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<h2 class="hd hd-2 unit-title">The classical 7-bit Hamming code</h2>
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<h3 class="hd hd-3 problem-header" id="s12-wk2-hamming-linear-problem-title" aria-describedby="block-v1:MITx+8.370.3x+1T2018+type@problem+block@s12-wk2-hamming-linear-problem-progress" tabindex="-1">
The classical 7-bit Hamming code
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<p>
Certain classical linear codes can be translated directly into quantum codes, and in this exercise we explore an example which illustrates the procedure, and also introduces the basic ideas of classical linear codes. </p>
<ol class="enumerate">
<li value="1">
<p>
A binary <a href="http://en.wikipedia.org/wiki/Linear_code" target="_blank"><em>linear code</em></a> [mathjaxinline]C[/mathjaxinline] encoding [mathjaxinline]k[/mathjaxinline] bits of information into an [mathjaxinline]n[/mathjaxinline] bit code space is a set of bit strings specified by an [mathjaxinline]n[/mathjaxinline] by [mathjaxinline]k[/mathjaxinline] <em>generator matrix</em> [mathjaxinline]G[/mathjaxinline] whose entries are zeroes and ones. The [mathjaxinline]2^ k[/mathjaxinline] codewords which comprise [mathjaxinline]C[/mathjaxinline] are given by [mathjaxinline]Gx[/mathjaxinline], where [mathjaxinline]x[/mathjaxinline] is a column vector specifying a [mathjaxinline]k[/mathjaxinline]-bit integer, ranging from [mathjaxinline]0[/mathjaxinline] to [mathjaxinline]2^ k-1[/mathjaxinline]. Note that arithmetic operaetions are all done modulo 2. </p>
<p>
The generator matrix </p>
<table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]G = \left[ \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 1 \end{array} \right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.1)</td>
</tr>
</table>
<p>
may be used to encode sixteen seven-bit codewords. </p>
<p>
Which of the bit strings below are valid codewords, produced by [mathjaxinline]G[/mathjaxinline]? </p>
<ul class="itemize">
<li>
<p>
<p style="display:inline"><big class="large"><tt class="tt">[ 0 0 0 0 0 0 1 ]</tt></big> is valid?&#160;</p>
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</div></div>
</p>
</li>
<li>
<p>
<p style="display:inline"><big class="large"><tt class="tt">[ 0 0 0 1 1 1 1 ]</tt></big> is valid?&#160;</p>
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</p>
</li>
<li>
<p>
<p style="display:inline"><big class="large"><tt class="tt">[ 1 0 0 1 1 1 1 ]</tt></big> is valid?&#160;</p>
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</li>
<li value="2">
<p>
Errors are detected by computing various parity checks, which are forumlated in terms of an [mathjaxinline]n-k[/mathjaxinline] by [mathjaxinline]n[/mathjaxinline] matrix [mathjaxinline]H[/mathjaxinline] satisfying [mathjaxinline]Hy=0[/mathjaxinline] for all codewords [mathjaxinline]y[/mathjaxinline]. Equivalently, since there exists [mathjaxinline]x[/mathjaxinline] such that [mathjaxinline]Gx=y[/mathjaxinline] for all valid [mathjaxinline]y[/mathjaxinline], then it must be true that [mathjaxinline]HGx=0[/mathjaxinline], so that in general we have [mathjaxinline]HG=0[/mathjaxinline]. We say that [mathjaxinline]H[/mathjaxinline] is a matrix whose the <a href="http://en.wikipedia.org/wiki/Kernel_(matrix)" target="_blank">null space</a> is the code [mathjaxinline]C[/mathjaxinline]. </p>
<p>
Which of the bit strings below are valid rows of the matrix [mathjaxinline]H[/mathjaxinline]? </p>
<ul class="itemize">
<li>
<p>
<p style="display:inline">[mathjaxinline]h_2 = \left[ \begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 \end{array}\right][/mathjaxinline] is valid?&#160;</p>
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<p>
<p style="display:inline">[mathjaxinline]h_4 = \left[ \begin{array}{ccccccc}1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 \end{array}\right][/mathjaxinline] is valid?&#160;</p>
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<p class="answer" id="answer_s12-wk2-hamming-linear_6_1"/>
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</p>
