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1 #+TITLE: Bugs in quantum mechanics
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2 #+AUTHOR: Dylan Holmes
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3 #+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics.
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4 #+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum
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5 #+SETUPFILE: ../../aurellem/org/setup.org
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6 #+INCLUDE: ../../aurellem/org/level-0.org
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7
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8
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9
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10 #Bugs in Quantum Mechanics
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11 #Bugs in the Quantum-Mechanical Momentum Operator
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12
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13
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14 I studied quantum mechanics the same way I study most subjects\mdash{}
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15 by collecting (and squashing) bugs in my understanding. One of these
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16 bugs persisted throughout two semesters of
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17 quantum mechanics coursework until I finally found
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18 the paper
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19 [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum
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20 mechanics/]], which helped me stamp out the bug entirely. I decided to
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21 write an article about the problem and its solution for a number of reasons:
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22
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23 - Although the paper was not unreasonably dense, it was written for
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24 teachers. I wanted to write an article for students.
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25 - I wanted to popularize the problem and its solution because other
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26 explanations are currently too hard to find. (Even Shankar's
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27 excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.)
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28 - Attempting an explanation is my way of making
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29 sure that the bug really /is/ gone.
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30 # entirely eradicated.
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31
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32 * COMMENT
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33 I recommend the
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34 paper not only for students who are learning
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35 quantum mechanics, but especially for teachers interested in debugging
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36 them.
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37
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38 * COMMENT
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39 On my first exam in quantum mechanics, my professor asked us to
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40 describe how certain measurements would affect a particle in a
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41 box. Many of these measurement questions required routine application
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42 of skills we had recently learned\mdash{}first, you recall (or
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43 calculate) the eigenstates of the quantity
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44 to be measured; second, you write the given state as a linear
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45 sum of these eigenstates\mdash{} the coefficients on each term give
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46 the probability amplitude.
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47
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48
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49 * Two methods of calculation that give different results.
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50
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51 In the infinitely deep well, there is a particle in the the
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52 normalized state
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53
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54 \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\)
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55
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56 This is apparently a perfectly respectable state: it is normalized ($A$ is a
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57 normalization constant), it is zero
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58 everywhere outside of the well, and it is moreover continuous.
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59
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60 Even so, we will find a problem if we attempt to calculate the average
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61 energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)).
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62
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63 ** First method
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64
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65 For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv
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66 H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a
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67 function of $x$ because we know how to express $H$ and $\psi$ in terms
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68 of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and
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69 $\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$
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70 is constant.
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71
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72 Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the
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73 following way.
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74
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75 \(\begin{eqnarray}
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76 \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H
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77 \psi\rangle\\
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78 &=& \langle \psi H | H\psi \rangle\\
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79 &=& \langle \bar\psi | \bar\psi \rangle\\
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80 &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\
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81 &=& \frac{A^2\hbar^4 a}{m^2}\\
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82 \end{eqnarray}\)
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83
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84 For future reference, observe that this value is nonzero
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85 (which makes sense).
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86
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87 ** Second method
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88 We can also calculate the average energy-squared of $|\psi\rangle$ in the
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89 following way.
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90
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91 \begin{eqnarray}
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92 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
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93 &=& \langle \psi |H \bar\psi \rangle\\
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94 &=&\int_0^a Ax(x-a)
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95 \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\
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96 &=& 0\quad (!)\\
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97 \end{eqnarray}
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98
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99 The second-to-last term must be zero because the second derivative
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100 of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero.
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101
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102 * What is the problem?
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103
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104 To recap: We used two different methods to calculate the average
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105 energy-squared of a state $|\psi\rangle$. For the first method, we
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106 used the fact that $H$ is a hermitian operator, replacing \(\langle
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107 \psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H
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108 \psi\rangle\). Using this substitution rule, we calculated the answer.
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109
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110 For the second method,
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111 #we didn't use the fact that $H$ was hermitian;
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112 we instead used the fact that we know how to represent $H$ and $\psi$
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113 as functions of $x$: $H$ is a differential operator
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114 \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic
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115 function of $x$. By applying $H$ to $\psi$, we took several
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116 derivatives and arrived at our answer.
