User:Justin545/Private Note

A Terrible, Mechanical Analog to Quantum Register

For a dynamic system ${\mathcal {K}}$ of $n$ variables (or parameters or degree of freedom) $v_{1}(t)$, $v_{2}(t)$, $\cdots$, $v_{n}(t)$, the state $s(t)$ of the system ${\mathcal {K}}$ at time $t$ can be described as a point in the phase space of dimension $n$. Where $s(t)=(v_{1}(t),v_{2}(t),\cdots ,v_{n}(t))$ is the coordinate of state $s(t)$ in the phase space. The phase space is spanned by the orthonormal basis

$\mathbb {B} \equiv \left\{\mathbf {e} _{i}\equiv {\begin{bmatrix}\delta _{i,1}&\delta _{i,2}&\cdots &\delta _{i,n}\end{bmatrix}}^{T}{\Big |}i\in \mathbb {D} \right\}$

of size $n$, where $\delta _{i,j}\equiv \delta _{ij}$ is the Kronecker delta, $\mathbb {Z}$ is the set of all integers, $\mathbb {D} \equiv [1,n]\cap \mathbb {Z}$. And we define sets $\mathbb {V} _{1}$, $\mathbb {V} _{2}$, ..., $\mathbb {V} _{n}$ as

$\forall i\in \mathbb {D} \left(\mathbb {V} _{i}\equiv \left\{v_{i}(t)|t\in \mathbb {T} \right\}\right)$

$\left[\forall i\!\in \!\mathbb {D} \,\,\forall t\!\in \!\mathbb {T} (v_{i}(t)\in \mathbb {V} _{i}\subseteq \mathbb {C} )\right]\land \left[\forall t\!\in \!\mathbb {T} \,\,\forall i\!\in \!\mathbb {D} (v_{i}(t)\in \mathbb {V} _{i}\subseteq \mathbb {C} )\right]$

where $\mathbb {T} \equiv [0,\infty )\cap \mathbb {R}$, $\mathbb {R}$ is the set of all real numbers, $\mathbb {C}$ is the set of all complex numbers.

The set of all state

$\mathbb {S} \equiv \left\{s(t)=(v_{1}(t),v_{2}(t),\cdots ,v_{n}(t)){\big |}t\in \mathbb {T} \right\}$

If we define

$\mathbb {W} \equiv \mathbb {V} _{1}\times \mathbb {V} _{2}\times \cdots \times \mathbb {V} _{n}=\left\{(x_{1},x_{2},\cdots ,x_{n}){\big |}x_{1}\in \mathbb {V} _{1}\land x_{2}\in \mathbb {V} _{2}\land \cdots \land x_{n}\in \mathbb {V} _{n}\right\}$

or briefly

$\mathbb {W} \equiv {\underset {i\in \mathbb {D} }{\times }}\mathbb {V} _{i}=\left\{(x_{1},x_{2},\cdots ,x_{n}){\big |}\forall i\in \mathbb {D} (x_{i}\in \mathbb {V} _{i})\right\}$,

where the sign $\times$ denotes the Cartesian product. We know that

$\mathbb {S} \subseteq \mathbb {W}$

In the most general case, it should satisfies

$\mathbb {S} =\mathbb {W}$ (1)

Now, we have an experiment or random trial whose goal is to measure the probability of the system ${\mathcal {K}}$ at a particular state $s(t_{r})$ at a random choosen time $t_{r}$. Which meas each state in $\mathbb {S}$ is an outcome or elementary event of the experiment. And the sample space $\Omega$ consists of all outcome of the experiment should be

$\Omega =\mathbb {S}$

In the most general case, (1) implies

$\Omega =\mathbb {W}$

Each outcome or elementary event in the sample space $\Omega$ is associated with a probability and there is a probability distribution associated to the sample space. And we can define a function which is similar to a joint probability density function

$f:\mathbb {V} _{1}\times \mathbb {V} _{2}\times \cdots \times \mathbb {V} _{n}\to [0,1]$

which maps each elementary event in $\Omega$ to the corresponding probability

$f:(x_{1},x_{2},\cdots ,x_{n})\mapsto \Pr \left(\left\{(x_{1},x_{2},\cdots ,x_{n})\right\}\right)$,

where $\left\{(x_{1},x_{2},\cdots ,x_{n})\right\}$ is an elementary event (a set) from $\Omega$. The function $f$ characterizing the probability distribution of the experiment for system ${\mathcal {K}}$ is similar to the the probability mass function or probability density function except its argument is a state of the system ${\mathcal {K}}$ rather then a value of a random variable. By the definition of probability, we know as well

