## Wednesday, January 6, 2016

### Some Introductory Quantum Mechanics: Theorems of the Formalism

Quantum mechanics (QM) has a number of curious and interesting associated phenomena. Some of these were hinted at in the first part of this series. The effects can be inferred from the mathematical formalism discussed in the previous post in this series. Here we will discuss several of these, again without reference to interpretation.

This is part of a multi-part series giving a general introduction to quantum theory. This is part 3.

## Heisenberg's Uncertainty Principle

The variance of any observable is defined as $\sigma_A^2=\left \langle A^2 \right \rangle-\left \langle A \right \rangle^2=\left \langle \left ( A-\left \langle A \right \rangle \right )^2 \right \rangle$ Where $\left \langle Q \right \rangle=\left. \langle \psi \right. | Q\left. |\psi \right \rangle$ is the expected value of the operator Q. Roughly speaking, $\sigma_A$ is the "width" of distribution of the potential values for A. We then define a new state vector as $\left. | a \right \rangle=\left (A-\left \langle A \right \rangle \right ) \left. |\psi \right \rangle$ So that $\sigma_A^2=\left \langle a \right. |\left. a \right \rangle$ We similarly define $\sigma_B^2=\left \langle B^2 \right \rangle-\left \langle B \right \rangle^2=\left \langle \left ( B-\left \langle B \right \rangle \right )^2 \right \rangle=\left \langle b \right. |\left. b \right \rangle$ Where $\left. | b \right \rangle=\left (B-\left \langle B \right \rangle \right ) \left. |\psi \right \rangle$ Then, by the Cauchy-Schwartz inequality: $\sigma_A^2\sigma_B^2=\left \langle a \right. |\left. a \right \rangle\left \langle b \right. |\left. b \right \rangle \geq \left | \left \langle a \right. |\left. b \right \rangle \right |^2$ Let $z= \left \langle a \right. |\left. b \right \rangle$. Then $\left | z \right |^2=[\mathrm{Re}(z)]^2+[\mathrm{Im}(z)]^2\geq[\mathrm{Im}(z)]^2=\left [\frac{z-\bar{z}}{2i} \right ]^2=\left [\frac{\left \langle a \right. |\left. b \right \rangle-\left \langle b \right. |\left. a \right \rangle}{2i} \right ]^2$ However, $\left \langle a \right. |\left. b \right \rangle=\left \langle \left ( A-\left \langle A \right \rangle \right ) \left ( B-\left \langle B \right \rangle \right ) \right \rangle=\left \langle AB \right \rangle-\left \langle A \right \rangle\left \langle B \right \rangle$ $\left \langle b \right. |\left. a \right \rangle=\left \langle \left ( B-\left \langle B \right \rangle \right ) \left ( A-\left \langle A \right \rangle \right ) \right \rangle=\left \langle BA \right \rangle-\left \langle B \right \rangle\left \langle A \right \rangle$ So $\left | z \right |^2\geq\left [\frac{\left \langle AB \right \rangle-\left \langle BA \right \rangle}{2i} \right ]^2=\left [\frac{\left \langle [A,B] \right \rangle}{2i} \right ]^2$ Where $[A,B]=AB-BA$ is the commutator of the two operators A and B (In general, two operators need not commute, and so the commutator will not vanish). Thus, we can state the general uncertainty principle for any two operators: $\sigma_A \sigma_B \geq\tfrac{1}{2}| \left \langle \left [A,B \right ] \right \rangle |$ For example, let us take the one-dimensional position and momentum operators: $A=x,\;B=\frac{\hbar}{i}\frac{\partial }{\partial x}$ $[x,p_x]\left. | \psi \right \rangle=xp_x\left. | \psi \right \rangle-p_xx\left. | \psi \right \rangle =\frac{\hbar}{i}\left (x\frac{\partial}{\partial x}\left. | \psi \right \rangle-\frac{\partial }{\partial x}x\left. | \psi \right \rangle \right )=i \hbar \left. | \psi \right \rangle$ Thus $\sigma_x \sigma_{p_x} \geq\frac{\hbar}{2}$ This is the famous position-momentum uncertainty relation.

