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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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The uncertainty principle (derivation)

Heisenberg’s uncertainty principle states that the position and momentum of a particle cannot be determined simultaneously with unlimited precision.

The uncertainty not only applies to the position and momentum of a particle, but to any pair of complementary observables, e.g. energy and time. In general, the uncertainty principle is expressed as:

\Delta A\Delta B\geq \frac{1}{2}\left\vert\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle\right\vert\; \; \; \; \; \; \; \; \; 12

where \hat{A} and \hat{B} are Hermitian operators and A and B are their respective observables.

The derivation of eq12 involves the following:

  1. Deriving the Schwarz inequality
  2. Proving the inequality \left ( \Delta A \right )^{2}\left ( \Delta B \right )^{2}\geq -\frac{1}{4}\left ( \left \langle \varphi \left | \left [ \hat{A},\hat{B} \right ] \right |\varphi \right \rangle \right )^{2}
  3. Showing that \left \langle \varphi \left | \left [ \hat{A},\hat{B} \right ] \right |\varphi \right \rangle =-\left \langle \varphi \left | \left [ \hat{A},\hat{B} \right ] \right |\varphi \right \rangle^{*}

 

Step 1

Let

f(\lambda)=\langle\phi-\lambda\psi\vert\phi-\lambda\psi\rangle\; \; \; \; \; \; \; \; 13

where \phi and \psi are arbitrary square integrable wavefunctions and \lambda is an arbitrary scalar.

Since \langle\phi-\lambda\psi\vert\phi-\lambda\psi\rangle=\int (\phi-\lambda\psi)^{*}(\phi-\lambda\psi)d\tau=\int \vert\phi-\lambda\psi\vert^{2}d\tau\geq 0

f(\lambda)\geq 0\; \; \; \; \; \; \; \; \; 14

Expanding eq13, we have

f(\lambda)=\langle\phi\vert\phi\rangle-\lambda\langle\phi\vert\psi\rangle-\lambda^{*}\langle\psi\vert\phi\rangle+\lambda^{*}\lambda\langle\psi\vert\psi\rangle\; \; \; \; \; \; \; \; 15

Since \lambda is an arbitrary scalar, substituting \lambda=\frac{\langle\psi\vert\phi\rangle}{\langle\psi\vert\psi\rangle} and \lambda^{*}=\frac{\langle\phi\vert\psi\rangle}{\langle\psi\vert\psi\rangle} in eq15 gives:

f(\lambda)=\langle\phi\vert\phi\rangle-\frac{\langle\phi\vert\psi\rangle}{\langle\psi\vert\psi\rangle}\langle\psi\vert\phi\rangle

Substituting eq14 in the above equation and rearranging yields \langle\phi\vert\psi\rangle\langle\psi\vert\phi\rangle \leq\langle\phi\vert\phi\rangle\langle\psi\vert\psi\rangle. Since \langle\phi\vert\psi\rangle= \langle\psi\vert\phi\rangle^{*}

\vert\langle\psi\vert\phi\rangle\vert^{2}\leq\langle\phi\vert\phi\rangle\langle\psi\vert\psi\rangle\; \; \; \; \; \; \; \; 16

Eq16 is called the Schwarz Inequality.

 

Step 2

Let \psi=(\hat{A}-\langle A\rangle)\varphi and \phi=(\hat{B}-\langle B\rangle)\varphi, where \varphi is normalised, and \hat{A} and \hat{B} are Hermitian operators, which implies that \hat{A}-\langle A\rangle and \hat{B}-\langle B\rangle are also Hermitian operators (see this article for proof). The variance of the observable of \hat{A} is

(\Delta A)^{2}=\frac{\sum_{i=1}^{N}(\hat{A}-\langle A\rangle)^{2}}{N}=\langle(\hat{A}-\langle A\rangle)^{2} \rangle

=\langle \varphi\vert(\hat{A}-\langle A\rangle)[(\hat{A}-\langle A\rangle)\varphi]\rangle=\langle [(\hat{A}-\langle A\rangle)\varphi]\vert[(\hat{A}-\langle A\rangle)\varphi]\rangle=\langle\psi\vert\psi \rangle\; \; \; \; \; \; \; \; 17