</li>
</ul>
<p>
Note that many bit strings may be found which can be valid rows of [mathjaxinline]H[/mathjaxinline], but of these, only [mathjaxinline]n-k[/mathjaxinline] bit strings will be linearly independent. Thus, [mathjaxinline]H[/mathjaxinline] is given by [mathjaxinline]n-k[/mathjaxinline] linearly independent rows, each with [mathjaxinline]n[/mathjaxinline] column entries. </p>
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<h2 class="hd hd-2 unit-title">7-bit Hamming code syndrome and distance</h2>
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7-bit Hamming code syndrome and distance
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<p>
Consider the classical [mathjaxinline]n=7[/mathjaxinline], [mathjaxinline]k=4[/mathjaxinline] code [mathjaxinline]C[/mathjaxinline] generated by </p>
<table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]G = \left[ \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 1 \end{array} \right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.2)</td>
</tr>
</table>
<p>
which encodes sixteen seven-bit codewords. Let us use as the parity check matrix </p>
<table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]H = \left[ \begin{array}{ccccccc} 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 \end{array} \right].[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.3)</td>
</tr>
</table>
<p>
Recall that for these binary matrices, addition is performed modulo two, and note that [mathjaxinline]HG=0[/mathjaxinline]. </p>
<ol class="enumerate">
<li value="1">
<p>
An error can be modeled as addition (modulo 2) of a random bit string [mathjaxinline]e[/mathjaxinline] to a codeword [mathjaxinline]x\in C[/mathjaxinline], giving [mathjaxinline]y = x + e[/mathjaxinline]. As long as [mathjaxinline]y[/mathjaxinline] is not a codeword, the error can be detected by computing [mathjaxinline]Hy = He \neq 0[/mathjaxinline]. Thus, [mathjaxinline]Hy[/mathjaxinline] is called the <em>error syndrome</em>. </p>
<p>
Give the error syndrome for the following bit strings. Please enter your answer as a vector, eg <tt class="tt">[0,1,1]</tt>. Keep in mind that the numbers should be just zero or one. </p>
<ul class="itemize">
<li>
<p>
[mathjaxinline]y_0^ T = \left[ \begin{array}{ccccccc} 0 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \end{array}\right][/mathjaxinline]: <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_s12-wk2-hamming-syndrome_2_1" class=" capa_inputtype textline">
<div class="unanswered ">
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
<p id="answer_s12-wk2-hamming-syndrome_2_1" class="answer"/>
</div>
</div></div> </p>
</li>
<li>
<p>
[mathjaxinline]y_1^ T = \left[ \begin{array}{ccccccc} 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 \end{array}\right][/mathjaxinline]: <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_s12-wk2-hamming-syndrome_3_1" class=" capa_inputtype textline">
<div class="unanswered ">
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<p id="answer_s12-wk2-hamming-syndrome_3_1" class="answer"/>
</div>
</div></div> </p>
</li>
</ul>
<p>
<div class="solution-span">
<span id="solution_s12-wk2-hamming-syndrome_solution_1"/>
</div></p>
</li>
<li value="2">
<p>
The maximum number of bit flip errors that can be tolerated is determined by the minimum Hamming distance between any two codewords, </p>
<table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]d(C) = \min _{x,y \in C, x\neq y} d(x,y)[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.4)</td>
</tr>
</table>
<p>
where [mathjaxinline]d(x,y)[/mathjaxinline] is the number of bits where [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]y[/mathjaxinline] differ. What is [mathjaxinline]d(C)[/mathjaxinline] for the above code? </p>
<p>
<p style="display:inline">[mathjaxinline]d(C) =[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">Dual classical code</h2>
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Dual classical code
</h3>
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<p>