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117
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118 These two methods gave different results. In the following sections,
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119 I'll describe and analyze the source of this difference.
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120
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121 ** Physical operators only act on physical wavefunctions
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122 :PROPERTIES:
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123 :ORDERED: t
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124 :END:
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125 #In quantum mechanics, an operator is a function that takes in a
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126 #physical state and produces another physical state as ouput. Some
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127 #operators correspond to physical quantities such as energy,
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128 #momentum, or position; the mathematical properties of these operators correspond to
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129 #physical properties of the system.
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130
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131 #Eigenstates are an example of this correspondence: an
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132
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133 Physical states are represented as wavefunctions in quantum
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134 mechanics. Just as we disallow certain physically nonsensical states
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135 in classical mechanics (for example, we consider it to be nonphysical
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136 for an object to spontaneously disappear from one place and reappear
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137 in another), we also disallow certain wavefunctions in quantum
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138 mechanics.
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139
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140 For example, since wavefunctions are supposed to correspond to
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141 probability amplitudes, we require wavefunctions to be normalized
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142 \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow
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143 wavefunctions that do not satisfy this property (although there are
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144 some exceptions[fn:2]).
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145
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146 As another example, we generally expect probability to vary smoothly\mdash{}if
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147 a particle is very likely or very unlikely to be found at a particular
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148 location, it should also be somewhat likely or somewhat unlikely to be
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149 found /near/ that location. In more precise terms, we expect that for
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150 physically meaningful wavefunctions, the probability
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151 \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of
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152 $x$ and, again, we disallow wavefunctions that do not satisfy this
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153 property because we consider them to be physically nonsensical.
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154
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155 So, physical wavefunctions must satisfy certain properties
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156 like the two just described. Wavefunctions that do not satisfy these properties are
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157 rejected for being physically nonsensical: even though we can perform
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158 calculations with them, the mathematical results we obtain do not mean
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159 anything physically.
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160
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161 Now, in quantum mechanics, an *operator* is a function that converts
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162 states into other states. Some operators correspond to
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163 physical quantities such as energy, momentum, or position, and as a
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164 result, the mathematical properties of these operators correspond to
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165 physical properties of the system. Such operators are called
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166 /hermitian operators/; one important property of hermitian operators
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167 is this rule:
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168
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169 #+begin_quote
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170 *Hermitian operator rule:* A hermitian operator must only operate on
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171 the wavefunctions we have deemed physical, and must only produce
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172 physical wavefunctions[fn:: If you require a hermitian operator to
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173 have physical eigenstates (which is reasonable), you get a very strong result: you guarantee
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174 that the operator will convert every physical wavefunction into
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175 another physical wavefunction:
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176
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177 For any linear operator $\Omega$, the eigenvalue equation is
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178 \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an
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179 eigenstate $|\omega\rangle$ is a physical wavefunction, the
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180 eigenvalue equation forces $\Omega|\omega\rangle$ to be a
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181 physical wavefunction as well. To elaborate, if the eigenstates of
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182 $\Omega$ are physical functions, then $\Omega$ is guaranteed to
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183 convert them into other physical functions. Even more is true if the
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184 operator $\Omega$ is also hermitian: there is a theorem which states
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185 that \ldquo{}If \Omega is hermitian, then every physical wavefunction
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186 can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This
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187 theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates
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188 of \Omega are physically allowed/, then \Omega is guaranteed to
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189 convert every physically allowed wavefunction into another physically
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190 allowed wavefunction.].
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191 #+end_quote
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192
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193 As you can see, this rule comes in two pieces. The first part is a
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194 constraint on *you*, the physicist: you must never feed a nonphysical
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195 state into a Hermitian operator, as it may produce nonsense. The
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196 second part is a constraint on the *operator*: the operator is
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197 guaranteed only to produce physical wavefunctions.
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198
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199 In fact, this rule for hermitian operators is the source of our
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200 problem, as we unknowingly violated it when applying our second
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201 method!