Note that the 2nd last equality uses the property of Hermitian operators (see eq36). Similarly,

(\Delta B)^{2}=\langle\phi\vert\phi\rangle\; \; \; \; \; \; \; \; 18

Substituting eq17 and eq18 in eq16 results in

(\Delta A)^{2}(\Delta B)^{2} \geq\vert \langle\psi\vert\phi\rangle\vert^{2} \; \; \; \; \; \; \; \; 19

Let z= \langle\psi\vert\phi\rangle where z=x+iy. So, \left | z \right |^{2}=x^{2}+y^{2}\geq y^{2}. Since y=\frac{z-z^{*}}{2i}, we have \left | z \right |^{2}\geq \left ( \frac{z-z^{*}}{2i} \right )^{2}, which is

\left | \langle\psi\vert\phi\rangle\right |^{2}\geq \left ( \frac{\langle\psi\vert\phi\rangle-\langle\psi\vert\phi\rangle^{*}}{2i} \right )^{2}=-\frac{1}{4}\left (\langle\psi\vert\phi\rangle-\langle\phi\vert\psi\rangle\right )^{2}\; \; \; \;\; \; \; \; 20

Substituting eq19 in eq20 gives

(\Delta A)^{2}(\Delta B)^{2}\geq -\frac{1}{4}\left ( \langle\psi\vert\phi\rangle-\langle\phi\vert\psi\rangle\right )^{2}\; \; \; \; \; \; \; \; 21

Next, we have

\langle\psi\vert\phi\rangle= \langle (\hat{A}-\langle A\rangle)\varphi\vert (\hat{B}-\langle B\rangle )\varphi\rangle=\langle\varphi\vert(\hat{A}-\langle A\rangle)(\hat{B}-\langle B\rangle)\varphi\rangle

=\langle\varphi\vert\hat{A}\hat{B}\vert\varphi\rangle-\langle B\rangle\langle\varphi\vert\hat{A}\vert\varphi\rangle-\langle A\rangle\langle\varphi\vert\hat{B}\vert\varphi\rangle+\langle A\rangle\langle B \rangle\langle\varphi\vert\varphi\rangle

=\langle\varphi\vert\hat{A}\hat{B}\vert\varphi\rangle-\langle A\rangle\langle B \rangle \; \; \; \; \; \; \; \; 22

Similarly,

\langle\phi\vert\psi\rangle=\langle\varphi\vert\hat{B}\hat{A}\vert\varphi\rangle-\langle B\rangle\langle A \rangle \; \; \; \; \; \; \; \; 23

Substituting eq22 and eq23 in eq21 yields

(\Delta A)^{2}(\Delta B)^{2}\geq -\frac{1}{4}(\langle\varphi\vert\hat{A}\hat{B}\vert\varphi\rangle-\langle\varphi\vert\hat{B}\hat{A}\vert\varphi\rangle )^{2}

=-\frac{1}{4}(\langle\varphi\vert\hat{A}\hat{B}-\hat{B}\hat{A}\vert\varphi\rangle )^{2}=-\frac{1}{4}(\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle )^{2}\; \; \; \; \; \; \; \; 24

 

Step 3

\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle=\langle\varphi\vert\hat{A}\hat{B}\vert\varphi\rangle-\langle\varphi\vert\hat{B}\hat{A}\vert\varphi\rangle=\langle\hat{A}\varphi\vert\hat{B}\varphi\rangle-\langle\hat{B}\varphi\vert\hat{A}\varphi\rangle

=\langle\varphi\vert\hat{B}(\hat{A}\varphi)\rangle^{*}-\langle\varphi\vert\hat{A}(\hat{B}\varphi)\rangle^{*}=- \{\langle\varphi\vert\hat{A}(\hat{B}\varphi)\rangle^{*}-\langle\varphi\vert\hat{B}(\hat{A}\varphi)\rangle^{*}\}

=-\{\langle\varphi\vert\hat{A}(\hat{B}\varphi)\rangle-\langle\varphi\vert\hat{B}(\hat{A}\varphi)\rangle\}^{*}=-\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle^{*}\; \; \; \; \; \; \; \; 25