$\sum _{e\in \mathbb {E} }\Pr(e)=1$,

where $\mathbb {E}$ is the set of all elementary event from $\Omega$. Since a function is an analogy to a vector, we can express $f$ as a vector which is a bra in bra-ket notation

$|\psi \rangle \simeq f$

As each elementary event $e$ has a corresponding probability value, we can treat each $e$ as a basis such that

$|\psi \rangle \equiv \sum _{\forall i\in \mathbb {D} (x_{i}\in \mathbb {V} _{i})}a_{(x_{1},x_{2},\cdots ,x_{n})}|x_{1}\rangle \otimes |x_{2}\rangle \otimes \cdots \otimes |x_{n}\rangle =\sum _{\forall i\in \mathbb {D} (x_{i}\in \mathbb {V} _{i})}a_{(x_{1},x_{2},\cdots ,x_{n})}\left(\bigotimes _{j=1}^{n}|x_{j}\rangle \right)$

where the sign $\otimes$ denotes the tensor product. Actually, function $f$ is pretty much like a square of wave function $\psi (x_{1},x_{2},\cdots ,x_{n})$ of $n$ quanta confined in a one-dimensional space where $x_{1}$, $x_{2}$, ..., $x_{n}$ are the positions of the quanta. Similarly, the square of the wave function $\psi (x_{1},x_{2},\cdots ,x_{n})$ maps combination of the variable values of the system to the probability of the occurrence of such combination. Therefore, a square of wave function is actually a joint probability density function. We may conclude that any joint probability density function has a corresponding tensor product vector space of infinite dimension.

When talking about quantum computer, we usually treat the set of qubit in a quantum register discrete rather than continuous. Similarly, joint probability density function is continuous and not suitable to be used with quantum register so we should find a discrete version of joint probability density function. We suppose the discrete version is called joint probability mass function. From the description above, we can understand that any joint probability mass function implies a quantum register. Which means if we can find any realistic system (not necessary to be a quantum system) can be modeled by a joint probability mass function, we will have found a quantum register.

For example, we can treat two spinning coins as a two-bit quantum register because when we try to stop them spinning, the probability of the outcome of such a system can be modeled by a joint probability mass function of two variables:

$f_{X_{1},X_{2}}(x_{1},x_{2})=a|tail\rangle |tail\rangle +b|tail\rangle |head\rangle +c|head\rangle |tail\rangle +d|head\rangle |head\rangle =|\Psi \rangle$

The next question is if we can find universal quantum gates for such a system. According to the book 'An Introduction to Quantum Computing':

Theorem 4.3.3: A set composed of any 2-qubit entangling gate, together with all 1-qubit gates, is universal.

According to the theorem, it seems not difficult to find universal quantum gates. But if the unitary properties of the universal quantum gates important? Or can we find universal quantum gates with unitary properties?

Example: Clock

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$F(V\times W)=\left\{\sum _{i=1}^{n}\alpha _{i}e_{(v_{i}\times w_{i})}\mid n\in \mathbb {N} ,\alpha _{i}\in K,(v_{i}\times w_{i})\in V\times W\right\},$

$f:(v_{1}(t),v_{2}(t),\cdots ,v_{n}(t))\mapsto \Pr \left(\left\{(v_{1}(t),v_{2}(t),\cdots ,v_{n}(t))\right\}\right)$,

$|\psi \rangle \equiv \sum _{t\in \mathbb {T} }a_{t}|v_{1}(t)\rangle \otimes |v_{2}(t)\rangle \otimes \cdots \otimes |v_{n}(t)\rangle$

composite system of v1, v2, ..., v_n

tensor product $\otimes$, size $|\mathbb {V} _{1}|\times |\mathbb {V} _{2}|\times \cdots \times |\mathbb {V} _{n}|=K^{n}$ if $|\mathbb {V} _{1}|=|\mathbb {V} _{2}|=\cdots =|\mathbb {V} _{n}|=K$

