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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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Completeness of a vector space

A complete vector space is one that has no “missing elements” (e.g. no missing coordinates).

In the previous article, we learned that the function d(\boldsymbol{\mathit{u}},\boldsymbol{\mathit{v}}) defines the distance between two elements of the vector space. Such a function is called a metric and it measures the ‘closeness’ of elements (or points) in a vector space. Since a vector space is a collection of elements, we can use a sequence, e.g. \left \{ x_n \right \}_{n=1}^{\infty}, to represent elements of a vector space X. If the distance between two members of the sequence gets closer as n gets larger (see diagram below), i.e.

\lim_{m,n\rightarrow \infty}d(x_n-x_m)=0

we call the sequence a Cauchy sequence.

Cauchy sequences are useful in determining the completeness of a vector space. A vector space V is complete if every Cauchy sequence in V converges to an element of V. For example, the sequence \left \{ x_n \right \}_{n=1}^{\infty}, where x_n=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}, is one of many Cauchy sequences of rational numbers in the rational number space \mathbb{Q}. However, the sequence converges to ln2, which is not an element of \mathbb{Q}. Therefore, \mathbb{Q} is not complete, and has “missing elements” or “gaps”, as compared to the real space number space \mathbb{R}, which is complete.

 

Question

Show that \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=ln2.

Answer

If \left | x \right |<1, then (1-x)(1+x+x^{2}+\cdots). So, \frac{1}{1-x}=1+x+x^{2}+\cdots or

\frac{1}{1-(-x)}=\frac{1}{1+x}=1-x+x^{2}+\cdots

Integrating the 2nd equality of the above equation on both sides gives

ln\left | 1+x \right |=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{k}}{k}

Substituting x=1  in the above equation yields ln2=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}.

 

A vector space with no “missing elements” is essential for scientists to work with to formulate theories (e.g. kinematics) and solving problems associated with those theories. Furthermore, the ability to compute limits in a complete vector space implies that we can apply calculus to solve problems defined by the space. For example, a complete inner product space called a Hilbert space, which will be discussed in the next article, is used to formulate the theories of quantum mechanics.

 

 

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Inner product space

An inner product space is a vector space with an inner product.

An inner product is an operation that assigns a scalar to a pair of vectors \langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle, or a scalar to a pair of functions \langle f\vert g\rangle. The way to assign the scalar may be through the matrix multiplication of the pair of vectors, for instance

\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle=\begin{pmatrix} u_{1}^{*} &u_{2}^{*}& \cdots &u_{N}^{*} \end{pmatrix}\begin{pmatrix} v_1\\v_2 \\ \vdots \\ v_N \end{pmatrix}=\sum_{i=1}^{N}u_{i}^{*}v_i\; \; \; \; \; \; \; \; 3

or it may be through an integral of the pair of functions:

\langle f\vert g\rangle=\int_{-\infty}^{\infty}f(x)^{*}g(x)dx

You may notice that eq3 resembles a dot product. The dot product pertains to vectors in \mathbb{R}^{3}, where \boldsymbol{\mathit{A}}\cdot\boldsymbol{\mathit{B}}=\sum_{i=1}^{3}A_iB_i, which can be extended to N-dimensions, where \langle\boldsymbol{\mathit{A}}\vert\boldsymbol{\mathit{B}}\rangle=\sum_{i=1}^{N}A_iB_i, and to include complex and real functions, \langle f\vert g\rangle=\int_{-\infty}^{\infty}f(x)^{*}g(x)dx. Therefore, an inner product is a generalisation of the dot product.

An inner product space has the following properties:

    1. Conjugate symmetry: \langle f\vert g\rangle=\langle g\vert f\rangle^{*}
    2. Additivity: \langle f+g\vert h\rangle=\langle f\vert h\rangle+\langle g\vert h\rangle
    3. Positive semi-definiteness: \langle f\vert f\rangle\geq 0, with \langle f\vert f\rangle= 0 if f=0

 

Question

i) Why is the inner product space positive semi-definite?
ii) Show that orthogonal vectors are linearly independent.
iii) Prove that .