We have used eq37 for the 2nd equality and eq35 for the 3rd equality. Substituting eq25 in one of the \langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi)\rangle in eq24 results in

(\Delta A)^{2}(\Delta B)^{2}\geq\frac{1}{4} \{\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle^{*}\}\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle

Therefore,

(\Delta A)^{2}(\Delta B)^{2}\geq \frac{1}{4}\vert\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle\vert^{2}

\Delta A\Delta B\geq \frac{1}{2}\vert\langle\varphi\vert[\hat{A},\hat{B}]\vert\varphi\rangle\vert\; \; \; \; \; \; \; \; 26

which is eq12, the general form of the uncertainty principle.

For, the observable pair of position x and momentum p, we have

\Delta x\Delta p\geq \frac{1}{2}\vert\langle\varphi\vert[\hat{x},\hat{p}]\vert\varphi\rangle\vert

Since [\hat{x},\hat{p}]\varphi=x\frac{\hbar}{i}\frac{d}{dx}\varphi-\frac{\hbar}{i}\frac{d}{dx}(x\varphi)=i\hbar\varphi

\Delta x\Delta p\geq \frac{1}{2}\vert\langle\varphi\vert i\hbar\varphi\rangle\vert=\frac{\hbar}{2}\vert i\vert

\Delta x\Delta p\geq \frac{\hbar}{2}\; \; \; \; \; \; \; \; 27

 

 

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Commuting operators

A pair of commuting operators that are Hermitian can have a common complete set of eigenfunctions.

Let \hat{O}_1 and \hat{O}_2 be two different operators, with observables \Omega_1 and \Omega_2 respectively.

\hat{O}_1(\hat{O}_2\psi)=\hat{O}_1(\Omega_2\psi)=\Omega_2\hat{O}_1\psi=\Omega_2\Omega_1\psi

\hat{O}_2(\hat{O}_1\psi)=\hat{O}_2(\Omega_1\psi)=\Omega_1\hat{O}_2\psi=\Omega_1\Omega_2\psi

So, \hat{O}_1(\hat{O}_2\psi)=\hat{O}_2(\hat{O}_1\psi)\; \; \; or\; \; \;\hat{O}_1(\hat{O}_2\psi)-\hat{O}_2(\hat{O}_1\psi)=0. If this is so, we say that the two operators commute. The short notation for \hat{O}_1(\hat{O}_2\psi)-\hat{O}_2(\hat{O}_1\psi) is \left [\hat{O}_1,\hat{O}_2\right ], where in the case of two commuting operators, \left [\hat{O}_1,\hat{O}_2\right ]=0.

When the effect of two operators depends on their order, we say that they do not commute, i.e. \left [\hat{O}_1,\hat{O}_2\right ]\neq 0. If this is the case, we say that the observables \Omega_1 and \Omega_2 are complementary.

One important concept in quantum mechanics is that we can select a common complete set of eigenfunctions for a pair of commuting Hermitian operators. The proof is as follows:

Let \left \{ \vert a_n\rangle \right \} and \left \{ \vert b_n\rangle \right \} be the complete sets of eigenfunctions of \hat{A} and \hat{B} respectively, such that \hat{A}\vert a_n\rangle=a_n\vert a_n\rangle and \hat{B}\vert b_m\rangle=b_m\vert b_m\rangle. If the two operators have a common complete set of eigenfunctions, we can express \vert a_n\rangle as a linear combination of \vert b_m\rangle:

\vert a_n\rangle=\sum_{m=1}^{k}c_{nm}\vert b_m\rangle\; \; \; \; \; \; \; \; 5

For example, the eigenfunction is:

\vert a_1\rangle=c_{11}\vert b_1\rangle+c_{12}\vert b_2\rangle+\cdots+c_{1k}\vert b_k\rangle\; \; \; \; \; \; \; \; 6

Since some of the eigenfunctions \vert b_m\rangle may describe degenerate states (i.e. some \vert b_m\rangle are associated with the same eigenvalue b_i), we can rewrite \vert a_n\rangle as:

\vert a_n\rangle=\sum_{i=1}^{j}\vert(a_n) b_i\rangle\; \; \; \; \; \; \; \; 7

where \vert(a_n) b_i\rangle=\sum_{m=1}^{k}d_{nm}\vert b_m\rangle\delta_{b_i,b_m} and b_i represents distinct eigenvalues of the complete set of eigenfunctions of \hat{B}.

