a constant in \int \psi^{*}\hat{O}\psi\, d\tau=\int \psi(\hat{O}\psi)^{*}\, d\tau of eq37 and simplify to give:

a^{*}\int f_{n}^{*}\hat{O}f_m\: d\tau+a\int f_{m}^{*}\hat{O}f_n\: d\tau=a\int f_n(\hat{O}f_{m})^{*}\: d\tau+a^{*}\int f_m(\hat{O}f_{n})^{*}\: d\tau\; \; \; \; \; \; \; \; 38

Since \psi can be expressed as any linear combination of the basis functions f_m and f_n, a can be any number. If we carry out the following:

  1. Substitute a=1 in eq38
  2. Subsitute a=i in eq38 and divide the resultant equation by i
  3. Sum the two resultant equations

we have eq36.

 

Hermitian operators have the following properties:

    1. Eigenfunctions of a Hermitian operator form a complete set
    2. Eigenvalues of a Hermitian operator are real
    3. Eigenfunctions of a Hermitian operator are orthogonal (if they have distinct eigenvalues) or can be chosen to be orthogonal (if they describe a degenerate state)

The first property is a postulate of quantum mechanics. For the 2nd property, consider two eigenvalue equations \hat{O}f_m=a_mf_m and \hat{O}f_n=a_nf_n. Multiplying the first equation on the left by f_{n}^{*}, multiplying the complex conjugate of the 2nd equation on the left by f_m, and then subtracting the two resultant equations and rearranging, yields:

f_{n}^{*}\hat{O}f_m-f_m\hat{O}^{*}f_{n}^{*}=(a_m-a_{n}^{*})f_{n}^{*}f_m

\int f_{n}^{*}\hat{O}f_m\, d\tau-\int f_m\hat{O}^{*}f_{n}^{*}\, d\tau=(a_m-a_{n}^{*})\int f_{n}^{*}f_m \, d\tau

Using the 2nd equality of eq36,

(a_m-a_{n}^{*})\int f_{n}^{*}f_m \, d\tau=0\; \; \; \; \; \; \; \; 39

To show that eigenvalues of a Hermitian operator are real, let m=n

(a_n-a_{n}^{*})\int \vert f_{n}\vert^{2}\, d\tau=0

The integral is zero if and only if f_n were zero over its entire domain. However, an eigenfunction is defined as a non-zero function. Therefore, a_n=a_{n}^{*}.

For the 3rd property, we rewrite eq39 as

(a_m-a_{n})\int f_{n}^{*}f_m \, d\tau=0

If m\neq n, we have a_m-a_n\neq 0 and so, \int f_{n}^{*}f_m\, d\tau=0.

Examples of Hermitian operators are \hat{x} and \hat{p}.

 

Question

Show that \hat{x} and \hat{p}_x are Hermitian.

Answer

We need to show that \int \psi^{*}\hat{O}\psi\, d\tau=\left ( \int \psi^{*}\hat{O}\psi\, d\tau\right )^{*}.

For \hat{x}, we have

\int \psi^{*}\hat{x}\psi\, d\tau=\int \psi^{*}x\psi\, d\tau=\int \psi x\psi^{*} \, d\tau=\left ( \int \psi^{*}x\psi\, d\tau\right )^{*}=\left ( \int \psi^{*}\hat{x}\psi\, d\tau\right )^{*}

where, for the 3rd equality, we used the fact that x is the position of a particle which must be real (x^{*}=x).

For \hat{p}_x, we integrate \int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx by parts to give:

\int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx=\left [\frac{\hbar}{i}\psi^{*}(x)\psi(x)\right ]_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\psi(x)\left (- \frac{\hbar}{i}\right )\frac{d}{dx}\psi^{*}(x)dx

Since a well-behaved wavefunction that is square-integrable is defined as \int_{-\infty}^{\infty}\left | \psi(x)\right |^{2}dx< \infty, i.e. it vanishes at x=\pm \infty, the 1st term on the RHS of the above equation vanishes and we have:

\int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx=\int_{-\infty}^{\infty}\psi(x)\left (- \frac{\hbar}{i}\right )\frac{d}{dx}\psi^{*}(x)dx=\left [ \int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx\right ]^{*}

 

Finally, two operators \hat{A} and \hat{B} are defined as a Hermitian conjugate pair if