Consider the classical [mathjaxinline]n=7[/mathjaxinline], [mathjaxinline]k=4[/mathjaxinline] code [mathjaxinline]C[/mathjaxinline] generated by </p>
<table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]G = \left[ \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 1 \end{array} \right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.5)</td>
</tr>
</table>
<p>
which encodes sixteen seven-bit codewords. Let us use as the parity check matrix </p>
<table id="a0000000007" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]H = \left[ \begin{array}{ccccccc} 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 \end{array} \right].[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.6)</td>
</tr>
</table>
<p>
Recall that for these binary matrices, addition is performed modulo two, and note that [mathjaxinline]HG=0[/mathjaxinline]. </p>
<p>
Define a new code [mathjaxinline]C^\perp[/mathjaxinline] which has generator matrix [mathjaxinline]G' = H^ T[/mathjaxinline] and parity check matrix [mathjaxinline]H' = G^ T[/mathjaxinline]. This code is known as the <em>dual</em> of [mathjaxinline]C[/mathjaxinline]. </p>
<p>
Recall that [mathjaxinline]n[/mathjaxinline] is the number of bits each codeword has, and [mathjaxinline]k[/mathjaxinline] is the number of bits which can be encoded by the codewords (ie [mathjaxinline]k=\log _2({\rm number~ of~ codewords})[/mathjaxinline]. </p>
<ul class="itemize">
<li>
<p>
<p style="display:inline">What is [mathjaxinline]n[/mathjaxinline] for [mathjaxinline]C^\perp[/mathjaxinline]?&#160;</p>
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<p>
<p style="display:inline">What is [mathjaxinline]k[/mathjaxinline] for [mathjaxinline]C^\perp[/mathjaxinline]?&#160;</p>
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<p>
Are all codewowrds of [mathjaxinline]C^\perp[/mathjaxinline] codewords of [mathjaxinline]C[/mathjaxinline]? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="inputtype option-input ">
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event.preventDefault();
go_to_search();
}
});
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'jump_to_url': jump_to_url,
'go_to_search': go_to_search,
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<h2 class="hd hd-2 unit-title">Dual classical code II</h2>
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data-problem-id="block-v1:MITx+8.370.3x+1T2018+type@problem+block@s12-wk2-hamming-dual2" data-url="/courses/course-v1:MITx+8.370.3x+1T2018/xblock/block-v1:MITx+8.370.3x+1T2018+type@problem+block@s12-wk2-hamming-dual2/handler/xmodule_handler"
data-problem-score="0"
data-problem-total-possible="3"
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data-content="
<h3 class="hd hd-3 problem-header" id="s12-wk2-hamming-dual2-problem-title" aria-describedby="block-v1:MITx+8.370.3x+1T2018+type@problem+block@s12-wk2-hamming-dual2-problem-progress" tabindex="-1">
Dual classical code II
</h3>
<div class="problem-progress" id="block-v1:MITx+8.370.3x+1T2018+type@problem+block@s12-wk2-hamming-dual2-problem-progress"></div>
<div class="problem">
<div>
<p>
Continuing with our consideratin of the classical [mathjaxinline]n=7[/mathjaxinline], [mathjaxinline]k=4[/mathjaxinline] code [mathjaxinline]C[/mathjaxinline] generated by </p>
<table id="a0000000011" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]G = \left[ \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 1 \end{array} \right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.9)</td>
</tr>
</table>
<p>
which encodes sixteen seven-bit codewords. Let us use as the parity check matrix </p>
<table id="a0000000012" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]H = \left[ \begin{array}{ccccccc} 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 \end{array} \right].[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.10)</td>
</tr>
</table>
<p>
Recall that for these binary matrices, addition is performed modulo two, and note that [mathjaxinline]HG=0[/mathjaxinline]. </p>
<p>