For example, if the linear combination of \vert a_1 \rangle in eq6 has \vert b_1\rangle and \vert b_2\rangle describing the same eigenstate with eigenvalue b_1, and \vert b_4\rangle and \vert b_5\rangle describing another common eigenstate with eigenvalue b_3,

\vert a_1\rangle=\vert(a_1) b_1\rangle+\vert(a_1) b_2\rangle+\vert(a_1) b_3\rangle+\cdots+\vert(a_1) b_j\rangle

where \vert(a_1) b_1\rangle=d_{11}\vert b_1\rangle+d_{12}\vert b_2\rangle, \vert(a_1) b_2\rangle=d_{13}\vert b_3\rangle, \vert(a_1) b_3\rangle=d_{14}\vert b_4\rangle+d_{15}\vert b_5\rangle and so on.

In other words, eq7 is a sum of eigenfunctions with distinct eigenvalues of \hat{B}. Since a linear combination of eigenfunctions describing a degenerate eigenstate is an eigenfunction of \hat{B}, we have

\hat{B}\vert(a_n)b_i\rangle=b_i\vert(a_n)b_i\rangle\; \; \; \; \; \; \; \; 8

i.e. \vert(a_n)b_i\rangle is an eigenfunction of \hat{B}. Furthermore, the set \left \{ \vert(a_n)b_i\rangle \right \} is complete, which is deduced from eq7, where the set \left \{ \vert a_n\rangle \right \} is complete.

From \hat{A}\vert a_n\rangle=a_n\vert a_n\rangle, we have:

(\hat{A}-a_n)\vert a_n\rangle=0

Substituting eq7 in the above equation, we have

(\hat{A}-a_n)\vert a_n\rangle=\sum_{i=1}^{j}(\hat{A}-a_n) \vert (a_n)b_i\rangle=0\; \; \; \; \; \; \; \; 9

By operating on the 1st term of the summation in the above equation with \hat{B}, and using the fact that \hat{A} commute with \hat{B},

\hat{B}(\hat{A}-a_n)\vert (a_n)b_1\rangle=(\hat{A}-a_n)\hat{B} \vert (a_n)b_1\rangle\; \; \; \; \; \; \; \; 10

Substituting eq8, where i=1 in the above equation,

\hat{B}(\hat{A}-a_n)\vert (a_n)b_1\rangle=b_1(\hat{A}-a_n) \vert (a_n)b_1\rangle\; \; \; \; \; \; \; \; 11

Repeating the operation of \hat{B} on the remaining terms of the summation in eq9, we obtain equations similar to eq11 and we can write:

\hat{B}(\hat{A}-a_n)\vert (a_n)b_i\rangle=b_i(\hat{A}-a_n) \vert (a_n)b_i\rangle

i.e. (\hat{A}-a_n)\vert (a_n)b_i\rangle is an eigenfunction of \hat{B} with distinct eigenvalues b_i. Since \hat{B} is Hermitian and (\hat{A}-a_n)\vert (a_n)b_i\rangle are associated with distinct eigenvalues, the eigenfunctions (\hat{A}-a_n)\vert (a_n)b_i\rangle are orthogonal and therefore linearly independent. Consequently, each term in the summation in eq9 must be equal to zero:

(\hat{A}-a_n)\vert (a_n)b_i\rangle=0\; \; \; \Rightarrow \; \; \;\hat{A}\vert (a_n)b_i\rangle=a_n\vert (a_n)b_i\rangle

This implies that \vert (a_n)b_i\rangle, which is a complete set as mentioned earlier, is also an eigenfunction of \hat{A}. Therefore, we can select a common complete set of eigenfunctions \left \{ \vert (a_n)b_i\rangle\right \} for a pair of commuting Hermitian operators. Conversely, if two Hermitian operators do not commute, eq10 is no longer valid and we cannot select a common complete set of eigenfunctions for them.