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202
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203 ** The Hamiltonian is nonphysical
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204 You'll remember that in the second method we had wavefunctions within
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205 the well
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206
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207 \(
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208 \begin{eqnarray}
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209 \psi(x) &=& A\;x(x-a)\\
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210 \bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\
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211 \end{eqnarray}
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212 \)
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213
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214 Using this, we wrote
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215
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216
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217 \(\begin{eqnarray}
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218 \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\
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219 &=& \langle \psi |H \bar\psi \rangle\\
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220 & \vdots&\\
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221 &=& 0\\
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222 \end{eqnarray}\)
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223
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224 However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction
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225 $|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar
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226 \psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does
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227 not approach zero at the edges of the well. By
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228 feeding $H$ a nonphysical wavefunction, we obtained nonsensical
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229 results.
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230
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231 Second, and more importantly, we were wrong to claim that $H$ was a
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232 physical operator\mdash{}that $H$ was hermitian. According to the
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233 rule, a hermitian operator must convert physical states into other
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234 physical states. But $|\psi\rangle$ is a physical state, as we said
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235 when we first introduced it \mdash{}it is a normalized, continuous
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236 function which approaches zero at the edges of the well and doesn't
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237 exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical
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238 because it does not go to zero at the edges of the well. It is
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239 therefore impermissible for $H$ to transform the physical state
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240 $|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$. Because
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241 $H$ converts some physical states into nonphysical states, it cannot
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242 be a hermitian operator as we assumed.
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243
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244 # Boundary conditions affect hermiticity
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245 ** Boundary conditions alter hermiticity
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246 It may surprise you (and it certainly surprised me) to find that the
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247 Hamiltonian is not hermitian. One of the fundamental principles of
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248 quantum mechanics is that hermitian operators correspond to physically
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249 observable quantities; for this reason, surely the
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250 Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian?
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251
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252 But we must understand the correspondence between physically
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253 observable quantities and hermitian operators: every hermitian
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254 operator corresponds to a physically observable quantity, but not
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255 every quantity that intuitively \ldquo{}ought\rdquo{} to be observable
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256 will correspond to a hermitian operator[fn::For a simple example,
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257 consider the differential operator \(D=\frac{d}{dx}\); although our
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258 intuitions might suggest that $D$ is observable which leads us to
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259 guess that $D$ is hermitian, it isn't. Still, the very closely related
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260 operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we
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261 ought to validate our intuitions by checking the definitions.]. The
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262 true definition of a hermitian operator imply that the Hamiltonian
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263 stops being hermitian in the infinitely deep well. Here we arrive at a
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264 crucial point:
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265
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266 Operators do not change /form/ between problems: the one-dimensional
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267 Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the
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268 one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$,
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269 and so on.
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270
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271 However, operators do change in this respect: hermitian operators must
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272 only take in physical states, and must only produce physical states; because
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273 in different problems we /do/ change the requirements for being a
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274 physical state, we also change what it takes for
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275 an operator to be called hermitian. As a result, an operator that
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276 is hermitian in one setting may fail to be hermitian in another.
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277
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278 Having seen how boundary conditions can affect hermiticity, we
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279 ought to be extra careful about which conditions we impose on our
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280 wavefunctions.
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281
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282 ** Choosing the right constraints
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283
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284 We have said already that physicists
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285 require wavefunctions to satisfy certain properties in order to be
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rlm@0
|
286 deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the
|
|
rlm@0
|
287 infinitely deep well
|
|
rlm@0
|
288 - Must be *normalizable*, because they correspond to
|
|
rlm@0
|
289 probability amplitudes.
|
|
rlm@0
|
290 - Must have *smoothly-varying probability*, because if a particle is very
|
|
rlm@0
|
291 likely to be at a location, it ought to be likely to be /near/
|
|
rlm@0
|
292 it as well.
|
|
rlm@0
|
293 - Must *not exist outside the well*, because it
|
|
rlm@0
|
294 would take an infinite amount of energy to do so.
|
|
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|
295
|
|
rlm@0
|
296 These conditions are surely reasonable. However, physicists sometimes
|
|
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|
297 assert that in order to satisfy the second and third conditions,
|
|
rlm@0
|
298 physical wavefunctions
|
|
rlm@0
|
299
|
|
rlm@0
|
300 - (?) Must *smoothly approach zero* towards the edges of the well.