## No Cloning and Related Theorems

Suppose we want to find an operator that takes a quantum state and produces a copy of it. That is, we feed in a state and a "blank" state, operate on the two of them, and the result is the original state and a copy of it. Let this operator be called C, and the blank state be called b. That is: $C \left. | \psi \right \rangle_A\left. | b \right \rangle_B= \left. | \psi \right \rangle_A\left. | \psi \right \rangle_B$ As C is a transformation/evolution operator, it must be unitary, so it preserves inner products, and $C^\dagger C=I$. Therefore $C \left. | \phi \right \rangle_A\left. | b \right \rangle_B=\left. | \phi \right \rangle_A\left. | \phi \right \rangle_B$ $\left \langle b \right.|_B \left \langle \phi \right.|_A C^\dagger=\left \langle \phi \right.|_B \left \langle \phi \right.|_A$ $\left \langle b \right.|_B \left \langle \phi \right.|_A \left. | \psi \right \rangle_A\left. | b \right \rangle_B = \left \langle b \right.|_B \left \langle \phi \right.|_A C^\dagger C \left. | \psi \right \rangle_A\left. | b \right \rangle_B =\left \langle \phi \right.|_B \left \langle \phi \right.|_A \left. | \psi \right \rangle_A\left. | \psi \right \rangle_B$ However, $\left \langle b|b \right \rangle=1$, so $\left \langle \phi|\psi \right \rangle=\left \langle \phi|\psi \right \rangle^2$ Thus $\left \langle \phi|\psi \right \rangle \in \left \{ 0,1 \right \}$, that is, the two wavefunctions are orthogonal or identical. But the two states can be chosen arbitrarily, and need not be identical or orthogonal (indeed we can always construct a wavefunction as a linear combination of an orthogonal state and an identical state, and so achieve any inner product).

Moreover,as C must be linear, if $\left. | \chi \right \rangle=\alpha \left. | \phi \right \rangle+\beta \left. | \psi \right \rangle$, then $C\left. | \chi \right \rangle_A \left. | b \right \rangle_B= C \left ( \alpha \left. | \phi \right \rangle_A+\beta \left. | \psi \right \rangle_B \right ) \left. | b \right \rangle_B =\alpha C \left. | \phi \right \rangle_A \left. | b \right \rangle_B+\beta C \left. | \psi \right \rangle_A \left. | b \right \rangle_B$ $C\left. | \chi \right \rangle_A \left. | b \right \rangle_B = \alpha \left. | \phi \right \rangle_A \left. | \phi \right \rangle_B + \beta \left. | \psi \right \rangle_A \left. | \psi \right \rangle_B$ However, $C\left. | \chi \right \rangle_A \left. | b \right \rangle_B=\left. | \chi \right \rangle_A \left. | \chi \right \rangle_B=\alpha^2 \left. | \phi \right \rangle_A \left. | \phi \right \rangle_B+\alpha\beta \left ( \left. | \phi \right \rangle_A \left. | \psi \right \rangle_B + \left. | \psi \right \rangle_A \left. | \phi \right \rangle_B \right )+\beta^2 \left. | \psi \right \rangle_A \left. | \psi \right \rangle_B$ And these two expressions clearly need not be equivalent. We are free to choose $\alpha, \beta, \phi$, and $\psi$ arbitrarily, and, in general, the two expressions will be unequal. Thus there cannot be a way to copy arbitrary quantum states.

Since there is no way to clone a quantum state, there is thus no way to go in the opposite direction, namely start with two identical states and transform that into a "blank" state and an original. The argument runs in much the same way, and can be seen as a dual of the no cloning theorem, called the no-deleting theorem.

Suppose it were possible to measure and communicate the state of an arbitrary quantum state as a sequence of classical bits. Since classical bits can be easily copied, it would then be possible to copy quantum states, in violation of the no cloning theorem. Thus it is not possible to measure and communicate the state of an arbitrary quantum state as a sequence of classical bits, and this is called the no teleportation theorem.

An extension of the no cloning theorem to mixed statesis the no broadcast theorem, which states that one can't convey a general quantum state to two or more recipients.