\langle\boldsymbol{\mathit{m}}\vert\hat{A}\vert\boldsymbol{\mathit{n}}\rangle=\langle\boldsymbol{\mathit{n}}\vert\hat{B}\vert\boldsymbol{\mathit{m}}\rangle^{*}

 

 

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Expectation value (quantum mechanics)

The expectation value of a quantum-mechanical operator \hat{O} is the weighted average value of its observable. It is defined as:

\langle O\rangle=\langle\psi\vert\hat{O}\vert\psi\rangle\; \; \; \; \; \; \; \; \; \; \; 34

The above equation has roots in probability theory, where the expectation value or expected value of an observable O is \langle O\rangle=\sum_{i=1}^{N}p_io_i, with p_i being the probability of observing the outcome o_i.

\langle O\rangle=\sum_{i=1}^{N}p_io_i=\sum_{i=1}^{N}\left | \langle\phi_i\vert\psi \rangle\right |^{2}o_i=\sum_{i=1}^{N} \langle\phi_i\vert\psi \rangle^{*}\langle\phi_i\vert\psi \rangle o_i

=\sum_{i=1}^{N} \langle\psi\vert\phi_i \rangle\langle\phi_i\vert\psi \rangle o_i=\langle\psi\vert\biggr\(\sum_{i=1}^{N}o_i\vert\phi_i \rangle\langle\phi_i\biggr\)\vert\psi \rangle

From eq30, \hat{O}=\sum_{i=1}^{N}o_i\vert\phi_i \rangle\langle\phi_i\vert, and so

\langle O\rangle=\langle\psi\vert\hat{O}\vert\psi \rangle

We further postulate that eq34 is valid for an infinite dimensional Hilbert space.

 

Question

Why is p_i=\vert\langle\phi_i\vert\psi\rangle\vert^{2} ?

Answer

Consider an operator with a complete set of orthonormal basis eigenfunctions \left \{ \phi_i \right \}. So, any eigenfunction \psi can be written as a linear combination of these basis eigenfunctions, i.e. \psi=\sum_{i=1}^{N}c_i\phi_i. According to the Born rule, the probability that a measurement will yield a given result is \vert\psi\vert^{2}=\psi^{*}\psi, where \int\psi^{*}\psi\: d \tau=1. So,

\int\psi^{*}\psi\: d \tau=\int \sum_{i=1}^{N}c_{i}^{*}\phi_{i}^{*}\sum_{i=1}^{N}c_{i}\phi_{i}d\tau=\sum_{i=1}^{N}\vert c_{i}\vert^{2}\int \phi_{i}^{*}\phi_i d\tau=\sum_{i=1}^{N}\vert c_{i}\vert^{2}=1

We have used the orthonormal property of  in the 2nd and 3rd equalities. \vert c_i\vert^2 is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the eigenfunction . Therefore,

\vert\langle\phi_i\vert\psi\rangle\vert^{2}=\vert\langle\phi_i\vert\sum_j c_j\phi_j\rangle\vert^{2}=\vert\langle\phi_i\vert\ c_i\phi_i\rangle\vert^{2}=\vert c_i\vert^{2}

 

Question

Using the Schrodinger equation, show that the expectation value of the Hamiltonian is E=\int \psi_i^{*}\hat{H}\psi_i\: d\tau.

Answer

Multiplying both sides of eq40 on the left by \psi_i^{*} and integrating over all space, we have

E=\frac{\int \psi_i^{*}\hat{H}\psi_i\: d\tau}{\int \psi_i^{*}\psi_i\: d\tau}

If the wavefunction is normalised, the above equation becomes E=\int \psi_i^{*}\hat{H}\psi_i\: d\tau.

 

 

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Projection operator (quantum mechanics)

A projection operator \hat{P} is a linear operator that transforms a vector in the direction of another vector, i.e., it projects one vector onto another.

In general,

\hat{P}=\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert

\hat{P}\vert\boldsymbol{\mathit{w}}\rangle=(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)\vert\boldsymbol{\mathit{w}}\rangle=\vert\boldsymbol{\mathit{u}}\rangle(\langle\boldsymbol{\mathit{v}}\vert\boldsymbol{\mathit{w}}\rangle) =c\vert\boldsymbol{\mathit{u}}\rangle

where c is a scalar.