Answer

i) A general vector space of \langle\boldsymbol{\mathit{a}}\vert\boldsymbol{\mathit{a}}\rangle can be positive or negative. The inner product space is defined such that \langle f\vert f\rangle\geq 0, with \langle f\vert f\rangle= 0 if f=0, which is useful in quantum mechanics.

ii) Let the set of vectors \left \{ \boldsymbol{\mathit{v_k}} \right \} in eq1 be orthogonal vectors. The dot product of eq1 with \boldsymbol{\mathit{v_i}} gives c_i\boldsymbol{\mathit{v_i}}\cdot\boldsymbol{\mathit{v_i}} =c_i\left |\boldsymbol{\mathit{v_i}} \right |^{2}=0. Since the magnitudes of orthogonal vectors are non-zero, c_i=0. Hence, orthogonal vectors are linearly independent.

iii) Let’s consider two vectors and as position vectors starting from the origin. Then the vector forms a triangle with them. According to the law of cosines, we have:

Substituting into the above equation gives:

which completes the proof.

 

Two functions (or two vectors) are orthogonal if \langle f\vert g\rangle= 0. Elements of a set of basis functions are orthonormal if \langle \phi_i\vert \phi_j\rangle=\delta_{ij} where

\delta_{ij}= \{\; \begin{matrix} 1 & for\; \; i=j\\ 0 & for\; \; i\neq j \end{matrix}

In other words, two functions (or two vectors) are orthonormal if they are orthogonal and normalised.

Finally, the norm (or length) of a vector \boldsymbol{\mathit{u}} is denoted by \left \|\boldsymbol{\mathit{u}}\right \| and is defined as \left \|\boldsymbol{\mathit{u}}\right \|=\sqrt{\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{u}}\rangle}=\sqrt{\left |\boldsymbol{\mathit{u}}\right |\left |\boldsymbol{\mathit{u}}\right |cos\: 0^{\circ} }=\left |\boldsymbol{\mathit{u}}\right |. With this association of inner product and the length of a vector, we can relate the inner product \langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle and the Euclidean distance d(\boldsymbol{\mathit{u}},\boldsymbol{\mathit{v}}) between 2 vectors \boldsymbol{\mathit{u}} and \boldsymbol{\mathit{v}}. Using the \mathbb{R}^{2} space as an example, where \boldsymbol{\mathit{u}}=\begin{pmatrix} 6\\9 \end{pmatrix} and \boldsymbol{\mathit{v}}=\begin{pmatrix} 3\\5 \end{pmatrix}, we have

d(\boldsymbol{\mathit{u}},\boldsymbol{\mathit{v}})=\left \|\boldsymbol{\mathit{u}}-\boldsymbol{\mathit{v}}\right \|=\sqrt{\langle\boldsymbol{\mathit{u}}- \boldsymbol{\mathit{v}}\vert\boldsymbol{\mathit{u}}- \boldsymbol{\mathit{v}}\rangle}=\sqrt{\langle\begin{pmatrix} 3\\4 \end{pmatrix}\vert \begin{pmatrix} 3\\4 \end{pmatrix}\rangle}

=\sqrt{\begin{pmatrix} 3 &4 \end{pmatrix}\begin{pmatrix} 3\\4 \end{pmatrix}}=5

Hence, the norm naturally comes from the inner product, i.e. every inner product space is a normed space, but not vice versa.

 

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Vector space of functions

As mentioned in the previous article, a vector space \textit{V} is a set of objects that follows certain rules of addition and multiplication. This implies that a set of functions \textit{V}=\left \{ f(x),g(x),h(x)\cdots \right \} that follows the same rules, forms a vector space of functions. The properties of a vector space of functions are:

1) Commutative and associative addition for all functions of the closed set \textit{V}.

f(x)+g(x)=g(x)+h(x)

\left [f(x)+g(x)\right ]+h(x)=f(x)+\left [g(x)+h(x)\right ]

2) Associativity and distributivity of scalar multiplication for all functions of the closed set.

\gamma\left [ \delta f(x) \right ]=(\gamma\delta)f(x)

\gamma\left [ f(x)+g(x) \right ]=\gamma f(x)+\gamma g(x)

(\gamma +\delta)f(x)=\gamma f(x)+\delta f(x)

where \gamma and \delta are scalars.

3) Scalar multiplication identity.

1f(x)=f(x)

4) Additive inverse.

f(x)+[-f(x)]=0

5) Existence of null vector \boldsymbol{\mathit{0}}, such that

\boldsymbol{\mathit{0}}+f(x)=f(x)

where \boldsymbol{\mathit{0}} in this case is a zero function that returns zero for any inputs.