## Correspondence Principle and the Ehrenfest Theorem

A rather clear demand on quantum mechanics is that its predictions tend to those of standard classical mechanics in the appropriate limits. Given that we do not observe macroscopic objects to display unusual, characteristically quantum phenomena, quantum mechanics must make the same predictions as classical mechanics, asymptotically. The probabilities for macroscopic objects to display such phenomena must be vanishingly small. In general, the observed classical parameters will correspond to the expected values of the quantum analogues.

As an example, let us find the rate of change of the expected value of a generic observable $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle A \right \rangle=\frac{\mathrm{d} }{\mathrm{d} t}\left \langle \psi|A|\psi \right \rangle =\left \langle \frac{\partial }{\partial t}\psi|A|\psi \right \rangle + \left \langle \psi|\frac{\partial }{\partial t}A|\psi \right \rangle + \left \langle \psi|A|\frac{\partial }{\partial t}\psi \right \rangle$ However, since the wavefunction satisfies the Schrodinger equation, we have $i \hbar \frac{\partial }{\partial t}\left. | \psi \right \rangle= H\left. | \psi \right \rangle$ And, moreover $-i \hbar \frac{\partial }{\partial t}\left \langle \psi | \right.= \left \langle \psi | \right. H$ Thus $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle A \right \rangle =-\frac{1}{i\hbar}\left \langle\psi|HA|\psi \right \rangle + \left \langle \psi|\frac{\partial }{\partial t}A|\psi \right \rangle + \frac{1}{i\hbar}\left \langle \psi|AH|\psi \right \rangle$ $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle A \right \rangle = \frac{1}{i\hbar}\left \langle[A,H] \right \rangle+\left \langle \frac{\partial }{\partial t}A \right \rangle$ Let $x=A$ $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle x \right \rangle = \frac{1}{i\hbar}\left \langle[x,H] \right \rangle+\left \langle \frac{\partial }{\partial t}x \right \rangle =\frac{1}{i\hbar}\left \langle[x,H] \right \rangle =\frac{1}{i\hbar}\left \langle[x,\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}+V] \right \rangle$ $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle x \right \rangle =\frac{1}{i\hbar}\left \langle[x,\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}+V] \right \rangle =\frac{\hbar}{im}\left \langle \frac{\partial }{\partial x} \right \rangle=\frac{\left \langle p \right \rangle}{m}$ Let $p=A$ $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle p \right \rangle = \frac{1}{i\hbar}\left \langle[p,H] \right \rangle+\left \langle \frac{\partial }{\partial t}p \right \rangle$ $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle p \right \rangle = \frac{1}{i\hbar}\left \langle[p,H] \right \rangle+\left \langle \frac{\partial }{\partial t}p \right \rangle =\frac{1}{i\hbar}\left \langle[p,\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}+V] \right \rangle = -\left \langle \frac{\partial V}{\partial x} \right \rangle$ These are the same as the classical dynamical equations for the position and momentum. Thus, as it is often the case that the wavefunctions are highly localized, at least compared to macroscopic scales, quantum mechanics predicts the same macroscopic behavior as classical mechanics.

Another fact derivable from the Ehrenfest theorem is the following. Suppose Q is an operator that does not depend explicitly on time. Then we have $\frac{\mathrm{d} }{\mathrm{d} t}\left \langle Q \right \rangle = \frac{1}{i\hbar}\left \langle[Q,H] \right \rangle$ From our discussion in the section of the Heisenberg uncertainty principle $\sigma_H\sigma_Q\geq\tfrac{1}{2}|\left \langle \left [ H,Q \right ] \right \rangle|=\frac{\hbar}{2}\left | \frac{\mathrm{d} }{\mathrm{d} t} \left \langle Q \right \rangle \right |$ Though time is not an observable, let us nevertheless define $\sigma_t=\frac{\sigma_Q}{|\mathrm{d}\left \langle Q \right \rangle/\mathrm{dt}|}$ We then have $\sigma_H\sigma_t\geq\frac{\hbar}{2}$ A result analogous to that of position and momentum.