|
|
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|
301
|
|
rlm@0
|
302 This final constraint is our reason for rejecting $|\bar\psi\rangle$
|
|
rlm@0
|
303 as nonphysical and is consequently the reason why $H$ is not hermitian. If
|
|
rlm@0
|
304 we can convince ourselves that the final constraint is unnecessary,
|
|
rlm@0
|
305 $H$ may again be hermitian. This will satisfy our intuitions that the
|
|
rlm@0
|
306 energy operator /ought/ to be hermitian.
|
|
rlm@0
|
307
|
|
rlm@0
|
308 But in fact, we have the following mathematical observation to save
|
|
rlm@0
|
309 us: a function $f$ does not need to be continuous in order for the
|
|
rlm@0
|
310 integral \(\int^x f\) to be continuous. As a particularly relevant
|
|
rlm@0
|
311 example, you may now notice that the function $\bar\psi(x)$ is not
|
|
rlm@0
|
312 itself continuous, although the integral $\int_0^x \bar\psi$ /is/
|
|
rlm@0
|
313 continuous. Evidently, it doesn't matter that the wavefunction
|
|
rlm@0
|
314 $\bar\psi$ itself is not continuous; the probability corresponding to
|
|
rlm@0
|
315 $\bar\psi$ /does/ manage to vary continuously anyways. Because the
|
|
rlm@0
|
316 probability corresponding to $\bar\psi$ is the only aspect of
|
|
rlm@0
|
317 $\bar\psi$ which we can detect physically, we /can/ safely omit the
|
|
rlm@0
|
318 final constraint while keeping the other three.
|
|
rlm@0
|
319
|
|
rlm@0
|
320 ** Symmetric operators look like hermitian operators, but sometimes aren't.
|
|
rlm@0
|
321
|
|
rlm@0
|
322
|
|
rlm@0
|
323 #+end_quote
|
|
rlm@0
|
324 ** COMMENT Re-examining physical constraints
|
|
rlm@0
|
325
|
|
rlm@0
|
326 We have now discovered a flaw: when applied to the state
|
|
rlm@0
|
327 $|\psi\rangle$, the second method violates the rule that physical
|
|
rlm@0
|
328 operators must only take in physical states and must only produce
|
|
rlm@0
|
329 physical states. Let's examine the problem more closely.
|
|
rlm@0
|
330
|
|
rlm@0
|
331 We have said already that physicists require wavefunctions to satisfy
|
|
rlm@0
|
332 certain properties in order to be deemed \ldquo{}physical\rdquo{}. To
|
|
rlm@0
|
333 be specific, wavefunctions in the infinitely deep well
|
|
rlm@0
|
334 - Must be *normalizable*, because they correspond to
|
|
rlm@0
|
335 probability amplitudes.
|
|
rlm@0
|
336 - Must have *smoothly-varying probability*, because if a particle is very
|
|
rlm@0
|
337 likely to be at a location, it ought to be likely to be /near/
|
|
rlm@0
|
338 it as well.
|
|
rlm@0
|
339 - Must *not exist outside the well*, because it
|
|
rlm@0
|
340 would take an infinite amount of energy to do so.
|
|
rlm@0
|
341
|
|
rlm@0
|
342 We now have discovered an important flaw in the second method: when
|
|
rlm@0
|
343 applied to the state $|\bar\psi\rangle$, the second method violates
|
|
rlm@0
|
344 the rule that physical operators must only take in
|
|
rlm@0
|
345 physical states and must only produce physical states. The problem is
|
|
rlm@0
|
346 even more serious, however
|
|
rlm@0
|
347
|
|
rlm@0
|
348
|
|
rlm@0
|
349
|
|
rlm@0
|
350 [fn:1] I'm defining a new variable just to make certain expressions
|
|
rlm@0
|
351 look shorter; this cannot affect the content of the answer we'll
|
|
rlm@0
|
352 get.