Cartesian product, size $1+1+\cdots +1=n$

probability, normalized $Pr(X)\in [0,1]$

$\{(x_{1},x_{2})|(x_{1}\in \mathbb {V} _{1})\land (x_{2}\in \mathbb {V} _{2})\}=\{x|x\in (\mathbb {V} _{1}\times \mathbb {V} _{2})\}$

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• probability distribution function or vector is actuall the state of quantum register
• sample space, event, outcome, elementary event, experiment/trial, measure, random variable
• clock
• isomorphic: vector and function? quantum register and this terrible analogy?
• multivariate random variable or random vector
• joint distribution
• correlated = dependent events ($\Pr(A|B)\neq \Pr(A)\land \Pr(A\cap B)=\Pr(A|B)\Pr(B)\neq \Pr(A)\Pr(B)$) $\approx$ entangled? independent events ($\Pr(A|B)=\Pr(A)\land \Pr(A\cap B)=\Pr(A|B)\Pr(B)=\Pr(A)\Pr(B)$) $\approx$ separable? any non-separable states = entangled state so making two events dependent or correlated imply entangled operation?
• dose the resulting probability related to manifold?

Wave Function Collapse of Separable State

Suppose we have two particles $A$ and $B$ their respective states are

$|\psi \rangle _{A}={\sqrt {0.2}}|0\rangle _{A}+{\sqrt {0.8}}|1\rangle _{A}$

$|\psi \rangle _{B}={\sqrt {0.3}}|0\rangle _{B}+{\sqrt {0.7}}|1\rangle _{B}$

The state of the composite system of $A$ and $B$ is

$|\psi \rangle _{A}\otimes |\psi \rangle _{B}=({\sqrt {0.2}}|0\rangle _{A}+{\sqrt {0.8}}|1\rangle _{A})\otimes ({\sqrt {0.3}}|0\rangle _{B}+{\sqrt {0.7}}|1\rangle _{B})$

$|\psi \rangle _{A}|\psi \rangle _{B}={\sqrt {0.06}}|0\rangle _{A}|0\rangle _{B}+{\sqrt {0.14}}|0\rangle _{A}|1\rangle _{B}+{\sqrt {0.24}}|1\rangle _{A}|0\rangle _{B}+{\sqrt {0.56}}|1\rangle _{A}|1\rangle _{B}$

If we try to measure the state of particle $A$ of $|\psi \rangle _{A}|\psi \rangle _{B}$ and get state $|1\rangle _{A}$, it means $|\psi \rangle _{A}|\psi \rangle _{B}$ collapses to either $|1\rangle _{A}|0\rangle _{B}$ or $|1\rangle _{A}|1\rangle _{B}$. Besides, the probability of finding particle $B$ in state $|1\rangle _{B}$ is

$P\left(|1\rangle _{B}{\Big |}|1\rangle _{A}\right)={\frac {P(|1\rangle _{A}\cap |1\rangle _{B})}{P(|1\rangle _{A})}}$ (according to $P(B\mid A)={\frac {P(A\cap B)}{P(A)}}$)

where

$P(|1\rangle _{A}\cap |1\rangle _{B})=|{\sqrt {0.56}}|^{2}=0.56$

$P(|1\rangle _{A})=|{\sqrt {0.24}}|^{2}+|{\sqrt {0.56}}|^{2}=0.8$

therefore,

$P\left(|1\rangle _{B}{\Big |}|1\rangle _{A}\right)={\frac {0.56}{0.8}}=0.7=|{\sqrt {0.14}}|^{2}+|{\sqrt {0.56}}|^{2}=P(|1\rangle _{B})$

Similarly, if we try to measure the state of particle $A$ of $|\psi \rangle _{A}|\psi \rangle _{B}$ and get state $|0\rangle _{A}$, it means $|\psi \rangle _{A}|\psi \rangle _{B}$ collapses to either $|0\rangle _{A}|0\rangle _{B}$ or $|0\rangle _{A}|1\rangle _{B}$. Besides, the probability of finding particle $B$ in state $|1\rangle _{B}$ is

$P\left(|1\rangle _{B}{\Big |}|0\rangle _{A}\right)={\frac {0.14}{0.06+0.14}}=0.7=P(|1\rangle _{B})$

The above illustration shows that we are not able to distinguish whether the state of particle $B$ has collapsed or not, because no matter the state of particle $A$ we measured is $|0\rangle _{A}$ or $|1\rangle _{A}$, the state of particle $B$ always collapses to $|1\rangle _{B}$ with probability $0.7$. Therefore, we can say the particle $B$ DOSE collapse when we measure particle $A$, but we just have no way to emphasize that.