## Bell's Theorem and the Kochen-Specker Theorem

Certain interpretations of quantum mechanics hold that the measurements and observations of the quantum systems are deterministic, and the only reason they seem indeterministic is because we lack full knowledge of the system. They hold that there are hidden variables in the system that we have not or maybe even can not uncover that govern the system, and it is only our ignorance of these that makes us unable to predict with certainty what we will observe. Models like these are called realistic, in the sense that, prior to the measurement, there is a definite, singular reality of what we will observe (or in some cases, counterfactually would observe).

Another principle typically regarded as fundamental is that the system is local, that is, causal effects cannot propagate faster than the speed of light. In principle, if the system could be appropriately manipulated, it would be possible to use non-local systems to send messages into the past.

Often, these interpretations are hard or impossible to test. However, certain versions can be tested, as they make predictions that would be inconsistent with those of standard quantum mechanics. Bell's inequality is one way to rule out certain types of local realistic models.

Let us take a source that produces a sequence of identical electron pairs in the entangled state $\tfrac{1}{\sqrt{2}}\left (\left. | \uparrow\downarrow \right \rangle+\left. | \downarrow\uparrow \right \rangle \right )$. That is, the two particles are perfectly anti-correlated in the z direction.

We then send the particles in opposite directions to two detectors, A and B. These detectors measure the spin along axes at angles $\alpha$ and $\beta$ with respect to the z-axis respectively. Let us define $p(\alpha,\beta)$ as +1 if the measured spins are the same (both up or both down) and -1 if they are different (one up, one down). $P(\alpha,\beta)$ we then define as the average of p over many trials. Standard quantum theory predicts that $P(\alpha,\beta)=-\cos(\alpha-\beta)$.

Suppose there are hidden variables that determine what will be measured. For simplicity, we consolidate them all, for the whole system, in the single variable $\textbf{v}$. Let $A(\alpha,\textbf{v})=1$ if the particle sent to A, which is set at angle $\alpha$, with variables $\textbf{v}$, will be found to have spin up, and similarly with $A(\alpha,\textbf{v})=-1$ for spin down. Likewise with $B(\beta,\textbf{v})=1$ and $B(\beta,\textbf{v})=-1$ for detector B. Clearly $p(\alpha,\beta,\textbf{v})=A(\alpha,\textbf{v})B(\beta,\textbf{v})$. Since the particles are perfectly anti-correlated when the detectors are aligned, we have $A(\alpha,\textbf{v})=-B(\alpha,\textbf{v})$.

To average over many trials, we merely average over the different hidden variables, which are assumed to follow some sort of distribution, $\rho(\textbf{v})$. Thus, we then have $P(\alpha,\beta)=\int_{\mathrm{all}\; \textbf{v}} \rho(\textbf{v})A(\alpha,\textbf{v})B(\beta,\textbf{v})d\textbf{v} =-\int_{\mathrm{all}\; \textbf{v}} \rho(\textbf{v})A(\alpha,\textbf{v})A(\beta,\textbf{v})d\textbf{v}$ $P(\alpha,\beta)-P(\alpha,\gamma)= -\int_{\mathrm{all}\; \textbf{v}} \rho(\textbf{v}) \left [A(\alpha,\textbf{v})A(\beta,\textbf{v})-A(\alpha,\textbf{v})A(\gamma,\textbf{v}) \right] d\textbf{v}$ As $A^2(\alpha,\textbf{v})$ for any input variables, we can write: $P(\alpha,\beta)-P(\alpha,\gamma)= -\int_{\mathrm{all}\; \textbf{v}} \rho(\textbf{v}) \left [1-A(\beta,\textbf{v})A(\gamma,\textbf{v}) \right]A(\alpha,\textbf{v})A(\beta,\textbf{v}) d\textbf{v}$ Given that $\left |\int_{R} f(\textbf{x})d\textbf{x} \right |\leq\int_{R} |f(\textbf{x})|d\textbf{x}, \;\;\; | A(\alpha,\textbf{v})|=1, \;\;\; \rho(\textbf{v}) \left [1-A(\beta,\textbf{v})A(\gamma,\textbf{v}) \right]\geq 0$ We then have $|P(\alpha,\beta)-P(\alpha,\gamma)|\leq \int_{\mathrm{all}\; \textbf{v}} \rho(\textbf{v}) \left [1-A(\beta,\textbf{v})A(\gamma,\textbf{v}) \right] d\textbf{v}$ $|P(\alpha,\beta)-P(\alpha,\gamma)|\leq1+P(\beta,\gamma)$ Which is Bell's Inequality. This equality should be satisfied for all local hidden variable interpretations (the reason locality is required is to preclude instantaneous or faster-than-light signals being sent from one detector to the other). However, it is incompatible with standard quantum mechanics. For instance, let $\alpha=0$, $\beta=\pi/2$ and $\gamma=\pi/4$. Then $|P(\alpha,\beta)-P(\alpha,\gamma) |=\tfrac{1}{\sqrt{2}}\nleq 1+P(\beta,\gamma) =1-\tfrac{1}{\sqrt{2}}$ As there have been experiments performed that violate Bell's inequality, this provides strong evidence against local hidden variables interpretations. However, there are some loopholes: for instance, superdeterminism, a sort of conspiracy theory that not only are the systems we study deterministically governed, but so are our experiments, including us, and are so as to make us observe statistical violations of Bell's inequality regardless.