Define a new code [mathjaxinline]C^\perp[/mathjaxinline] which has generator matrix [mathjaxinline]G' = H^ T[/mathjaxinline] and parity check matrix [mathjaxinline]H' = G^ T[/mathjaxinline]. This code is known as the <em>dual</em> of [mathjaxinline]C[/mathjaxinline]. </p>
<p>
Recall that [mathjaxinline]n[/mathjaxinline] is the number of bits each codeword has, and [mathjaxinline]k[/mathjaxinline] is the number of bits which can be encoded by the codewords (ie [mathjaxinline]k=\log _2({\rm number~ of~ codewords})[/mathjaxinline]. </p>
<p>
What is the relationship between codewords in [mathjaxinline]C[/mathjaxinline] and [mathjaxinline]C^\perp[/mathjaxinline]? </p>
<ul class="itemize">
<li>
<p>
If [mathjaxinline]x\in C^\perp[/mathjaxinline] and [mathjaxinline]y\in C[/mathjaxinline], then what is [mathjaxinline]x\cdot y[/mathjaxinline]? Recall that addition is modulo 2. </p>
<p>
<p style="display:inline">[mathjaxinline]x\cdot y =[/mathjaxinline] </p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_s12-wk2-hamming-dual2_2_1" class="text-input-dynamath capa_inputtype inline textline">
<div class="unanswered inline">
<input type="text" name="input_s12-wk2-hamming-dual2_2_1" id="input_s12-wk2-hamming-dual2_2_1" aria-describedby="status_s12-wk2-hamming-dual2_2_1" value="" class="math"/>
<span class="trailing_text" id="trailing_text_s12-wk2-hamming-dual2_2_1"/>
<span class="status unanswered" id="status_s12-wk2-hamming-dual2_2_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
<p id="answer_s12-wk2-hamming-dual2_2_1" class="answer"/>
<div id="display_s12-wk2-hamming-dual2_2_1" class="equation">`{::}`</div>
<textarea style="display:none" id="input_s12-wk2-hamming-dual2_2_1_dynamath" name="input_s12-wk2-hamming-dual2_2_1_dynamath"/>
</div>
</div></div>
</p>
</li>
<li>
<p>
If [mathjaxinline]x \in C^\perp[/mathjaxinline], then what is [mathjaxinline]\sum _{y\in C} (-1)^{x\cdot y}[/mathjaxinline]? </p>
<p>
<p style="display:inline">[mathjaxinline]\sum _{y\in C} (-1)^{x\cdot y} =[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_s12-wk2-hamming-dual2_3_1" class=" capa_inputtype inline textline">
<div class="unanswered inline">
<input type="text" name="input_s12-wk2-hamming-dual2_3_1" id="input_s12-wk2-hamming-dual2_3_1" aria-describedby="status_s12-wk2-hamming-dual2_3_1" value=""/>
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
<p id="answer_s12-wk2-hamming-dual2_3_1" class="answer"/>
</div>
</div></div>
</p>
</li>
<li>
<p>
If [mathjaxinline]x\notin C^\perp[/mathjaxinline] but [mathjaxinline]x\in C[/mathjaxinline], then what is [mathjaxinline]\sum _{y\in C} (-1)^{x\cdot y}[/mathjaxinline]? </p>
<p>
<p style="display:inline">[mathjaxinline]\sum _{y\in C} (-1)^{x\cdot y} =[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_s12-wk2-hamming-dual2_4_1" class=" capa_inputtype inline textline">
<div class="unanswered inline">
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
<p id="answer_s12-wk2-hamming-dual2_4_1" class="answer"/>
</div>
</div></div>
</p>
</li>
</ul>
<p>
<div class="solution-span">
<span id="solution_s12-wk2-hamming-dual2_solution_1"/>
</div></p>
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<div class="xblock xblock-public_view xblock-public_view-html xmodule_display xmodule_HtmlBlock" data-block-type="html" data-runtime-class="LmsRuntime" data-usage-id="block-v1:MITx+8.370.3x+1T2018+type@html+block@site_search_box004116" data-course-id="course-v1:MITx+8.370.3x+1T2018" data-has-score="False" data-graded="True" data-runtime-version="1" data-init="XBlockToXModuleShim" data-request-token="80b2212e026211efa50f0afff417eba9">
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<span><a href="/asset-v1:MITx+8.370.3x+1T2018+type@asset+block/NONE" id="dummy_course_static_link" style="display:none"/><a href="/courses/course-v1:MITx+8.370.3x+1T2018/jump_to_id/NONE" id="dummy_jump_link" style="display:none"/><script type="text/javascript">
var add_site_search = function(){
course_static_url = $('#dummy_course_static_link').attr('href').replace('/NONE', '');
jump_to_url = $('#dummy_jump_link').attr('href').replace('/NONE', '');
if (typeof String.prototype.startsWith != 'function') {
// see below for better implementation!
