dylan: org/quandary.org comparison

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initial commit of dylan's stuff

author	Robert McIntyre <rlm@mit.edu>
date	Mon, 17 Oct 2011 23:17:55 -0700
parents
children	10c30f787f4b

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+#+TITLE: Bugs in quantum mechanics
+#+AUTHOR: Dylan Holmes
+#+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
+#+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
+#+SETUPFILE: ../../aurellem/org/setup.org
+#+INCLUDE:   ../../aurellem/org/level-0.org
+#Bugs in Quantum Mechanics
+#Bugs in the Quantum-Mechanical Momentum Operator
+I studied quantum mechanics the same way I study most subjects\mdash{}
+by collecting (and squashing) bugs in my understanding. One of these
+bugs persisted throughout two semesters of
+quantum mechanics coursework until I finally found
+the paper
+[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
+mechanics/]], which helped me stamp out the bug entirely. I decided to
+write an article about the problem and its solution for a number of reasons:
+- Although the paper was not unreasonably dense, it was written for
+teachers. I wanted to write an article for students.
+- I wanted to popularize the problem and its solution because other
+explanations are currently too hard to find. (Even Shankar's
+excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
+- Attempting an explanation is my way of making
+sure that the bug really /is/ gone.
+# entirely eradicated.
+* COMMENT
+I recommend the
+paper not only for students who are learning
+quantum mechanics, but especially for teachers interested in debugging
+them.
+* COMMENT
+On my first exam in quantum mechanics, my professor asked us to
+describe how certain measurements would affect a particle in a
+box. Many of these measurement questions required routine application
+of skills we had recently learned\mdash{}first, you recall (or
+calculate) the eigenstates of the quantity
+to be measured; second, you write the given state as a linear
+sum of these eigenstates\mdash{} the coefficients on each term give
+the probability amplitude.
+* Two methods of calculation that give different results.
+In the infinitely deep well, there is a particle in the the
+normalized state
+\(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
+This is apparently a perfectly respectable state: it is normalized ($A$ is a
+normalization constant), it is zero
+everywhere outside of the well, and it is moreover continuous.
+Even so, we will find a problem if we attempt to calculate the average
+energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)).
+** First method
+For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
+H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
+function of $x$ because we know how to express $H$ and $\psi$ in terms
+of  $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
+$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
+is constant.
+Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
+following way.
+\(\begin{eqnarray}
+\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
+\psi\rangle\\
+&=& \langle \psi H | H\psi \rangle\\
+&=& \langle \bar\psi | \bar\psi \rangle\\
+&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
+&=& \frac{A^2\hbar^4 a}{m^2}\\
+\end{eqnarray}\)
+For future reference, observe that this value is  nonzero
+(which makes sense).
+** Second method
+We can also calculate the average energy-squared of $|\psi\rangle$ in the
+following way.
+\begin{eqnarray}
+\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
+&=& \langle \psi |H \bar\psi \rangle\\
+&=&\int_0^a Ax(x-a)
+\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
+&=& 0\quad (!)\\
+\end{eqnarray}
+The second-to-last term must be zero because the second derivative
+of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
+* What is the problem?
+To recap: We used two different methods to calculate the average
+energy-squared of a state $|\psi\rangle$. For the first method, we
+used the fact that $H$ is a hermitian operator, replacing \(\langle
+\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
+\psi\rangle\). Using this substitution rule, we calculated the answer.
+For the second method,
+#we didn't use the fact that $H$ was hermitian;
+we instead used the fact that we know how to represent $H$ and $\psi$
+as functions of $x$: $H$ is a differential operator
+\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
+function of $x$. By applying $H$ to $\psi$, we took several
+derivatives and arrived at our answer.
+These two methods gave different results. In the following sections,
+I'll describe and analyze the source of this difference.
+** Physical operators only act on physical wavefunctions
+:PROPERTIES:
+:ORDERED:  t
+:END:
+#In quantum mechanics, an operator is a function that takes in a
+#physical state and produces another physical state as ouput. Some
+#operators correspond to physical quantities such as energy,
+#momentum, or position; the mathematical properties of these operators correspond to
+#physical properties of the system.