Quantum: Difference between an operator and a measurement

Suppose there is a qubit whose state is

$|\psi \rangle ={\frac {1}{\sqrt {3}}}|0\rangle +{\frac {\sqrt {2}}{\sqrt {3}}}|1\rangle ={\begin{bmatrix}{\tfrac {1}{\sqrt {3}}}\\{\tfrac {\sqrt {2}}{\sqrt {3}}}\end{bmatrix}}$

After we measure the qubit, the state of the qubit will change from $|\psi \rangle$ to $|1\rangle$ with probability

${\big |}\langle 1|\psi \rangle {\big |}^{2}=\left({\frac {\sqrt {2}}{\sqrt {3}}}\right)^{2}={\frac {2}{3}}\approx 0.666\dots$.

This process is called wave function collapse. If $|1\rangle$ is observed after the measurement, the qubit becomes

$|\psi \rangle =|1\rangle$

Instead of measuring the qubit, a Hadamard gate

$H={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}$

operates on the qubit will be

$H|\psi \rangle ={\frac {1}{\sqrt {2}}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}{\begin{bmatrix}{\tfrac {1}{\sqrt {3}}}\\{\tfrac {\sqrt {2}}{\sqrt {3}}}\end{bmatrix}}={\frac {1}{\sqrt {2}}}{\begin{bmatrix}{\tfrac {1+{\sqrt {2}}}{\sqrt {3}}}\\{\tfrac {1-{\sqrt {2}}}{\sqrt {3}}}\end{bmatrix}}={\frac {1+{\sqrt {2}}}{\sqrt {6}}}|0\rangle +{\frac {1-{\sqrt {2}}}{\sqrt {6}}}|1\rangle$

as I know, the process of the operation is 'not' a wave function collapse.

My problem is why an operator acts on a qubit doesn't cause a wave function collapse? As I know, any subtle interaction with the qubit will cause the wave function to collapse. The Hadamard gate operator which is apparatus when acts on the qubit should also interact with the qubit. So how the process of an operation can circumvent the wave function collapse?

any subtle change of environment or any kind of interaction should cause wave function collapse including the operation of the operator, won't it? why a measurement will causes the wave function collapse but an operator won't? How the process of an operator circumvent the wave function collapse? An operator will interact with the qubit like a measurement, won't it? - Justin545 (talk) 01:49, 7 October 2008 (UTC)

Function Expansion, Basis Function of Dirac Delta

Dirac delta $\delta (t)$ can be viewed as the limit of

$\delta (t)=\lim _{a\to 0}\delta _{a}(t)$ (1)

where $\delta _{a}(t)$ is sometimes called a nascent delta function. There are many kind of definitions of $\delta _{a}(t)$. For example, it can be defined as

$\delta _{a}(t)={\begin{cases}{\frac {1}{a}},&-{\frac {a}{2}}<t<{\frac {a}{2}}\\0,&{\mbox{otherwise}}\end{cases}}$ (2)

Using Fourier transforms, one finds that

$\int _{-\infty }^{\infty }1\cdot e^{-i2\pi ft}\,dt=\delta (f)$ (3)

and therefore

$\int _{-\infty }^{\infty }e^{i2\pi f_{1}t}\left(e^{i2\pi f_{2}t}\right)^{*}\,dt=\int _{-\infty }^{\infty }e^{-i2\pi (f_{2}-f_{1})t}\,dt=\delta (f_{2}-f_{1})$ (4)

which is a statement of the orthogonality property for the Fourier kernel.