 

 

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Operators (quantum mechanics)

An operator \hat{O} in a vector space V maps a set of vectors \boldsymbol{\mathit{v}}_i to another set of vectors \boldsymbol{\mathit{v}}_j (or transforms one function into another function), where \boldsymbol{\mathit{v}}_i,\boldsymbol{\mathit{v}}_j\in V. For example, the operator \frac{d^{2}}{dx^{2}} transforms the function f(x) into f^{''}(x):

\frac{d^{2}}{dx^{2}}f(x)=f^{''}(x)

Linear operators have the following properties:

    1. \hat{O}(\boldsymbol{\mathit{u}}+\boldsymbol{\mathit{v}})=\hat{O}\boldsymbol{\mathit{u}}+\hat{O}\boldsymbol{\mathit{v}}
    2. \hat{O}(c\boldsymbol{\mathit{u}})=c\hat{O}\boldsymbol{\mathit{u}}
    3. (\hat{O}_1\hat{O}_2)\boldsymbol{\mathit{u}}=\hat{O}_1(\hat{O}_2\boldsymbol{\mathit{u}})

Two operators commonly encountered in quantum mechanics are the position and linear momentum operators. To construct these operators, we refer to probability theory, where the expectation value of the position x of a particle in a 1-D box of length L is

\langle x\rangle=\int_{0}^{L}xP(x)dx=\int_{0}^{L}x\left |\psi(x) \right |^{2}dx=\int_{0}^{L}\psi(x)\, x\, \psi(x)dx

where P(x) is the probability of observing the particle at a particular position between 0 and L, and \psi(x) is the particle’s wavefunction, which is assumed to be real.

Comparing the above equation with the expression of the expectation value of a quantum-mechanical operator, \hat{x}=x.

One may infer that the linear momentum operator is \hat{p}_x=p_x. However, we must find a form of p_x that is a function of x so that we can compute \int_{0}^{L}\psi(x)p_x\psi(x)dx. If we compare the time-independent Schrodinger equation \left [ -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x) \right ]\psi(x)=E\psi(x) with the total energy equation \frac{p_x^{\: 2}}{2m}+V(x)=E, we have \hat{p}_x=\frac{\hbar}{i}\frac{d}{dx}.

To test the validity of \hat{x}=x and \hat{p}_x=\frac{\hbar}{i}\frac{d}{dx}, we compute \langle x\rangle and \langle p_x\rangle using the 1-D box wavefunction of \sqrt\frac{2}{L}sin\frac{n\pi x}{L} and check if the results are reasonable with respect to classical mechanics.

Integrating \langle x\rangle=\frac{2}{L}\int_{0}^{L}xsin^{2}\left ( \frac{n\pi x}{L} \right )dx by parts, we have \langle x\rangle=\frac{L}{2}. In classical mechanics, the particle can be anywhere in the 1-D box with equal probability. Therefore, the average position of \langle x\rangle=\frac{L}{2} is reasonable.

For the linear momentum operator, we have \langle p_x\rangle=\frac{2\hbar}{iL}\int_{0}^{L}sin\left ( \frac{n\pi x}{L} \right )\frac{d}{dx}sin\left ( \frac{n\pi x}{L} \right )dx=0. Since \langle x\rangle=\frac{L}{2}, we expect \langle p_x\rangle=m\frac{d\langle x\rangle}{dt}=0. Therefore, x and \frac{\hbar}{i}\frac{d}{dx} are reasonable assignments of the position and linear momentum operators respectively. In 3-D, the position and linear momentum operators are:

\hat{x}=x\; \; \; \; \;\hat{y}=y\; \; \; \; \;\hat{z}=z

\hat{p}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}\; \; \; \; \;\hat{p}_y=\frac{\hbar}{i}\frac{\partial}{\partial y}\; \; \; \; \;\hat{p}_z=\frac{\hbar}{i}\frac{\partial}{\partial z}\; \; \; \; \;\; \; \; 4

To see the proof that the position and linear momentum operators are Hermitian, read this article.