An associated result called the Kochen-Specker (KS) Theorem shows that non-contextual hidden variable interpretations are incompatible with quantum mechanics. That is, interpretations in which the observables measured have a single definite value independent of how they are measured are incompatible with quantum mechanics. However, it leaves open the possibility for contextual hidden-variables interpretations, in which the manner of measurement is relevant to the obtained result.

One might think that one could use entanglements to send messages faster than light, given that the effects are instantaneous (The moment Alice's electron is observed to have spin up on the z-axis, Bob's electron will have spin down on the z-axis). However, a theorem called the no communication theorem shows that it is not possible for one observer, by measuring some subset of a system, to communicate information to another observer. While the effects may be instantaneous, they do not carry information, and it is only after the two observers meet up and compare results that they note that they have correlations that defy local realism.

## The Quantum Zeno Effect

Suppose we have a particle that can be in one of two states (spinning up or down, decayed or not decayed). We can represent it as a 2 by 1 matrix. Suppose it begins in the state $\left. |\psi(0) \right \rangle=\begin{bmatrix} 1\\ 0 \end{bmatrix}$ If it is allowed to evolve by itself, its time dependent state is given by $\left. |\psi(t) \right \rangle=\begin{bmatrix} \alpha(t)\\ \beta(t) \end{bmatrix}$ Where the functions satisfy the condition stated above at t=0, and the state is properly normalized. Suppose that the other state is stable, i.e., that once it "flips" it stays "flipped".

Suppose $|\beta(t)|^2\approx (t/\tau)^n$ for t close to 0, where $\tau$ is some characteristic time of the system. Suppose we allow the state to evolve unperturbed for a length of time T (small relative to $\tau$), and then measure it. The probability that it will be found in the original state is simply $P_1=|\alpha(T)|^2\approx 1- (T/\tau)^n$ However, suppose, instead, that we measure it N times, after each time of length T/N. Then the chance that it will be found in the original state is the chance that it hadn't been found to have changed after any interval. That can be found by the usual methods of probability theory: $P_N=(|\alpha(T/N)|^2)^N\approx (1- \left (\tfrac{T}{N\tau} \right)^n)^N\approx e^{- \left(\tfrac{T}{\tau}\right)^n N^{1-n}}$ Thus, if $n>1$, the probability tends to 1 as N increases, and if $n<1$ the probability tends to 0 as N increases (if $n=1$, the probability tends to an exponential function of time). Thus, if the probability changes in the appropriate way, watching a system repeatedly tends to keep it in the same state. Moreover, if the system is measured continuously, it would never change at all. This has lead some to remark that a quantum watched pot never boils. This phenomenon is called the quantum Zeno effect, after the philosophical paradoxes of a similar nature.