Similarly, we can show the orthogonality property for the Dirac delta. Consider the property of Dirac delta

$\int _{-\infty }^{\infty }f(t)\delta (t-T)\,dt=f(T)$ (5)

Replace $T$ by $x$ in (5)

$\int _{-\infty }^{\infty }f(t)\delta (t-x)\,dt=f(x)$ (6)

Let

$f(t)\equiv \delta (t-y)$ (7)

Replace $t$ by $x$ in (7)

$f(x)=\delta (x-y)$ (8)

Replace (7) and (8) into (6)

$\int _{-\infty }^{\infty }\delta (t-y)\delta (t-x)\,dt=\delta (x-y)$ (9)

$\int _{-\infty }^{\infty }\delta (t-x)\delta (t-y)\,dt=\delta (x-y)$ (10)

Thus, Dirac delta are orthogonal eigenfunctions. According to Sturm-Liouville theory, a given function $f(t)$, satisfying suitable conditions, can be expanded in an infinite series of eigenfunctions $\phi _{n}(t)$ of the more general Sturm–Liouville problem of

$\left[p(x)y'\right]'-q(x)y+\lambda r(x)y=0$ (11)

$a_{1}y(0)+a_{2}y'(0)=0$ (12)

$b_{1}y(1)+b_{2}y'(1)=0$ (13)

such that

$f(t)=\sum _{n=1}^{\infty }c_{n}\phi _{n}(t)$ (14)

Each element of the set of eigenfunctions $\{\phi _{n}(t)\}_{n=1}^{\infty }$ is a solution satisfying the more general Sturm–Liouville problem (11), (12) and (13).

could be expressed as a series of eigenfunctions $\phi _{n}(t)$ such that

as a series of eigenfunctions of Dirac delta?

p.s. I don't know how to classify this question, so I put it here rather than Wikipedia:Reference_desk/Science because I think more math is involved than quantum mechanics. - Justin545 (talk) 05:24, 24 March 2008 (UTC)

Data Compression by Specifying The Space and Time

cosmology, big bang, initial state, absolute time, relativity - Justin545 (talk) 03:17, 26 March 2008 (UTC)

Quantum Viewpoint: Optic Filter & Measurement

optic filter - Justin545 (talk) 08:54, 19 April 2008 (UTC)

(temporary)

From Inelastic_collision#Perfectly_inelastic_collision we have

$\mathbf {v} _{f}={\frac {m_{1}\mathbf {v} _{1,i}+m_{2}\mathbf {v} _{2,i}}{m_{1}+m_{2}}}$

If $m_{1}=m_{2}=m$ and let $\mathbf {v} _{1}=\mathbf {v} _{1,i}$, $\mathbf {v} _{2}=\mathbf {v} _{2,i}$ we have

$\mathbf {v} _{f}={\frac {m\mathbf {v} _{1}+m\mathbf {v} _{2}}{m+m}}={\frac {m(\mathbf {v} _{1}+\mathbf {v} _{2})}{2m}}={\frac {\mathbf {v} _{1}+\mathbf {v} _{2}}{2}}$

---

$|\psi \rangle _{A}=|0\rangle _{A}+0|1\rangle _{A}$

$|\psi \rangle _{B}={\sqrt {0.3}}|0\rangle _{B}+{\sqrt {0.7}}|1\rangle _{B}$

$|\psi \rangle _{A}|\psi \rangle _{B}={\sqrt {0.3}}|0\rangle _{A}|0\rangle _{B}+{\sqrt {0.7}}|0\rangle _{A}|1\rangle _{B}+0|1\rangle _{A}|0\rangle _{B}+0|1\rangle _{A}|1\rangle _{B}$

$P\left(|1\rangle _{B}{\Big |}|0\rangle _{A}\right)={\frac {P(|0\rangle _{A}\cap |1\rangle _{B})}{P(|0\rangle _{A})}}={\frac {0.7}{1}}$

$P\left(|1\rangle _{B}{\Big |}|1\rangle _{A}\right)={\frac {P(|1\rangle _{A}\cap |1\rangle _{B})}{P(|1\rangle _{A})}}={\frac {0}{0}}$

$P\left(|0\rangle _{A}{\Big |}|0\rangle _{B}\right)={\frac {P(|0\rangle _{A}\cap |0\rangle _{B})}{P(|0\rangle _{B})}}={\frac {0.3}{0.3}}$

$P\left(|0\rangle _{A}{\Big |}|1\rangle _{B}\right)={\frac {P(|0\rangle _{A}\cap |1\rangle _{B})}{P(|1\rangle _{B})}}={\frac {0.7}{0.7}}$