|
|
rlm@0
|
353
|
|
rlm@0
|
354 [fn:2] For example, in vaccuum (i.e., when the potential of the
|
|
rlm@0
|
355 physical system is $V(x)=0$ throughout all space), the momentum
|
|
rlm@0
|
356 eigenstates are not normalizable\mdash{}the relevant integral blows
|
|
rlm@0
|
357 up to infinity instead of converging to a number. Physicists modify
|
|
rlm@0
|
358 the definition of normalization slightly so that
|
|
rlm@0
|
359 \ldquo{}delta-normalizable \rdquo{} functions like these are included
|
|
rlm@0
|
360 among the physical wavefunctions.
|
|
rlm@0
|
361
|
|
rlm@0
|
362
|
|
rlm@0
|
363
|
|
rlm@0
|
364 * COMMENT: What I thought I knew
|
|
rlm@0
|
365
|
|
rlm@0
|
366 The following is a list of things I thought were true of quantum
|
|
rlm@0
|
367 mechanics; the catch is that the list contradicts itself.
|
|
rlm@0
|
368
|
|
rlm@0
|
369 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal.
|
|
rlm@0
|
370 2. For any hermitian operator: Any physically allowed state can be
|
|
rlm@0
|
371 written as a linear sum of eigenstates of the operator.
|
|
rlm@0
|
372 3. The momentum operator and energy operator are hermitian, because
|
|
rlm@0
|
373 momentum and energy are measureable quantities.
|
|
rlm@0
|
374 4. In the vacuum potential, the momentum and energy operators have these eigenstates:
|
|
rlm@0
|
375 - the momentum operator has an eigenstate
|
|
rlm@0
|
376 \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$.
|
|
rlm@0
|
377 - the energy operator has an eigenstate \(|E\rangle =
|
|
rlm@0
|
378 \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and
|
|
rlm@0
|
379 the particular choice of momentum $p=\sqrt{2mE}$.
|
|
rlm@0
|
380 5. In the infinitely deep potential well, the momentum and energy
|
|
rlm@0
|
381 operators have these eigenstates:
|
|
rlm@0
|
382 - The momentum eigenstates and energy eigenstates have the same form
|
|
rlm@0
|
383 as in the vacuum potential: $p(x) =
|
|
rlm@0
|
384 \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$.
|
|
rlm@0
|
385 - Even so, because of the boundary conditions on the
|
|
rlm@0
|
386 well, we must make the following modifications:
|
|
rlm@0
|
387 + Physically realistic states must be impossible to find outside the well. (Only a state of infinite
|
|
rlm@0
|
388 energy could exist outside the well, and infinite energy is not
|
|
rlm@0
|
389 realistic.) This requirement means, for example, that momentum
|
|
rlm@0
|
390 eigenstates in the infinitely deep well must be
|
|
rlm@0
|
391 \(p(x)
|
|
rlm@0
|
392 = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;
|
|
rlm@0
|
393 \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
|
|
rlm@0
|
394 + Physically realistic states must vary smoothly throughout
|
|
rlm@0
|
395 space. This means that if a particle in some state is very unlikely to be
|
|
rlm@0
|
396 /at/ a particular location, it is also very unlikely be /near/
|
|
rlm@0
|
397 that location. Combining this requirement with the above
|
|
rlm@0
|
398 requirement, we find that the momentum operator no longer has
|
|
rlm@0
|
399 an eigenstate for each value of $p$; instead, only values of
|
|
rlm@0
|
400 $p$ that are integer multiples of $\pi \hbar/a$ are physically
|
|
rlm@0
|
401 realistic. Similarly, the energy operator no longer has an
|
|
rlm@0
|
402 eigenstate for each value of $E$; instead, the only energy
|
|
rlm@0
|
403 eigenstates in the infinitely deep well
|
|
rlm@0
|
404 are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$.
|
|
rlm@0
|
405
|
|
rlm@0
|
406 * COMMENT:
|
|
rlm@0
|
407
|
|
rlm@0
|
408 ** Eigenstates with different eigenvalues are orthogonal
|
|
rlm@0
|
409
|
|
rlm@0
|
410 #+begin_quote
|
|
rlm@0
|
411 *Theorem:* Eigenstates with different eigenvalues are orthogonal.