+#Eigenstates are an example of this correspondence: an
+Physical states are represented as wavefunctions in quantum
+mechanics. Just as we disallow certain physically nonsensical states
+in classical mechanics (for example, we consider it to be nonphysical
+for an object to spontaneously disappear from one place and reappear
+in another), we also disallow certain wavefunctions in quantum
+mechanics.
+For example, since wavefunctions are supposed to correspond to
+probability amplitudes, we require wavefunctions to be normalized
+\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
+wavefunctions that do not satisfy this property (although there are
+some exceptions[fn:2]).
+As another example, we generally expect probability to vary smoothly\mdash{}if
+a particle is very likely or very unlikely to be found at a particular
+location, it should also be somewhat likely or somewhat unlikely to be
+found /near/ that location. In more precise terms, we expect that for
+physically meaningful wavefunctions, the probability
+\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
+$x$ and, again, we disallow wavefunctions that do not satisfy this
+property because we consider them to be physically nonsensical.
+So, physical wavefunctions must satisfy certain properties
+like the two just described. Wavefunctions that do not satisfy these properties are
+rejected for being physically nonsensical: even though we can perform
+calculations with them, the mathematical results we obtain do not mean
+anything physically.
+Now, in quantum mechanics, an *operator* is a function that converts
+states into other states. Some operators correspond to
+physical quantities such as energy, momentum, or position, and as a
+result, the mathematical properties of these operators correspond to
+physical properties of the system. Such operators are called
+/hermitian operators/; one important property of hermitian operators
+is this rule:
+#+begin_quote
+*Hermitian operator rule:* A hermitian operator must only operate on
+the wavefunctions we have deemed physical, and must only produce
+physical wavefunctions[fn:: If you require a hermitian operator to
+have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
+that the operator will convert every physical wavefunction into
+another physical wavefunction:
+For any linear operator $\Omega$, the eigenvalue equation is
+\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
+eigenstate $|\omega\rangle$ is a physical wavefunction, the
+eigenvalue equation forces $\Omega|\omega\rangle$ to be a
+physical wavefunction as well. To elaborate, if the eigenstates of
+$\Omega$ are physical functions, then $\Omega$ is guaranteed to
+convert them into other physical functions.  Even more is true if the
+operator $\Omega$ is also hermitian: there is a theorem which states
+that \ldquo{}If \Omega is hermitian, then every physical wavefunction
+can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
+theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
+of \Omega are physically allowed/, then \Omega is guaranteed to
+convert every physically allowed wavefunction into another physically
+allowed wavefunction.].
+#+end_quote
+As you can see, this rule comes in two pieces. The first part is a
+constraint on *you*, the physicist: you must never feed a nonphysical
+state into a Hermitian operator, as it may produce nonsense. The
+second part is a constraint on the *operator*: the operator is
+guaranteed only to produce physical wavefunctions.
+In fact, this rule for hermitian operators is the source of our
+problem, as we unknowingly violated it when applying our second
+method!
+** The Hamiltonian is nonphysical
+You'll remember that in the second method we had wavefunctions within
+the well
+\(
+\begin{eqnarray}
+\psi(x) &=& A\;x(x-a)\\
+\bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
+\end{eqnarray}
+\)
+Using this, we wrote
+\(\begin{eqnarray}
+\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
+&=& \langle \psi |H \bar\psi \rangle\\
+& \vdots&\\
+&=& 0\\
+\end{eqnarray}\)
+However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
+$|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
+\psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
+not approach zero at the edges of the well. By
+feeding $H$ a nonphysical wavefunction, we obtained nonsensical
+results.
+Second, and more importantly, we were wrong to claim that $H$ was a
+physical operator\mdash{}that $H$ was hermitian. According to the
+rule, a hermitian operator must convert physical states into other
+physical states. But $|\psi\rangle$ is a physical state, as we said
+when we first introduced it \mdash{}it is a normalized, continuous
+function which approaches zero at the edges of the well and doesn't
+exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
+because it does not go to zero at the edges of the well. It is
+therefore impermissible for $H$ to transform the physical state
+$|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$.  Because
+$H$ converts some physical states into nonphysical states, it cannot
+be a hermitian operator as we assumed.