String.prototype.startsWith = function (str){
return this.indexOf(str) === 0;
};
}
if(typeof(String.prototype.trim) === "undefined")
{
String.prototype.trim = function()
{
return String(this).replace(/^\s+|\s+$/g, '');
};
}
var lb = String.fromCharCode(60);
var rb = String.fromCharCode(62);
var amp = String.fromCharCode(38);
var rlb = rb + lb;
var mke = function(x){ return lb + x + rb; }
var search_module_url = "";
var get_search_module_ficus = function(){
var cid = $('div.xblock').data('course-id');
if (cid){
console.log("cid = ", cid);
// search_module_url = "/courses/course-v1:MITx+8.370.3x+1T2018/" + cid + "/courseware/welcome/Search_this_course/";
search_module_url = "/courses/course-v1:MITx+8.370.3x+1T2018/courseware/welcome/Search_this_course/"; // automatically rewritten
console.log("3. search_module_url = ", search_module_url);
return;
}
var course_root_link = $('span.nav-item-course').find('a').attr('href');
if (course_root_link){
console.log("course_root_link = ", course_root_link);
search_module_url = course_root_link.replace("course/", "courseware/welcome/Search_this_course/");
console.log("2. search_module_url = ", search_module_url);
return
}
console.log("cannot determine search module url");
}
var get_search_module = function(){
// find search this module link
if (!($('div.course-index').length)){
return get_search_module_ficus();
}
$('div.course-index').find('nav').find('a').each(function(){
if ($(this).text().trim().startsWith("Search this course")){
search_module_url = $(this).attr('href');
console.log("search_module_url = ", search_module_url);
}
});
}
var go_to_search = function(){
get_search_module();
var sterm = $('#site-search-box').val();
// new_url = jump_to_url + "/Search_this_module/?q=" + sterm;
new_url = search_module_url + "?q=" + sterm;
console.log("sterm = ", sterm, " ; going to ", new_url);
window.location.href = new_url;
}
if (!$('#site-search-box').length){
$("nav.courseware").find("ol").append(lb + "section style='float:right'" + rlb + "input size='20'"
+ " id='site-search-box'"
+ rlb + "img src='" + course_static_url
+ "/images_search_glass.png'/" + rlb + "/input" + rlb + "/section" + rb);
}
$("#site-search-box").keypress(function(event) {
if (event.which == 13) {
event.preventDefault();
go_to_search();
}
});
// $('#site-search-box').bind("enterKey", go_to_search);
var get = function(x){
return eval(x);
}
return {'course_static_url': course_static_url,
'jump_to_url': jump_to_url,
'go_to_search': go_to_search,
'get_search_module': get_search_module,
'get_search_module_ficus': get_search_module_ficus,
'get': get,
}
}
var the_site_search = add_site_search();
var add_fix_transcript = function(){
if ($('div.wrap-instructor-info').length==0){
return;
}
$('div.xblock-student_view-video').each(function(key, vblock_e){
var vblock = $(vblock_e);
var vuid = vblock.data('usage-id').split('@');
var vid;
if (vuid.length==1){
vuid = vblock.data('usage-id').split(';_')
vid = vuid[5];
}else{
vid = vuid[2];
}
var mfnpre = vid.split("_video",1)[0];
var mfnid = mfnpre; // no periods
mfnpre = mfnpre.replace('8_370', '8.370'); // periods in gh filename
var lb = String.fromCharCode(60);
var rb = String.fromCharCode(62);
var mke = function(x){ return lb + x + rb; }
var ftid = "fix_transcript_" + mfnid;
if (!$('#' + ftid).length){
var html = lb + "span id='" + ftid + "' style='float:right'" + rb + lb + "a href='#'" + rb;
html += "contribute transcript fix" + mke("/a") + mke("/span");
console.log("html = ", html);
vblock.after(html)
}
$('#' + ftid).click(function(){
var cst = $('ol.subtitles').find('li.current');
var cindex = Number(cst.data('index'));
var gurl;
if (mfnpre.endsWith('_cq_sol')){
gurl = "https://github.com/mitocw/content-mit-8370x-cq-sol-subtitles/blob/master/";
}else{
gurl = "https://github.com/mitocw/content-mit-8370x-subtitles/blob/master/";
}
gurl += mfnpre + ".txt#L" + String(cindex + 10 + 1);
console.log("going to ", gurl);
window.open(gurl, "MITx 8.370x subtitle source");
});
});
}
try{
add_fix_transcript();
}
catch(err){
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<h2 class="hd hd-2 unit-title">Density matrices and unravelings</h2>
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Density matrices and unravelings
</h3>
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<p>
This problem illustrates the many possible physical interpretations of a density matrix, based on the infinite number possible <em>unravelings</em> of density matrices into different possible statistical mixtures of pure states. </p>
<p>
Consider the bipartite quantum state </p>