It is useful in quantum mechanics to have a projection operator that maps a vector onto another vector, which is part of a complete set of orthonormal basis vectors \left \{\boldsymbol{\mathit{i}} \right \} in a Hilbert space. We define the operator as:

\hat{P}_{\boldsymbol{\mathit{i}=\boldsymbol{\mathit{m}}}}=\vert\boldsymbol{\mathit{m}}\rangle\langle\boldsymbol{\mathit{m}}\vert

This allows us to project a vector \boldsymbol{\mathit{w}} onto the basis vector \boldsymbol{\mathit{m}}:

\hat{P}_{\boldsymbol{\mathit{i}=\boldsymbol{\mathit{m}}}}\vert\boldsymbol{\mathit{w}}\rangle=\vert\boldsymbol{\mathit{m}}\rangle\langle\boldsymbol{\mathit{m}}\vert\boldsymbol{\mathit{w}}\rangle=c\vert\boldsymbol{\mathit{m}}\rangle

If \boldsymbol{\mathit{w}} is a wavefunction \psi that is a linear combination of a complete set of orthonormal basis functions, i.e., \psi=\sum_{i=1}^{N}c_i\phi_i, then

\hat{P}_i\psi=\vert\phi_i\rangle\langle\phi_i\vert\psi\rangle=c_i\phi_i

When we measure an observable of a system whose state is described by \psi, we get an eigenvalue corresponding to an eigenfunction, which is one of the orthonormal basis functions in the complete set. We say that the wavefunction \psi is projected onto (or collapsed into) the eigenfunction \phi_i.

 

 

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Matrix elements of an operator

The matrix elements of an operator are the entries of the matrix representation of the operator.

Consider a linear map from a vector space \vert\psi\rangle to the same vector space, i.e. \vert\phi\rangle=\hat{O}\vert\psi\rangle, where and the orthonormal basis states \vert\varphi_n\rangle span . The matrix representation of the equation is

\begin{pmatrix} \phi_1\\\phi_2 \\ \vdots \end{pmatrix}=\begin{pmatrix} O_{11} &O_{12} &\cdots \\ O_{21} &O_{22} &\cdots \\ \vdots & \vdots &\ddots \end{pmatrix}\begin{pmatrix} \psi_1\\\psi_2 \\ \vdots \end{pmatrix}

where \phi_m and \psi_n are the coefficients of the vectors \vert\phi\rangle and \vert\psi\rangle respectively.

The matrix elements of \vert\phi\rangle are given by

\phi_m=\sum_{n}O_{mn}\psi_n\; \; \; \; \; \; \; \; 32a

Since the orthonormal basis states \vert\varphi_n\rangle span , we have . So,

\vert\phi\rangle=\hat{O} \sum_{n}\vert\varphi_n\rangle\langle\varphi_n\vert\psi\rangle =\sum_{n}\hat{O}\vert\varphi_n\rangle\langle\varphi_n\vert\psi\rangle

\langle\varphi_m\vert\phi\rangle=\sum_{n}\langle\varphi_m\vert\hat{O}\vert\varphi_n\rangle\langle\varphi_n\vert\psi\rangle

Similarly, and so

Comparing eq33 with eq32a, O_{mn}=\langle\varphi_m\vert\hat{O}\vert\varphi_n\rangle. Therefore, O_{mn} are the matrix elements of \hat{O} with respect to the basis states of \vert\varphi_n\rangle.

 

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Spectral decomposition of an operator

The spectral decomposition (also known as eigendecomposition or diagonalisation) of an operator is the transformation of an operator in a given basis to one in another basis, such that the resultant operator is represented by a diagonal matrix.

There are 2 main reasons for diagonalising an operator, especially a Hermitian operator. One is to find its eigenvalues and the other is to convert it into a form that is easier to multiply with.

 

Question

What is a spectrum with respect to linear algebra?

Answer

A spectrum is a collection of all eigenvalues of a matrix. If the matrix represents an operator, its spectral decomposition transforms it to a diagonal matrix with the eigenvalues as its diagonal elements.