+# Boundary conditions affect hermiticity
+** Boundary conditions alter hermiticity
+It may surprise you (and it certainly surprised me) to find that the
+Hamiltonian is not hermitian. One of the fundamental principles of
+quantum mechanics is that hermitian operators correspond to physically
+observable quantities; for this reason, surely the
+Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
+But we must understand the correspondence between physically
+observable quantities and hermitian operators: every hermitian
+operator corresponds to a physically observable quantity, but not
+every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
+will correspond to a hermitian operator[fn::For a simple example,
+consider the differential operator \(D=\frac{d}{dx}\); although our
+intuitions might suggest that $D$ is observable which leads us to
+guess that $D$ is hermitian, it isn't. Still, the very closely related
+operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
+ought to validate our intuitions by checking the definitions.]. The
+true definition of a hermitian operator imply that the Hamiltonian
+stops being hermitian in the infinitely deep well. Here we arrive at a
+crucial point:
+Operators do not change /form/ between problems: the one-dimensional
+Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
+one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
+and so on.
+However, operators do change in this respect: hermitian operators must
+only take in physical states, and must only produce physical states; because
+in different problems we /do/ change the requirements for being a
+physical state, we also change what it takes for
+an operator to be called hermitian.  As a result, an operator that
+is hermitian in one setting may fail to be hermitian in another.
+Having seen how boundary conditions can affect hermiticity, we
+ought to be extra careful about which conditions we impose on our
+wavefunctions.
+** Choosing the right constraints
+We have said already that physicists
+require wavefunctions to satisfy certain properties in order to be
+deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
+infinitely deep well
+- Must be *normalizable*, because they correspond to
+probability amplitudes.
+- Must have *smoothly-varying probability*, because if a particle is very
+likely to be at a location, it ought to be likely to be /near/
+it as well.
+- Must *not exist outside the well*, because it
+would take an infinite amount of energy to do so.
+These conditions are surely reasonable. However, physicists sometimes
+assert that in order to satisfy the second and third conditions,
+physical wavefunctions
+- (?) Must *smoothly approach zero* towards the edges of the well.
+This final constraint is our reason for rejecting $|\bar\psi\rangle$
+as nonphysical and is consequently the reason why $H$ is not hermitian. If
+we can convince ourselves that the final constraint is unnecessary,
+$H$ may again be hermitian. This will satisfy our intuitions that the
+energy operator /ought/ to be hermitian.
+But in fact, we have the following mathematical observation to save
+us: a function $f$ does not need to be continuous in order for the
+integral \(\int^x f\) to be continuous. As a particularly relevant
+example, you may now notice that the function $\bar\psi(x)$ is not
+itself continuous, although the integral $\int_0^x \bar\psi$ /is/
+continuous. Evidently, it doesn't matter that the wavefunction
+$\bar\psi$ itself is not continuous; the probability corresponding to
+$\bar\psi$ /does/ manage to vary continuously anyways. Because the
+probability corresponding to $\bar\psi$ is the only aspect of
+$\bar\psi$ which we can detect physically, we /can/ safely omit the
+final constraint while keeping the other three.
+** Symmetric operators look like hermitian operators, but sometimes aren't.
+#+end_quote
+** COMMENT Re-examining physical constraints
+We have now discovered a flaw: when applied to the state
+$|\psi\rangle$, the second method violates the rule that physical
+operators must only take in physical states and must only produce
+physical states. Let's examine the problem more closely.
+We have said already that physicists require wavefunctions to satisfy
+certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
+be specific, wavefunctions in the infinitely deep well
+- Must be *normalizable*, because they correspond to
+probability amplitudes.
+- Must have *smoothly-varying probability*, because if a particle is very
+likely to be at a location, it ought to be likely to be /near/
+it as well.
+- Must *not exist outside the well*, because it
+would take an infinite amount of energy to do so.
+We now have discovered an important flaw in the second method: when
+applied to the state $|\bar\psi\rangle$, the second method violates
+the rule that physical operators must only take in
+physical states and must only produce physical states. The problem is
+even more serious, however
+[fn:1] I'm defining a new variable just to make certain expressions
+look shorter; this cannot affect the content of the answer we'll
+get.