<table id="a0000000016" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]|\psi _{AB}\rangle = \sqrt{\frac{3}{4}} |00\rangle + \sqrt{\frac{1}{4}} |11\rangle[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.11)</td>
</tr>
</table>
<p>
jointly held by Alice and Bob (each has one qubit). Suppose Bob measures his qubit in the computational basis. We may depict this with the quantum circuit: </p>
<p>
<center>
<img src="/assets/courseware/v1/884624d0e94ac1af1cf3db8c44449694/asset-v1:MITx+8.370.3x+1T2018+type@asset+block/images_denmat-measure-basis1.png" width="420"/>
</center>
</p>
<p>
What state does Alice have? Answer this by writing down, at first, Alice's state [mathjaxinline]|\psi '_{A,k}\rangle[/mathjaxinline] (after Bob's measurement) conditioned on Bob obtaining result [mathjaxinline]k[/mathjaxinline] (either [mathjaxinline]k=0[/mathjaxinline] or [mathjaxinline]k=1[/mathjaxinline]) from his measurement; include the probability of this result occuring. </p>
<p>
Be sure that [mathjaxinline]|\psi '_{A,k}\rangle[/mathjaxinline] is properly normalized. </p>
<p>
Recall that states should be entered using &#8220;ket" notation, e.g. [mathjaxinline]|0\rangle[/mathjaxinline] is [mathjaxinline]{\tt |0\rangle }[/mathjaxinline]. </p>
<ul class="itemize">
<li>
<p>
<p style="display:inline">Alice's state [mathjaxinline]|\psi '_{A,0}\rangle[/mathjaxinline] when Bob measures [mathjaxinline]k=0[/mathjaxinline]:</p>
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<p style="display:inline">Probability of Bob measuring [mathjaxinline]k=0[/mathjaxinline]:</p>
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<p style="display:inline">Alice's state [mathjaxinline]|\psi '_{A,1}\rangle[/mathjaxinline] when Bob measures [mathjaxinline]k=1[/mathjaxinline]:</p>
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<p>
<p style="display:inline">Probability of Bob measuring [mathjaxinline]k=1[/mathjaxinline]:</p>
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<h2 class="hd hd-2 unit-title">Density matrices and unravelings II</h2>
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Density matrices and unravelings II
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<p>
Now suppose that Bob did the measurement of his qubit in a different basis, say by first applying a Hadamard gate then measuring in the computational basis. We may depict this with the quantum circuit (recall [mathjaxinline]|\psi _{AB}\rangle = \sqrt{\frac{3}{4}} |00\rangle + \sqrt{\frac{1}{4}} |11\rangle[/mathjaxinline]): </p>
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<img src="/assets/courseware/v1/383828b46ac92af8574748f0252be53b/asset-v1:MITx+8.370.3x+1T2018+type@asset+block/images_denmat-measure-basis2.png" width="420"/>
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<p>
What state does Alice have? Again, answer this by writing down Alice's state [mathjaxinline]|\psi "_{A,k}\rangle[/mathjaxinline] (after Bob's measurement) conditioned on Bob obtaining result [mathjaxinline]k[/mathjaxinline] (either [mathjaxinline]k=0[/mathjaxinline] or [mathjaxinline]k=1[/mathjaxinline]) from his measurement; include the probability of this result occuring. Be sure that [mathjaxinline]|\psi "_{A,k}\rangle[/mathjaxinline] is properly normalized: </p>
<ul class="itemize">
<li>
<p>
<p style="display:inline">Alice's state [mathjaxinline]|\psi "_{A,0}\rangle[/mathjaxinline] when Bob measures [mathjaxinline]k=0[/mathjaxinline]:</p>
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<p>
<p style="display:inline">Probability of Bob measuring [mathjaxinline]k=0[/mathjaxinline]:</p>
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<h2 class="hd hd-2 unit-title">Density matrices and unravelings III</h2>
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Density matrices and unravelings III
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<p>
The two sets of states [mathjaxinline]\{ |\psi '_{A,k}\rangle \}[/mathjaxinline] and [mathjaxinline]\{ |\psi "_{A,k}\rangle \}[/mathjaxinline] look quite different, somehow implying that the state Alice holds after Bob's measurement is different, and depends on what Bob's measurement result is, and also how Bob chooses to measure his qubit. </p>
<p>
However, it is conceptually problematic to think that Alice's state depends in any way on Bob's measurement. What if Bob chose how to do his measurement long after Alice inquires about the state of her qubit? What if Alice and Bob were far apart &#8211; so fat away from each other that they could each do operations on their own qubits faster than information (eg carried at the speed of light) could travel between them? It should not be possible for Bob to be able to signal Alice faster than the speed of light! </p>