|
|
rlm@0
|
412 #+end_quote
|
|
rlm@0
|
413
|
|
rlm@0
|
414 ** COMMENT :
|
|
rlm@0
|
415 I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$
|
|
rlm@0
|
416 and $|b\rangle$ are eigenstates of $\Lambda$. This means that
|
|
rlm@0
|
417
|
|
rlm@0
|
418
|
|
rlm@0
|
419 \(
|
|
rlm@0
|
420 \begin{eqnarray}
|
|
rlm@0
|
421 \Lambda |a\rangle&=& a|a\rangle,\\
|
|
rlm@0
|
422 \Lambda|b\rangle&=& b|b\rangle.\\
|
|
rlm@0
|
423 \end{eqnarray}
|
|
rlm@0
|
424 \)
|
|
rlm@0
|
425
|
|
rlm@0
|
426 If we take the difference of these eigenstates, we find that
|
|
rlm@0
|
427
|
|
rlm@0
|
428 \(
|
|
rlm@0
|
429 \begin{eqnarray}
|
|
rlm@0
|
430 \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle
|
|
rlm@0
|
431 \qquad \text{(because $\Lambda$ is linear.)}\\
|
|
rlm@0
|
432 &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and
|
|
rlm@0
|
433 $|b\rangle$ are eigenstates of $\Lambda$)}
|
|
rlm@0
|
434 \end{eqnarray}\)
|
|
rlm@0
|
435
|
|
rlm@0
|
436
|
|
rlm@0
|
437 which means that $a\neq b$.
|
|
rlm@0
|
438
|
|
rlm@0
|
439 ** Eigenvectors of hermitian operators span the space of solutions
|
|
rlm@0
|
440
|
|
rlm@0
|
441 #+begin_quote
|
|
rlm@0
|
442 *Theorem:* If $\Omega$ is a hermitian operator, then every physically
|
|
rlm@0
|
443 allowed state can be written as a linear sum of eigenstates of
|
|
rlm@0
|
444 $\Omega$.
|
|
rlm@0
|
445 #+end_quote
|
|
rlm@0
|
446
|
|
rlm@0
|
447
|
|
rlm@0
|
448
|
|
rlm@0
|
449 ** Momentum and energy are hermitian operators
|
|
rlm@0
|
450 This ought to be true because hermitian operators correspond to
|
|
rlm@0
|
451 observable quantities. Since we expect momentum and energy to be
|
|
rlm@0
|
452 measureable quantities, we expect that there are hermitian operators
|
|
rlm@0
|
453 to represent them.
|
|
rlm@0
|
454
|
|
rlm@0
|
455
|
|
rlm@0
|
456 ** Momentum and energy eigenstates in vacuum
|
|
rlm@0
|
457 An eigenstate of the momentum operator $P$ would be a state
|
|
rlm@0
|
458 \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\).
|
|
rlm@0
|
459
|
|
rlm@0
|
460 ** Momentum and energy eigenstates in the infinitely deep well
|
|
rlm@0
|
461
|
|
rlm@0
|
462
|
|
rlm@0
|
463
|
|
rlm@0
|
464 * COMMENT Can you measure momentum in the infinitely deep well?
|
|
rlm@0
|
465 In summary, I thought I knew:
|
|
rlm@0
|
466 1. For any hermitian operator: eigenstates with different eigenvalues
|
|
rlm@0
|
467 are orthogonal.
|
|
rlm@0
|
468 2. For any hermitian operator: any physically realistic state can be
|
|
rlm@0
|
469 written as a linear sum of eigenstates of the operator.
|
|
rlm@0
|
470 3. The momentum operator and energy operator are hermitian, because
|
|
rlm@0
|
471 momentum and energy are observable quantities.
|
|
rlm@0
|
472 4. (The form of the momentum and energy eigenstates in the vacuum potential)
|
|
rlm@0
|
473 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential)
|
|
rlm@0
|
474
|
|
rlm@0
|
475 Additionally, I understood that because the infinitely deep potential
|
|
rlm@0
|
476 well is not realistic, states of such a system are not necessarily
|
|
rlm@0
|
477 physically realistic. Instead, I understood
|
|
rlm@0
|
478 \ldquo{}realistic states\rdquo{} to be those that satisfy the physically
|
|
rlm@0
|
479 unrealistic Schr\ouml{}dinger equation and its boundary conditions.