## Quantum Teleportation and Indirect Entanglement

Suppose Alice and Bob are in separate locations, but connected by classical communication channels. They also each have one of a pair of entangled particles in the state $\tfrac{1}{\sqrt{2}}\left. |\uparrow \right \rangle_A\left. |\uparrow \right \rangle_B+\tfrac{1}{\sqrt{2}}\left. |\downarrow \right \rangle_A\left. |\downarrow \right \rangle_B$ Where the subscripts denote whose particle it is. Alice also has another particle in the arbitrary state $\alpha\left. |\uparrow \right \rangle_C+\beta\left. |\downarrow \right \rangle_C$ The state of the entire system can be written as $\tfrac{\alpha}{\sqrt{2}}\left. |\uparrow \uparrow \uparrow \right \rangle_{ABC}+ \tfrac{\alpha}{\sqrt{2}}\left. |\downarrow \downarrow \uparrow \right \rangle_{ABC}+ \tfrac{\beta}{\sqrt{2}}\left. |\uparrow \uparrow \downarrow \right \rangle_{ABC}+ \tfrac{\beta}{\sqrt{2}}\left. |\downarrow \downarrow \downarrow \right \rangle_{ABC}$ This can also be written in the form $\frac{1}{2} \begin{pmatrix} \tfrac{\left. |\uparrow \uparrow \right \rangle_{AC}+\left. |\downarrow \downarrow \right \rangle_{AC}}{\sqrt{2}}\left (\alpha\left. |\uparrow \right \rangle_{B}+\beta\left. |\downarrow \right \rangle_{B} \right ) + \tfrac{\left. |\uparrow \uparrow \right \rangle_{AC}-\left. |\downarrow \downarrow \right \rangle_{AC}}{\sqrt{2}}\left (\alpha\left. |\uparrow \right \rangle_{B}-\beta\left. |\downarrow \right \rangle_{B} \right ) \\ + \tfrac{\left. |\uparrow \downarrow \right \rangle_{AC}+\left. |\downarrow \uparrow \right \rangle_{AC}}{\sqrt{2}} \left (\beta\left. |\uparrow \right \rangle_{B}+\alpha\left. |\downarrow \right \rangle_{B} \right ) +\tfrac{\left. |\uparrow \downarrow \right \rangle_{AC}-\left. |\downarrow \uparrow \right \rangle_{AC}}{\sqrt{2}}\left (\beta\left. |\uparrow \right \rangle_{B}-\alpha\left. |\downarrow \right \rangle_{B} \right ) \end{pmatrix}$ Thus, if Alice measures her pair of particles to be in any of the four entangled states (all of which are mutually orthogonal, and so are completely distinguishable) $\tfrac{\left. |\uparrow \uparrow \right \rangle_{AC}+\left. |\downarrow \downarrow \right \rangle_{AC}}{\sqrt{2}},\; \tfrac{\left. |\uparrow \uparrow \right \rangle_{AC}-\left. |\downarrow \downarrow \right \rangle_{AC}}{\sqrt{2}},\; \tfrac{\left. |\uparrow \downarrow \right \rangle_{AC}+\left. |\downarrow \uparrow \right \rangle_{AC}}{\sqrt{2}},\; \tfrac{\left. |\uparrow \downarrow \right \rangle_{AC}-\left. |\downarrow \uparrow \right \rangle_{AC}} {\sqrt{2}}$ Bob's state will become, respectively $\left (\alpha\left. |\uparrow \right \rangle_{B}+\beta\left. |\downarrow \right \rangle_{B} \right ),\; \left (\alpha\left. |\uparrow \right \rangle_{B}-\beta\left. |\downarrow \right \rangle_{B} \right ),\; \left (\beta\left. |\uparrow \right \rangle_{B}+\alpha\left. |\downarrow \right \rangle_{B} \right ),\; \left (\beta\left. |\uparrow \right \rangle_{B}-\alpha\left. |\downarrow \right \rangle_{B} \right )$ It then suffices for Alice to communicate to Bob which entangled state she measured, and then Bob can apply an appropriate operator to put his particle in the state in which particle C was originally. Thus, the state has been teleported from Alice to Bob. Indeed, neither Alice nor Bob need know what particle C's original state was, though they can know that it was perfectly teleported. Note that the entangelment between Alice's and Bob's particles is, in the end, broken, and Alice's two particles are left entangled.