+[fn:2] For example, in vaccuum (i.e., when the potential of the
+physical system is $V(x)=0$ throughout all space), the momentum
+eigenstates are not normalizable\mdash{}the relevant integral blows
+up to infinity instead of converging to a number. Physicists modify
+the definition of normalization slightly so that
+\ldquo{}delta-normalizable \rdquo{} functions like these are included
+among the physical wavefunctions.
+* COMMENT: What I thought I knew
+The following is a list of things I thought were true of quantum
+mechanics; the catch is that the list contradicts itself.
+1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
+2. For any hermitian operator: Any physically allowed state can be
+written as a linear sum of eigenstates of the operator.
+3. The momentum operator and energy operator are hermitian, because
+momentum and energy are measureable quantities.
+4. In the vacuum potential, the momentum and energy operators have these eigenstates:
+- the momentum operator has an eigenstate
+\(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
+- the energy operator has an eigenstate \(|E\rangle =
+\alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
+the particular choice of momentum $p=\sqrt{2mE}$.
+5. In the infinitely deep potential well, the momentum and energy
+operators have these eigenstates:
+- The momentum eigenstates and energy eigenstates have the same form
+as in the vacuum potential: $p(x) =
+\exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
+- Even so, because of the boundary conditions on the
+well, we must make the following modifications:
++ Physically realistic states must be impossible to find outside the well. (Only a state of infinite
+energy could exist outside the well, and infinite energy is not
+realistic.) This requirement means, for example, that momentum
+eigenstates in the infinitely deep well must be
+\(p(x)
+= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
+\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
++ Physically realistic states must vary smoothly throughout
+space. This means that if a particle in some state is very unlikely to be
+/at/ a particular location, it is also very unlikely be /near/
+that location. Combining this requirement with the above
+requirement, we find that the momentum operator no longer has
+an eigenstate for each value of $p$; instead, only values of
+$p$ that are integer multiples of $\pi \hbar/a$ are physically
+realistic. Similarly, the energy operator no longer has an
+eigenstate for each value of $E$; instead, the only energy
+eigenstates in the infinitely deep well
+are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
+* COMMENT:
+** Eigenstates with different eigenvalues are orthogonal
+#+begin_quote
+*Theorem:* Eigenstates with different eigenvalues are orthogonal.
+#+end_quote
+** COMMENT :
+I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
+and $|b\rangle$ are eigenstates of $\Lambda$. This means that
+\(
+\begin{eqnarray}
+\Lambda |a\rangle&=& a|a\rangle,\\
+\Lambda|b\rangle&=& b|b\rangle.\\
+\end{eqnarray}
+\)
+If we take the difference of these eigenstates, we find that
+\(
+\begin{eqnarray}
+\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
+\qquad \text{(because $\Lambda$ is linear.)}\\
+&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
+$|b\rangle$ are eigenstates of $\Lambda$)}
+\end{eqnarray}\)
+which means that $a\neq b$.
+** Eigenvectors of hermitian operators span the space of solutions
+#+begin_quote
+*Theorem:* If $\Omega$ is a hermitian operator, then every physically
+allowed state can be written as a linear sum of eigenstates of
+$\Omega$.
+#+end_quote
+** Momentum and energy are hermitian operators
+This ought to be true because hermitian operators correspond to
+observable quantities. Since we expect momentum and energy to be
+measureable quantities, we expect that there are hermitian operators
+to represent them.
+** Momentum and energy eigenstates in vacuum
+An eigenstate of the momentum operator $P$ would be a state
+\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
+** Momentum and energy eigenstates in the infinitely deep well
+* COMMENT Can you measure momentum in the infinitely deep well?
+In summary, I thought I knew:
+1. For any hermitian operator: eigenstates with different eigenvalues
+are orthogonal.
+2. For any hermitian operator: any physically realistic state can be
+written as a linear sum of eigenstates of the operator.
+3. The momentum operator and energy operator are hermitian, because
+momentum and energy are observable quantities.
+4. (The form of the momentum and energy eigenstates in the vacuum potential)
+5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
+Additionally, I understood that because the infinitely deep potential
+well is not realistic, states of such a system  are not necessarily
+physically realistic. Instead, I understood
+\ldquo{}realistic states\rdquo{} to be those that satisfy the physically
+unrealistic Schr\ouml{}dinger equation and its boundary conditions.