---

Proof

$-{\frac {d}{dx}}\left[p(x){\frac {dy}{dx}}\right]+q(x)y=\lambda w(x)y.$ (1)

$y(a)\cos \alpha -p(a)y^{\prime }(a)\sin \alpha =0,$ (2)

$y(b)\cos \beta -p(b)y^{\prime }(b)\sin \beta =0,$ (3)

$Lu=-{d \over dx}\left[p(x){du \over dx}\right]+q(x)u$ (xx4)

Rearrange terms of (2) and (3), we have

$y'(a)={\frac {y(a)\cos \alpha }{p(a)\sin \alpha }}$ (?)

$y'(b)={\frac {y(b)\cos \beta }{p(b)\sin \beta }}$ (6)

If the boundary conditions is

$a_{1}y(0)+a_{2}y'(0)=0$

$b_{1}y(1)+b_{2}y'(1)=0$

Rearrange terms, we get

$y'(0)=-{\frac {a_{1}}{a_{2}}}y(0)$ (1)

$y'(1)=-{\frac {b_{1}}{b_{2}}}y(1)$ (1)

---

$y_{1}'(b)={\frac {y_{1}(b)\cos \beta }{p(b)\sin \beta }}$

$y_{2}'(b)={\frac {y_{2}(b)\cos \beta }{p(b)\sin \beta }}$

$y_{1}'(a)={\frac {y_{1}(a)\cos \alpha }{p(a)\sin \alpha }}$

$y_{2}'(a)={\frac {y_{2}(a)\cos \alpha }{p(a)\sin \alpha }}$

$Ly_{1}=\lambda _{1}w(x)y_{1}$

$Ly_{2}=\lambda _{2}w(x)y_{2}$

$\int _{c}^{d}(Lu)v-u(Lv)\,dx=-p(u'v-uv'){\bigg |}_{c}^{d}$ (6)

$\int _{a}^{b}(Ly_{1})y_{2}-y_{1}(Ly_{2})\,dx=-p(y_{1}'y_{2}-y_{1}y_{2}'){\bigg |}_{a}^{b}$

$\int _{a}^{b}[Ly_{1}(x)]y_{2}(x)-y_{1}(x)[Ly_{2}(x)]\,dx=-p(x)[y_{1}'(x)y_{2}(x)-y_{1}(x)y_{2}'(x)]{\bigg |}_{a}^{b}$

$\int _{a}^{b}[Ly_{1}(x)]y_{2}(x)-y_{1}(x)[Ly_{2}(x)]\,dx=-p(b)\left[{\frac {y_{1}(b)\cos \beta }{p(b)\sin \beta }}y_{2}(b)-y_{1}(b){\frac {y_{2}(b)\cos \beta }{p(b)\sin \beta }}\right]+p(a)\left[{\frac {y_{1}(a)\cos \alpha }{p(a)\sin \alpha }}y_{2}(a)-y_{1}(a){\frac {y_{2}(a)\cos \alpha }{p(a)\sin \alpha }}\right]$

$\int _{a}^{b}[Ly_{1}(x)]y_{2}(x)-y_{1}(x)[Ly_{2}(x)]\,dx=-p(b)\left[{\frac {\cos \beta }{p(b)\sin \beta }}y_{1}(b)y_{2}(b)-y_{1}(b)y_{2}(b){\frac {\cos \beta }{p(b)\sin \beta }}\right]+p(a)\left[{\frac {\cos \alpha }{p(a)\sin \alpha }}y_{1}(a)y_{2}(a)-y_{1}(a)y_{2}(a){\frac {\cos \alpha }{p(a)\sin \alpha }}\right]$

---

In order to render the theory as simple as possible while retaining considerable generality, we assume w(x) is a real-valued function and w(x) > 0 for all x on the interval [a,b]. In terms of linear boundary value problems, Lagrange's identity is

$\int _{0}^{1}(Lu)v-u(Lv)\,dx=-p(u'v-uv'){\bigg |}_{0}^{1}$

Retrace the steps in the proof of Lagrange's identity, we can also proof that

$\int _{c}^{d}(Lu)v-u(Lv)\,dx=-p(u'v-uv'){\bigg |}_{c}^{d}$ (6)