|
|
rlm@0
|
480
|
|
rlm@0
|
481 With that final caveat, here is the problem:
|
|
rlm@0
|
482
|
|
rlm@0
|
483 According to (5), the momentum eigenstates in the well are
|
|
rlm@0
|
484
|
|
rlm@0
|
485 \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\)
|
|
rlm@0
|
486
|
|
rlm@0
|
487 (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.)
|
|
rlm@0
|
488
|
|
rlm@0
|
489 However, /these/ states are not orthogonal, which contradicts the
|
|
rlm@0
|
490 assumption that (3) the momentum operator is hermitian and (2)
|
|
rlm@0
|
491 eigenstates of a hermitian are orthogonal if they have different eigenvalues.
|
|
rlm@0
|
492
|
|
rlm@0
|
493 #+begin_quote
|
|
rlm@0
|
494 *Problem 1. The momentum eigenstates of the well are not orthogonal*
|
|
rlm@0
|
495
|
|
rlm@0
|
496 /Proof./ If $p_1\neq p_2$, then
|
|
rlm@0
|
497
|
|
rlm@0
|
498 \(\begin{eqnarray}
|
|
rlm@0
|
499 \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\
|
|
rlm@0
|
500 &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{
|
|
rlm@0
|
501 outside the well.}\\
|
|
rlm@0
|
502 &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\
|
|
rlm@0
|
503 &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\
|
|
rlm@0
|
504 \end{eqnarray}\)
|
|
rlm@0
|
505 $\square$
|
|
rlm@0
|
506
|
|
rlm@0
|
507 #+end_quote
|
|
rlm@0
|
508
|
|
rlm@0
|
509
|
|
rlm@0
|
510
|
|
rlm@0
|
511 ** COMMENT Momentum eigenstates
|
|
rlm@0
|
512
|
|
rlm@0
|
513 In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the
|
|
rlm@0
|
514 momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\).
|
|
rlm@0
|
515
|
|
rlm@0
|
516 In the infinitely deep potential well, the Hamiltonian is the same but
|
|
rlm@0
|
517 there is a new condition in order for states to qualify as physically
|
|
rlm@0
|
518 allowed: the states must not exist anywhere outside of well, as it
|
|
rlm@0
|
519 takes an infinite amount of energy to do so.
|
|
rlm@0
|
520
|
|
rlm@0
|
521 Notice that the momentum eigenstates defined above do /not/ satisfy
|
|
rlm@0
|
522 this condition.
|
|
rlm@0
|
523
|
|
rlm@0
|
524
|
|
rlm@0
|
525
|
|
rlm@0
|
526 * COMMENT
|
|
rlm@0
|
527 For each physical system, there is a Schr\ouml{}dinger equation that
|
|
rlm@0
|
528 describes how a particle's state $|\psi\rangle$ will change over
|
|
rlm@0
|
529 time.
|
|
rlm@0
|
530
|
|
rlm@0
|
531 \(\begin{eqnarray}
|
|
rlm@0
|
532 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
|
|
rlm@0
|
533 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
|
|
rlm@0
|
534
|
|
rlm@0
|
535 This is a differential equation; each solution to the
|
|
rlm@0
|
536 Schr\ouml{}dinger equation is a state that is physically allowed for
|
|
rlm@0
|
537 our particle. Here, physically allowed states are
|
|
rlm@0
|
538 those that change in physically allowed ways. However, like any differential
|
|
rlm@0
|
539 equation, the Schr\ouml{}dinger equation can be accompanied by
|
|
rlm@0
|
540 /boundary conditions/\mdash{}conditions that further restrict which
|
|
rlm@0
|
541 states qualify as physically allowed.