 

 

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Kronecker product

The Kronecker product, denoted by \otimes, is a multiplication method for generating a new vector space from existing vector spaces, and therefore, new vectors from existing vectors.

Consider 2 vectors spaces, e.g. V=\mathbb{R}^{2} and W=\mathbb{R}^{3}. For \boldsymbol{\mathit{v}}=\begin{pmatrix} a_1\\a_2 \end{pmatrix} in V and \boldsymbol{\mathit{w}}=\begin{pmatrix} b_1\\b_2\\b_3 \end{pmatrix} in W, we can define a new vector space, V\otimes W, which consists of the vector \boldsymbol{\mathit{v}}\otimes\boldsymbol{\mathit{w}}, where:

\boldsymbol{\mathit{v}}\otimes\boldsymbol{\mathit{w}}=\begin{pmatrix} a_1\\a_2 \end{pmatrix}\otimes \begin{pmatrix} b_!\\b_2 \\ b_3 \end{pmatrix}=\begin{pmatrix} a_1b_1\\a_1b_2 \\a_1b_3 \\ a_2b_1 \\ a_2b_2 \\ a_2b_3 \end{pmatrix}

If the basis vectors for V and W are V=\left \{\boldsymbol{\mathit{e_1}},\boldsymbol{\mathit{e_2}}\right \} and W=\left \{\boldsymbol{\mathit{f_1}},\boldsymbol{\mathit{f_2}},\boldsymbol{\mathit{f_3}}\right \} respectively, the basis for V\otimes W is:

 

Question

Why is a new vector space?

Answer

An -dimensional vector space is spanned by  linearly independent basis vectors. The basis vectors for V=\left \{\boldsymbol{\mathit{e_1}},\boldsymbol{\mathit{e_2}}\right \} and W=\left \{\boldsymbol{\mathit{f_1}},\boldsymbol{\mathit{f_2}},\boldsymbol{\mathit{f_3}}\right \} are

and consequently, the basis vectors for  are

These 6 linearly independent basis vectors therefore span a 6-dimensional space.

 

This implies that V\otimes W is nm dimensional if V is n-dimensional and W is m-dimensional. Since V\otimes W is a vector space, the vectors \boldsymbol{\mathit{v}}\otimes\boldsymbol{\mathit{w}} must follow the rules of addition and multiplication of a vector space. Each vector \boldsymbol{\mathit{v}}\otimes\boldsymbol{\mathit{w}} in the new vector space can then be written as a linear combination of the basis vectors \boldsymbol{\mathit{e_i}}\otimes\boldsymbol{\mathit{f_j}}, i.e. \sum c_{i,j}\boldsymbol{\mathit{e_i}}\otimes\boldsymbol{\mathit{f_j}}.

In general, if

then

Since the pair  in  is distinct for each \boldsymbol{\mathit{e_i}}\otimes\boldsymbol{\mathit{f_j}} vector, the Kronecker product \boldsymbol{\mathit{e_i}}\otimes\boldsymbol{\mathit{f_j}} results in  basis vectors, which span an  vector space.

As mentioned in an earlier article, a vector space is a set of objects that follows certain rules of addition and multiplication. If the objects are matrices, we have a vector space of matrices. For example, the vector spaces of matrices and generates a new vector space of matrices , where

Similarly, if the objects are functions, we have a vector space of functions and the Kronecker product of two vector spaces of functions  and generates a new vector space of functions . If and are spanned by basis functions and basis functions respectively, is spanned by basis functions.

A vector space that is generated from two separate vector spaces has applications in quantum composite systems and in group theory.

Question

What is the relation between the matrix entries of A, B and C in ?

Answer

Let the matrix entries of A, B and C be , and respectively, where

Using the ordering convention called dictionary order, where  is determined by  and , and is determined by and , such that  and are given by

For example, if and ,

We can then express the matrix entries of as .

 

 

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Hilbert space

A Hilbert space H is a complete inner product space. It allows the application of linear algebra and calculus techniques in a space that may have an infinite dimension.