 

Consider an operator with a complete set of orthonormal eigenvectors \{\boldsymbol{\mathit{e_i}}\} that is represented by the eigenvalue equation \hat{O}\vert\boldsymbol{\mathit{e_i}}\rangle=o_i\vert\boldsymbol{\mathit{e_i}}\rangle, where i\in \mathbb{N} and o_i are eigenvalues of \hat{O}. Since the eigenvectors form a complete set, any vector \boldsymbol{\mathit{u}} can be written as a linear combination of the basis eigenvectors:

\vert\boldsymbol{\mathit{u}}\rangle=\sum_{i=1}^{N}c_i\vert\boldsymbol{\mathit{e_i}}\rangle \; \; \; \; \; \; \; \; 28

where c_i is the coefficient of the basis eigenvector.

Letting \hat{O} act on eq28, \hat{O}\vert\boldsymbol{\mathit{u}}\rangle=\sum_{i=1}^{N}c_io_i\vert\boldsymbol{\mathit{e_i}}\rangle. As we have a complete set of orthonormal eigenvectors,  \langle\boldsymbol{\mathit{e_i}}\vert\boldsymbol{\mathit{u}}\rangle=c_i and \hat{O}\vert\boldsymbol{\mathit{u}}\rangle=\sum_{i=1}^{N}\langle\boldsymbol{\mathit{e_i}}\vert\boldsymbol{\mathit{u}}\rangle o_i\vert\boldsymbol{\mathit{e_i}}\rangle. Furthermore, \langle\boldsymbol{\mathit{e_i}}\vert\boldsymbol{\mathit{u}}\rangle is a scalar and matrix multiplication is associative. Therefore,

\hat{O}\vert\boldsymbol{\mathit{u}}\rangle=\sum_{i=1}^{N}o_i\vert\boldsymbol{\mathit{e_i}}\rangle\langle\boldsymbol{\mathit{e_i}}\vert\boldsymbol{\mathit{u}}\rangle =\left\(\sum_{i=1}^{N}o_i\vert\boldsymbol{\mathit{e_i}}\rangle\langle\boldsymbol{\mathit{e_i}}\vert\right\)\vert\boldsymbol{\mathit{u}}\rangle\; \; \; \; \; \; \; \; 29

and

\hat{O}=\sum_{i=1}^{N}o_i\vert\boldsymbol{\mathit{e_i}}\rangle\langle\boldsymbol{\mathit{e_i}}\vert\; \; \; \; \; \; \; \; 30

We call eq30 the spectral decomposition of \hat{O}. Since  is the projection operator onto the eigenspace corresponding to o_i, we can say that the spectral decomposition of a quantum operator represents the operator as a sum of projections onto its eigenstates, weighted by its eigenvalues.

 

Question

Show that \hat{O} in eq30, where N=3, is represented by a diagonal matrix.

Answer

\hat{O}=o_1 \begin{pmatrix} 1\\ 0\\0 \end{pmatrix} \begin{pmatrix} 1 &0 &0 \end{pmatrix}+o_2 \begin{pmatrix} 0\\ 1\\0 \end{pmatrix} \begin{pmatrix} 0 &1 &0 \end{pmatrix}+o_3 \begin{pmatrix} 0\\ 0\\1 \end{pmatrix} \begin{pmatrix} 0 &0 &1 \end{pmatrix}

\hat{O}=o_1 \begin{pmatrix} 1 &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \end{pmatrix}+o_2 \begin{pmatrix} 0 &0 &0 \\ 0 &1 &0 \\ 0 &0 &0 \end{pmatrix}+o_3 \begin{pmatrix} 0 &0 &0 \\ 0 &0 &0 \\ 0 &0 &1 \end{pmatrix}=\begin{pmatrix} o_1 &0 &0 \\ 0 &o_2 &0 \\ 0 &0 &o_3 \end{pmatrix}

Each o_i is a diagonal element of the operator, as well as an eigenvalue of the operator.

 

In other words, any operator can be expressed in the form of a diagonal matrix if it has the following properties:

    1. Eigenvectors of the operator form a complete set, i.e. the eigenvectors span the vector space.
    2. Eigenvectors of the operator are orthogonal or can be chosen to be orthogonal.

If the eigenvalues of \hat{O} are real,

This implies that a Hermitian operator can also be expressed in the form of a diagonal matrix because the properties of a Hermitian matrix are:

    1. Eigenvectors of the operator form a complete set, i.e. the eigenvectors span the vector space.
    2. Eigenvectors of the operator are orthogonal or can be chosen to be orthogonal.
    3. Eigenvalues of the operator are real.
    4. \hat{O}.

 

 

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