Similarly, a set of linearly independent functions y_0(x),y_1(x),\cdots,y_n(x) forms a set of basis functions. We have a complete set of basis functions if any well-behaved function in the domain a\leq x\leq b can be written as a linear combination of these basis functions, i.e.

f(x)=\sum_{n=0}^{\infty}c_ny_n(x)

In quantum chemistry, physical states of a system are expressed as functions called wavefunctions.

 

 

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Vector space

A vector space is a set of objects  that follows specific rules of addition and multiplication.

These objects are called vectors and the rules are:

1) Commutative and associative addition for all elements of the closed set.

\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}}=\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{1}}}

(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})+\boldsymbol{\mathit{v_{3}}}=\boldsymbol{\mathit{v_{1}}}+(\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{3}}})

2) Associativity and distributivity of scalar multiplication for all elements of the closed set

c_1(c_2\boldsymbol{\mathit{v_{1}}})=(c_1c_2)\boldsymbol{\mathit{v_{1}}}

c_1(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})=c_1\boldsymbol{\mathit{v_{1}}}+c_1\boldsymbol{\mathit{v_{2}}}

(c_1+c_2)\boldsymbol{\mathit{v_{1}}}=c_1\boldsymbol{\mathit{v_{1}}}+c_2\boldsymbol{\mathit{v_{1}}}

where c_1 and c_2 are scalars.

3) Scalar multiplication identity.

\boldsymbol{\mathit{1}}\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}

4) Additive inverse.

\boldsymbol{\mathit{v_{1}}}+(-\boldsymbol{\mathit{v_{1}}})=0

5) Existence of null vector , such that

\boldsymbol{\mathit{0}}+\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}

In a vector space V=\left \{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\cdots ,\boldsymbol{\mathit{v_{k}}}\right \}, one vector can be expressed as a linear combination of other vectors in the set, e.g.:

\boldsymbol{\mathit{z}}=c_1\boldsymbol{\mathit{v_{1}}}+c_2\boldsymbol{\mathit{v_{2}}}+\cdots+c_k\boldsymbol{\mathit{v_{k}}}

The span of a set of vectors V is the set of vectors that can be written as a linear combination of vectors in the set . For example, the span of the set of unit vectors \boldsymbol{\mathit{\hat{i}}} and \boldsymbol{\mathit{\hat{j}}} in the \mathbb{R}^{2} space is the set of all vectors (including the null vector) in the \mathbb{R}^{2} space. Alternative, we say that \boldsymbol{\mathit{\hat{i}}} and \boldsymbol{\mathit{\hat{j}}} span \mathbb{R}^{2}.

If we vary c_1,c_2,\cdots c_k (but not the trivial case where all scalars are zero) such that \boldsymbol{\mathit{z}} is equal to \boldsymbol{\mathit{0}},

c_1\boldsymbol{\mathit{v_1}}+c_2\boldsymbol{\mathit{v_2}}+\cdots+c_k\boldsymbol{\mathit{v_k}}=\boldsymbol{\mathit{0}}\; \; \; \; \; \; \;\; 1

the set of vectors \boldsymbol{\mathit{v_1}},\boldsymbol{\mathit{v_2}},\cdots\boldsymbol{\mathit{v_k}} is said to be linearly dependent because any vector can be written as a linear combination of the others:

\boldsymbol{\mathit{v_1}}=\frac{-c_2}{c_1}\boldsymbol{\mathit{v_2}}+\cdots+\frac{-c_k}{c_1}\boldsymbol{\mathit{v_k}}\; \; \; \; \; \; \; \; 2

If the only way to satisfy eq1 is when c_k=0 for all k, the set of vectors is said to be linearly independent. In this case, we can no longer express any vector as a linear combination of the other vectors (as c_1=0, resulting in RHS of eq2 being undefined). An example of a set of linearly independent vectors is the set of unit vectors \boldsymbol{\mathit{\hat{i}}}, \boldsymbol{\mathit{\hat{j}}} in the \mathbb{R}^{2} space.

 

Question

Can a set of linearly independent vectors include the zero vector?

Answer

No, because if and in eq1, then can be any number. Since is not necessarily 0, it contradicts the definition of linear independence.

 

A set of \textit{N} linearly independent vectors in an \textit{N}-dimensional vector space \textit{V} forms a set of basis vectors, \boldsymbol{\mathit{e_1}},\boldsymbol{\mathit{e_2}},\cdots,\boldsymbol{\mathit{e_N}}. A complete basis set is formed by a set of basis vectors of \textit{V} if any vector \boldsymbol{\mathit{x}} in the span of \textit{V} can be written as a linear combination of those basis vectors, i.e.