The inner product in a Hilbert space has the following properties:

  1. Conjugate symmetry: \langle\phi_1\vert\phi_2\rangle=\langle\phi_2\vert\phi_1\rangle^{*}
  2. Linearity with respect to the 2nd argument: \langle\phi_1\vert c_2\phi_2+c_3\phi_3\rangle=c_2\langle\phi_1\vert \phi_2\rangle+c_3\langle\phi_1\vert \phi_3\rangle
  3. Antilinearity with respect to the first argument: \langle c_1\phi_1+c_2\phi_2\vert \phi_3\rangle=c_1^{*}\langle\phi_1\vert \phi_3\rangle+c_2^{*}\langle\phi_2\vert \phi_3\rangle
  4. Positive semi-definiteness: \langle\phi_1\vert \phi_1\rangle\geq 0, with \langle\phi_1\vert \phi_1\rangle= 0 if \phi_1=0

The last property can be illustrated using the \mathbb{R}^{2} space that is equipped with an inner product. Such a space is an example of a real finite-dimensional Hilbert space. The inner product of the vector \boldsymbol{\mathit{u}} with itself is:

\boldsymbol{\mathit{u}}\cdot\boldsymbol{\mathit{u}}=\begin{pmatrix} u_1 &u_2 \end{pmatrix}\begin{pmatrix} u_1\\u_2 \end{pmatrix}=u{_{1}}^{2}+u{_{2}}^{2}=\{\begin{matrix} >0 &if\; \boldsymbol{\mathit{u}}\neq 0 \\ 0&if \; \boldsymbol{\mathit{u}}=0\end{matrix}

We define a complete Hilbert space as one where every Cauchy sequence in H converges to an element of H. If you recall, a Cauchy sequence is a sequence, e.g. \left \{ x_n \right \}_{n=1}^{\infty} where x_n=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}, for which

\lim_{m,n\rightarrow \infty}\left | x_n-x_m \right |=0

We can also define the completeness of a Hilbert space in terms of a sequence of vectors \left \{ \boldsymbol{\mathit{v_n}} \right \}_{n=1}^{\infty}, where \boldsymbol{\mathit{v_n}} =\sum_{k=1}^{n}\boldsymbol{\mathit{u_k}}. Each element \boldsymbol{\mathit{v_n}} is represented by a series of vectors, which converges absolutely (i.e. \sum_{k=1}^{\infty}\left \| \boldsymbol{\mathit{u_k}} \right \|< \infty) and converges to an element of H. In other words, the series of vectors in H converges to some limit vector \boldsymbol{\mathit{L}} in H:

\lim_{n\rightarrow \infty}\left \|\boldsymbol{\mathit{L}}-\sum_{k=1}^{n}\boldsymbol{\mathit{u_k}} \right \|=0

Generally, every element of a vector space can be a point, a vector or a function. In quantum mechanics, we are interested in a Hilbert space called the L^{2} space, where the eigenfunctions of a Hermitian operator are square integrable, i.e. \int_{-a}^{b}\left | \phi(x) \right |^{2}dx< \infty.

Not to be confused with the completeness of a Hilbert space, the completeness of a set of basis eigenfunctions refers to the property that any eigenfunction of the Hilbert space can be expressed as a linear combination of the basis eigenfunctions. An example is the \mathbb{R}^{2} space, where the set of basis vectors \left \{\boldsymbol{\mathit{\hat i}},\boldsymbol{\mathit{\hat j}}\right \} is complete, with a linear combination of \boldsymbol{\mathit{\hat i}} and \boldsymbol{\mathit{\hat j}} spanning \mathbb{R}^{2}. In H, the number of basis vectors \boldsymbol{\mathit{u_k}} may be infinite. If the set of \boldsymbol{\mathit{u_k}} is complete, we say that it spans H, which is itself complete.

Just as the orthonormal vectors \boldsymbol{\mathit{\hat i}} and \boldsymbol{\mathit{\hat j}} form a complete set of basis vectors in the \mathbb{R}^{2} space, where any vector in \mathbb{R}^{2} can be expressed as a linear combination of \boldsymbol{\mathit{\hat i}} and \boldsymbol{\mathit{\hat j}}, we postulate the existence of a complete basis set of orthonormal wavefunctions of any Hermitian operator in L^{2}.

 

 

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