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Another example of the odd nature of entanglement can be demonstrated with the following. Suppose we have two independent sources that produce the entangled particle pairs $\tfrac{\left. |\uparrow \uparrow \right \rangle_{AB}+\left. |\downarrow \downarrow \right \rangle_{AB}}{\sqrt{2}},\;\; \tfrac{\left. |\uparrow \uparrow \right \rangle_{CD}+\left. |\downarrow \downarrow \right \rangle_{CD}}{\sqrt{2}}$ Particle A is sent to Alice, D to Dave, and B and C to Becca. We can write the total state of the system as follows $\tfrac{1}{2}\left ( \left. |\uparrow \uparrow\uparrow \uparrow \right \rangle_{ABCD}+ \left. |\uparrow \uparrow\downarrow \downarrow \right \rangle_{ABCD}+ \left. |\downarrow \downarrow\uparrow \uparrow \right \rangle_{ABCD}+ \left. |\downarrow \downarrow\downarrow \downarrow \right \rangle_{ABCD} \right )$ Alternatively, we could write it the following, equivalent way $\frac{1}{2} \begin{pmatrix} \tfrac{\left. |\uparrow \uparrow \right \rangle_{BC}+\left. |\downarrow \downarrow \right \rangle_{BC}}{\sqrt{2}} \tfrac{\left. |\uparrow \uparrow \right \rangle_{AD}+\left. |\downarrow \downarrow \right \rangle_{AD}}{\sqrt{2}}+ \tfrac{\left. |\uparrow \uparrow \right \rangle_{BC}-\left. |\downarrow \downarrow \right \rangle_{BC}}{\sqrt{2}} \tfrac{\left. |\uparrow \uparrow \right \rangle_{AD}-\left. |\downarrow \downarrow \right \rangle_{AD}}{\sqrt{2}} \\ + \tfrac{\left. |\uparrow \downarrow \right \rangle_{BC}+\left. |\downarrow \uparrow \right \rangle_{BC}}{\sqrt{2}} \tfrac{\left. |\uparrow \downarrow \right \rangle_{AD}+\left. |\downarrow \uparrow \right \rangle_{AD}}{\sqrt{2}} + \tfrac{\left. |\uparrow \downarrow \right \rangle_{BC}-\left. |\downarrow \uparrow \right \rangle_{BC}}{\sqrt{2}} \tfrac{\left. |\uparrow \downarrow \right \rangle_{AD}-\left. |\downarrow \uparrow \right \rangle_{AD}}{\sqrt{2}} \end{pmatrix}$ Thus, if Becca measures to see if her two particles are in any of the standard entangled states, as in the quantum teleportation setup, Alice's and Dave's particles will become entangled, and in the same entangled state as Becca's particles, no less. Becca can disentangle her particles from Alice's and Dave's, while entangling Alice's and Dave's particles, which initially bore no relation to one another. In this way, two particles can become entangled without ever having interacted, so entanglement need not require interaction.

## Spatial Phenomena

Let us look at the case of an electron in an infinite quantum well. That is, an electron in the potential that has the form $V(x)=\left\{\begin{matrix} 0\;\;\;0\leq x \leq L \\ \infty\;\;\mathrm{o.w.} \end{matrix}\right.$ As the wavefunction must be continuous, and clearly the wavefunction is zero outside the well, we have $\psi(0)=0$ and $\psi(L)=0$. Let us suppose the wavefunction is in an energy eigenstate. In that case, we solve the time-independent Schrodinger equation inside the well: $E\psi(x)=\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}\psi(x)$ This has the solutions $\psi(x)=A\cos(\lambda x)+B\sin(\lambda x)$ Where $\lambda=\sqrt{2mE}/\hbar$. From the condition that $\psi(0)=0$, $A=0$. From the condition that $\psi(L)=0$, $\lambda=n\pi/L$, where n is a positive integer. From the normalization condition, we have $B=\sqrt{2/L}$. Thus $\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left ( \frac{n\pi x}{L} \right )$ And the corresponding energy is $E_n=n^2\frac{\pi^2 \hbar^2}{2m L^2}$ Note that the energy is quantized, that is, it is only ever found to have a value in this discrete set of values. This feature is common in quantum mechanics: the boundary conditions, or conditions for convergence will restrict certain observables to fall into a discrete set. A related phenomenon is when the set of possible values for an observable falls into a fragmented set, of the form $[a_1,a_2]\cup[a_3,a_4]\cup...$ where the a's are strictly increasing. In such a case, the system will have allowed bands, and will need sizable "kicks" to get over the gaps. This is the basis for how transistors work.