\boldsymbol{\mathit{x}}=\sum_{i=1}^{N}x_i\boldsymbol{\mathit{e_i}}

 

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Dirac bra-ket notation

The Dirac bra-ket notation is a concise way to represent objects in a complex vector space \mathbb{C}^{n}.

A ket, denoted by \vert \textbf{\textit{v}} \rangle, is a vector \textbf{\textit{v}}. Since a linear operator \hat{O} maps a vector to another vector, we have \hat{O}\vert \boldsymbol{\mathit{v_{1}}}\rangle=\vert \boldsymbol{\mathit{v_{2}}} \rangle.

A bra, denoted by \langle\boldsymbol{\mathit{u}}\vert , is often associated with a ket in the form of an inner product, denoted by \langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle. If a ket is expressed as a column vector, the corresponding bra is the conjugate transpose of its ket, i.e. \langle\boldsymbol{\mathit{u}}\vert=\vert\boldsymbol{\mathit{u}}\rangle^{\dagger}. The inner product can therefore be written as the following matrix multiplication:

or in the case of functions:

\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle=\int_{-\infty}^{\infty}f_{u}^{*}(x)f_{v}(x)dx

Since a linear operator acting on a ket is another ket, we can express an inner product as:

\langle\phi_{i}\vert\phi_{k}\rangle=\langle\phi_{i}\vert\hat{O}\vert\phi_{j}\rangle=\int \phi_{i}^{*}\hat{O}\phi_{j}d\tau

where \hat{O}\vert\phi_{j}\rangle=\vert\phi_{k}\rangle.

If i=j, then \langle\phi\vert\hat{O}\vert\phi\rangle is the expectation value (or average value) of the operator \hat{O}.

As mentioned above, bras and kets can be represented by matrices. Therefore, the multiplication of a bra and a ket that involves a linear operator is associative, e.g.:

\langle\boldsymbol{\mathit{u}}\vert(\hat{O}\vert\boldsymbol{\mathit{v}}\rangle)=(\langle\boldsymbol{\mathit{u}}\vert\hat{O})\vert\boldsymbol{\mathit{v}}\rangle\equiv\langle\boldsymbol{\mathit{u}}\vert\hat{O}\vert\boldsymbol{\mathit{v}}\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)

(\hat{O}\vert\boldsymbol{\mathit{u}}\rangle)\langle\boldsymbol{\mathit{v}}\vert=\hat{O}(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)\equiv\hat{O}\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert

You can verify the above examples using a 2×2 matrix with complex elements to represent the operator acting on a vector in \mathbb{C}^{2}. The three examples reveal that:

    1. \hat{O}\vert\boldsymbol{\mathit{v}}\rangle produces another ket.
    2. \langle\boldsymbol{\mathit{u}}\vert\hat{O} results in another bra. This is because (\langle\boldsymbol{\mathit{u}}\vert\hat{O})\vert\boldsymbol{\mathit{v}}\rangle=\langle\boldsymbol{\mathit{u}}\vert(\hat{O}\vert\boldsymbol{\mathit{v}}\rangle)=\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v'}}\rangle=c, where c is a scalar; and if (\langle\boldsymbol{\mathit{u}}\vert\hat{O})\vert\boldsymbol{\mathit{v}}\rangle=c, the only possible identity of (\langle\boldsymbol{\mathit{u}}\vert\hat{O}) is a bra.
    3. \vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert , which is called an outer product, is an operator because (\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, i.e. \vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert maps the ket \vert\boldsymbol{\mathit{w}}\rangle to another ket c\vert\boldsymbol{\mathit{u}}\rangle. In other words, the operator \vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert transforms the vector \vert\boldsymbol{\mathit{w}}\rangle in the direction of the vector \vert\boldsymbol{\mathit{u}}\rangle, i.e. \vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert projects \vert\boldsymbol{\mathit{w}}\rangle onto \vert\boldsymbol{\mathit{u}}\rangle.
    4. The product of two linear operators is another linear operator: \hat{O}\hat{O'}=\hat{O}(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)=(\hat{O}\vert\boldsymbol{\mathit{u}}\rangle)\langle\boldsymbol{\mathit{v}}\vert=\vert\boldsymbol{\mathit{u'}}\rangle\langle\boldsymbol{\mathit{v}}\vert=\hat{O''}.

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