#+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.