If v=u, the identity (6) becomes

$\int _{c}^{d}(Lu)u-u(Lu)\,dx=-p(u'u-uu'){\bigg |}_{c}^{d}$

$\int _{c}^{d}(Lu)u-u(Lu)\,dx=-p\cdot 0{\bigg |}_{c}^{d}=0$

$\int _{c}^{d}(Lu)u\,dx=\int _{c}^{d}u(Lu)\,dx$ (7)

Actually, it can be shown that the identity (7) becomes

$\int _{c}^{d}(Lu){\overline {u}}\,dx=\int _{c}^{d}u{\overline {(Lu)}}\,dx$ (8)

when u is a complex-valued function of x. Whereas the overlines denote the complex conjugate. Suppose λ_n is the n-th eigenvalue of the problem (1)-(2)-(3) and y_n is the corresponding eigenfunction. Because λ_n and y_n are possibly complex-valued, we presume that they have the forms λ_n = A + iB and y_n = C(x) + iD(x), where A, B, C(x) and D(x) are real. Replace c=a, d=b and u=y_n into (8), we have

$\int _{a}^{b}(Ly_{n}){\overline {y_{n}}}\,dx=\int _{a}^{b}y_{n}{\overline {(Ly_{n})}}\,dx$ (9)

Replace u=y_n into (xx4), we have

$Ly_{n}=-{d \over dx}\left[p(x){dy_{n} \over dx}\right]+q(x)y_{n}$ (10)

Since y_n is an eigenfunction, it also satisfies (1), that is

$-{\frac {d}{dx}}\left[p(x){\frac {dy_{n}}{dx}}\right]+q(x)y_{n}=\lambda _{n}w(x)y_{n}$ (11)

Replace (11) into (10), we have

$Ly_{n}=\lambda _{n}w(x)y_{n}$ (12)

All of eigenvalues are real

Replace (12) into (9), we have

$\int _{a}^{b}[\lambda _{n}w(x)y_{n}]{\overline {y_{n}}}\,dx=\int _{a}^{b}y_{n}{\overline {[\lambda _{n}w(x)y_{n}]}}\,dx$

$\int _{a}^{b}\lambda _{n}w(x)y_{n}{\overline {y_{n}}}\,dx=\int _{a}^{b}y_{n}{\overline {\lambda _{n}}}\,{\overline {w(x)}}{\overline {y_{n}}}\,dx$ (13)

Since w(x) is real, (13) becomes

$\int _{a}^{b}\lambda _{n}w(x)y_{n}{\overline {y_{n}}}\,dx=\int _{a}^{b}y_{n}{\overline {\lambda _{n}}}w(x){\overline {y_{n}}}\,dx$ (14)

Rearrange terms of (14), we have

$(\lambda _{n}-{\overline {\lambda _{n}}})\int _{a}^{b}w(x)y_{n}{\overline {y_{n}}}\,dx=0$

$(\lambda _{n}-{\overline {\lambda _{n}}})\int _{a}^{b}w(x)[C^{2}(x)+D^{2}(x)]\,dx=0$ (15)

Because all eigenfunctions are not trivial solution which means y_n ≠ 0. Also w(x) > 0, we conclude the integration part of (15) is not zero and

$\lambda _{n}-{\overline {\lambda _{n}}}=0$

$(A+iB)-(A-iB)=2iB=0$

which means B = 0. So the eigenvalue λ_n is real. Q.E.D.

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south park

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   /'._  --  _.'\  /'._  --  _.'\   /'. `'--'` .'\/   '._-.__--__.-_.'
 \/_   `""""`   _\/_   `""""`   _\ /_  `-./\.-'  _\'.    `""""""""`'`\
 (__/    '|    \ _)_|           |_)_/            \__)|        '        
   |_____'|_____|   \__________/|;                  `_________'________`;-'
   s'----------'    '----------'   '--------------'`--------------------`
      S at N          K Y L E        K E N N Y         C A R T M A N
omg it's south park
omg no f-ing kidding.

I MEI, Gummy, strawberry milk choco ball

8 / 20 = 0.4 serving/package

0.4 serving/package * 118 kcal/serving = 47.2 kcal/package