|
|
rlm@0
|
542
|
|
rlm@0
|
543
|
|
rlm@0
|
544
|
|
rlm@0
|
545
|
|
rlm@0
|
546 ** Eigenstates of momentum
|
|
rlm@0
|
547
|
|
rlm@0
|
548
|
|
rlm@0
|
549
|
|
rlm@0
|
550
|
|
rlm@0
|
551 #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger
|
|
rlm@0
|
552
|
|
rlm@0
|
553 #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\)
|
|
rlm@0
|
554
|
|
rlm@0
|
555
|
|
rlm@0
|
556
|
|
rlm@0
|
557
|
|
rlm@0
|
558
|
|
rlm@0
|
559
|
|
rlm@0
|
560
|
|
rlm@0
|
561 * COMMENT
|
|
rlm@0
|
562
|
|
rlm@0
|
563 #* The infinite square well potential
|
|
rlm@0
|
564
|
|
rlm@0
|
565 A particle exists in a potential that is
|
|
rlm@0
|
566 infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the
|
|
rlm@0
|
567 particle exists in a potential[fn:coords][fn:infinity]
|
|
rlm@0
|
568
|
|
rlm@0
|
569
|
|
rlm@0
|
570 \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for
|
|
rlm@0
|
571 }\;x<0\text{ or }x>a.\end{cases}\)
|
|
rlm@0
|
572
|
|
rlm@0
|
573 The Schr\ouml{}dinger equation describes how the particle's state
|
|
rlm@0
|
574 \(|\psi\rangle\) will change over time in this system.
|
|
rlm@0
|
575
|
|
rlm@0
|
576 \(\begin{eqnarray}
|
|
rlm@0
|
577 i\hbar \frac{\partial}{\partial t}|\psi\rangle &=&
|
|
rlm@0
|
578 H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\)
|
|
rlm@0
|
579
|
|
rlm@0
|
580 This is a differential equation; each solution to the
|
|
rlm@0
|
581 Schr\ouml{}dinger equation is a state that is physically allowed for
|
|
rlm@0
|
582 our particle. Here, physically allowed states are
|
|
rlm@0
|
583 those that change in physically allowed ways. However, like any differential
|
|
rlm@0
|
584 equation, the Schr\ouml{}dinger equation can be accompanied by
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585 /boundary conditions/\mdash{}conditions that further restrict which
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586 states qualify as physically allowed.
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587
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588
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589 Whenever possible, physicists impose these boundary conditions:
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590 - A physically allowed state ought to be a /smoothly-varying function of position./ This means
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591 that if a particle in the state is likely to be /at/ a particular location,
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592 it is also likely to be /near/ that location.
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593
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594 These boundary conditions imply that for the square well potential in
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595 this problem,
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596
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597 - Physically allowed states must be totally confined to the well,
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598 because it takes an infinite amount of energy to exist anywhere
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599 outside of the well (and physically allowed states ought to have
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600 only finite energy).
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601 - Physically allowed states must be increasingly unlikely to find very
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602 close to the walls of the well. This is because of two conditions: the above
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603 condition says that the particle is /impossible/ to find
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604 outside of the well, and the smoothly-varying condition says
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605 that if a particle is impossible to find at a particular location,
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606 it must be unlikely to be found nearby that location.
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607
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608 #; physically allowed states are those that change in physically
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609 #allowed ways.
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610
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611
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612 #** Boundary conditions
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613 Because the potential is infinite everywhere except within the well,
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614 a realistic particle must be confined to exist only within the
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615 well\mdash{}its wavefunction must be zero everywhere beyond the walls
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616 of the well.
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617
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618
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619 [fn:coords] I chose my coordinate system so that the well extends from
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620 \(0<x<a\). Others choose a coordinate system so that the well extends from
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621 \(-\frac{a}{2}<x<\frac{a}{2}\). Although both coordinate systems describe the same physical
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622 situation, they give different-looking answers.
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623
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624 [fn:infinity] Of course, infinite potentials are not
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625 realistic. Instead, they are useful approximations to finite
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626 potentials when \ldquo{}infinity\rdquo{} and \ldquo{}the actual height
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627 of the well\rdquo{} are close enough for your own practical
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628 purposes. Having introduced a physical impossibility into the problem
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629 already, we don't expect to get physically realistic solutions; we
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630 just expect to get mathematically consistent ones. The forthcoming
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631 trouble is that we don't.
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