Note also that in the case of the quantum well, all the eigenfunctions are orthogonal and form a complete set. An arbitrary initial wavefunction $\psi(x,0)$ will, at time t be equal to $\psi(x,t)=\frac{2}{L}\sum_{n=1}^{\infty} c_n \sin\left ( \frac{n\pi x}{L} \right ) e^{-itE_n/\hbar} , \;\; c_n=\int_{0}^{L}\psi(x,0)\sin\left ( \frac{n\pi x}{L}\right )dx$ We can also see something of the correspondence principle, namely, that for high energies, the probability distribution is nearly uniform in the well (it oscillates, as it goes above and below its average value, but the scale of the oscillations, for high enough energy, is imperceptible at macroscopic scales). Classically, for a particle bouncing back and forth in such a well, we would expect a uniform distribution (supposing we didn't know where the particle began).

Another result from this, which is true of quantum systems in general, is that even in the lowest energy state, when the most energy possible has been removed (the system is as "cold" as possible), the energy is non-zero. This is called the zero-point energy. Thus, even at "absolute" zero, the electron would still not be motionless, since, in this case $0< E_1 =\left \langle p^2 \right \rangle/2m$, and so the root-mean-square momentum would be non-zero.

An important case of a sort of quantum well is the atom, in which the nucleus attracts the electrons and so confines them. In the case of the atom, there are likewise quantized energy states. Since these are stationary states, the wavefunction does not vary with time, and so the effective charge density likewise is constant. This explains why the atom does not radiate energy, as it would in the classical case. However, in the case of the atom, which is necessarily a three-dimensional system, the states are also quantized with respect to angular momentum.

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Another interesting phenomenon is that of quantum tunneling. Suppose we have a particle moving in the +x direction impinging on a finite barrier of the form $V(x)=\left\{\begin{matrix} \tfrac{\hbar^2 q}{2m}\;\;\;0\leq x \leq L \\ 0\;\;\;\;\; \mathrm{o.w.} \end{matrix}\right.$ Let us call the regions before in and beyond the barrier regions I, II, and III respectively. Suppose it initially has momentum $\hbar k$. Its energy will be given by $\hbar^2 k^2/2m$, and further suppose that this energy is less than the potential barrier. Solving the schrodinger equation inside the barrier, we easily find that the wavefunction will be of the form $A e^{\lambda x}+Be^{-\lambda x}$, where $\lambda=\sqrt{q-k^2}$.

Then we can write the wavefunction (ignoring normalization) in the three regions as $\psi(x)=\left\{\begin{matrix} A_1 e^{ikx}+B_1 e^{-ikx} \;\;\;\, \mathrm{ I} \\ A_2 e^{\lambda x}+B_2 e^{-\lambda x} \;\;\;\;\; \mathrm{ II} \\ A_3 e^{ikx}+B_3 e^{-ikx} \;\;\;\;\; \mathrm{III} \end{matrix}\right.$ However, $B_3=0$, since that term corresponds to a wave moving to the left, which would not happen in the case of an incident wave going in the +x direction. The other coefficients can be found by ensuring that the wavefunction and its derivative are continuous. In particular, we find that $T=|A_1/A_3|^2=\frac{1}{1+\tfrac{q^2}{4k^2(q-k^2)}\sinh(\lambda L)}$ This represents the probability that the particle will be found on the opposite side of the barrier. Note that, contrary to classical mechanics, there is a definite, non-zero probability of finding the particle on the opposite side of the barrier. This feature of particles doing classically impossible things is a frequent characteristics of quantum mechanics. This phenomenon helps explain why the sun continues to fuse hydrogen even though it is not hot enough for the atoms to overcome the electrostatic repulsion, as the particles have a probability of tunneling through the classically forbidden region.

Similarly, we can see that, even if the particle did have enough energy to cross the barrier, there is not a 100% chance of finding it on the other side of the barrier. Just as a particle may sometimes cross a classically forbidden barrier, sometimes it fails to cross a classically allowed barrier.

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