+With that final caveat, here is the problem:
+According to (5), the momentum eigenstates in the well are
+\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
+(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
+However, /these/ states are not orthogonal, which contradicts the
+assumption that (3) the momentum operator is hermitian and (2)
+eigenstates of a hermitian are orthogonal if they have different eigenvalues.
+#+begin_quote
+*Problem 1. The momentum eigenstates of the well are not orthogonal*
+/Proof./ If $p_1\neq p_2$, then
+\(\begin{eqnarray}
+\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
+&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
+outside the well.}\\
+&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
+&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
+\end{eqnarray}\)
+$\square$
+#+end_quote
+** COMMENT  Momentum eigenstates
+In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
+momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
+In the infinitely deep potential well, the Hamiltonian is the same but
+there is a new condition in order for states to qualify as physically
+allowed: the states must not exist anywhere outside of well, as it
+takes an infinite amount of energy to do so.
+Notice that the momentum eigenstates defined above do /not/ satisfy
+this condition.
+* COMMENT
+For each physical system, there is a Schr\ouml{}dinger equation that
+describes how a particle's state $|\psi\rangle$  will change over
+time.
+\(\begin{eqnarray}
+i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
+H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
+This is a differential equation; each solution to the
+Schr\ouml{}dinger equation is a state that is physically allowed for
+our particle. Here, physically allowed states are
+those that change in physically allowed ways. However, like any differential
+equation, the Schr\ouml{}dinger equation can be accompanied by
+/boundary conditions/\mdash{}conditions that further restrict which
+states qualify as physically allowed.
+** Eigenstates of momentum
+#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
+#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
+* COMMENT
+#* The infinite square well potential
+A particle exists in a potential that is
+infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
+particle exists in a potential[fn:coords][fn:infinity]
+\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
+}\;x<0\text{ or }x>a.\end{cases}\)
+The Schr\ouml{}dinger equation describes how the particle's state
+\(|\psi\rangle\) will change over time in this system.
+\(\begin{eqnarray}
+i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
+H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
+This is a differential equation; each solution to the
+Schr\ouml{}dinger equation is a state that is physically allowed for
+our particle. Here, physically allowed states are
+those that change in physically allowed ways. However, like any differential
+equation, the Schr\ouml{}dinger equation can be accompanied by
+/boundary conditions/\mdash{}conditions that further restrict which
+states qualify as physically allowed.
+Whenever possible, physicists impose these boundary conditions:
+- A physically allowed state ought to be a /smoothly-varying function of position./ This means
+that if a particle in the state  is likely to be /at/ a particular location,
+it is also likely to be /near/ that location.
+These boundary conditions imply that for the square well potential in
+this problem,
+- Physically allowed states must be totally confined to the well,
+because it takes an infinite amount of energy to exist anywhere
+outside of the well (and physically allowed states ought to have
+only finite energy).
+- Physically allowed states must be increasingly unlikely to find very
+close to the walls of the well. This is because of two conditions: the above
+condition says that the particle is /impossible/ to find
+outside of the well, and the smoothly-varying condition says
+that if a particle is impossible to find at a particular location,
+it must be unlikely to be found nearby that location.
+#; physically allowed states are those that change in physically
+#allowed ways.
+#** Boundary conditions
+Because the potential is infinite everywhere except within the well,
+a realistic particle must be confined to exist only within the
+well\mdash{}its wavefunction must be zero everywhere beyond the walls
+of the well.
+[fn:coords] I chose my coordinate system so that the well extends from
+\(0<x<a\). Others choose a coordinate system so that the well extends from
+\(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
+situation, they give different-looking answers.
+[fn:infinity] Of course, infinite potentials are not
+realistic. Instead, they are useful approximations to finite
+potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
+of the well\rdquo{} are close enough for your own practical
+purposes. Having introduced a physical impossibility into the problem
+already, we don't expect to get physically realistic solutions; we
+just expect to get mathematically consistent ones. The forthcoming
+trouble is that we don't.

Mercurial > dylan

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