<p>
This apparant paradox can be manifestly resolved (or more accurtately, perhaps, <em>avoided</em>) by using a better mathematical tool to represent the <em>statistical mixture</em> which describes each of the ensemble of states Alice may hold, in the two scenarios above. For an ensemble of states [mathjaxinline]|\psi _ k\rangle[/mathjaxinline] which occur with probability [mathjaxinline]p_ k[/mathjaxinline], the density matrix is </p>
<table id="a0000000017" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\rho = \sum p_ k |\psi _ k\rangle \langle \psi _ k|[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.12)</td>
</tr>
</table>
<p>
where [mathjaxinline]|\psi _ k\rangle \langle \psi _ k|[/mathjaxinline] is an <a href="http://en.wikipedia.org/wiki/Outer_product" target="_blank"><em>outer product</em></a> of [mathjaxinline]|\psi _ k\rangle[/mathjaxinline] and [mathjaxinline]\langle \psi _ k|[/mathjaxinline]. For example, </p>
<table id="a0000000018" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]|0\rangle \langle 0| = \left[ \begin{array}{cc}{1}&amp; {0}\\ {0}&amp; {0}\end{array}\right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.13)</td>
</tr>
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<p>
and </p>
<table id="a0000000019" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]|0\rangle \langle 1| = \left[ \begin{array}{cc}{0}&amp; {1}\\ {0}&amp; {0}\end{array}\right][/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.14)</td>
</tr>
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<p>
Think of the column labels as the inputs, and the row labels as the outputs. </p>
<p>
Give the density matrices for the ensembles resulting in the two scenarios above. Recall that matrices are specified using an input which is a list of row vectors, eg [mathjaxinline]|0\rangle \langle 1|[/mathjaxinline] is <tt class="tt">[[0,1],[0,0]]</tt>: </p>
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<p style="display:inline">Density matrix [mathjaxinline]\rho '[/mathjaxinline] for Alice's state [mathjaxinline]|\psi '_{A,k}\rangle[/mathjaxinline]:</p>
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<span><a href="/asset-v1:MITx+8.370.3x+1T2018+type@asset+block/NONE" id="dummy_course_static_link" style="display:none"/><a href="/courses/course-v1:MITx+8.370.3x+1T2018/jump_to_id/NONE" id="dummy_jump_link" style="display:none"/><script type="text/javascript">
var add_site_search = function(){
course_static_url = $('#dummy_course_static_link').attr('href').replace('/NONE', '');
jump_to_url = $('#dummy_jump_link').attr('href').replace('/NONE', '');
if (typeof String.prototype.startsWith != 'function') {
// see below for better implementation!
String.prototype.startsWith = function (str){
return this.indexOf(str) === 0;
};
}
if(typeof(String.prototype.trim) === "undefined")
{
String.prototype.trim = function()
{
return String(this).replace(/^\s+|\s+$/g, '');
};
}
var lb = String.fromCharCode(60);
var rb = String.fromCharCode(62);
var amp = String.fromCharCode(38);
var rlb = rb + lb;
var mke = function(x){ return lb + x + rb; }
var search_module_url = "";
var get_search_module_ficus = function(){
var cid = $('div.xblock').data('course-id');
if (cid){
console.log("cid = ", cid);
// search_module_url = "/courses/course-v1:MITx+8.370.3x+1T2018/" + cid + "/courseware/welcome/Search_this_course/";
search_module_url = "/courses/course-v1:MITx+8.370.3x+1T2018/courseware/welcome/Search_this_course/"; // automatically rewritten
console.log("3. search_module_url = ", search_module_url);
return;
}
var course_root_link = $('span.nav-item-course').find('a').attr('href');
if (course_root_link){
console.log("course_root_link = ", course_root_link);
search_module_url = course_root_link.replace("course/", "courseware/welcome/Search_this_course/");
console.log("2. search_module_url = ", search_module_url);
return
}
console.log("cannot determine search module url");
}
var get_search_module = function(){
// find search this module link
if (!($('div.course-index').length)){
return get_search_module_ficus();
}
$('div.course-index').find('nav').find('a').each(function(){
if ($(this).text().trim().startsWith("Search this course")){
search_module_url = $(this).attr('href');
console.log("search_module_url = ", search_module_url);
}
});
}
var go_to_search = function(){
get_search_module();
var sterm = $('#site-search-box').val();
// new_url = jump_to_url + "/Search_this_module/?q=" + sterm;
new_url = search_module_url + "?q=" + sterm;
console.log("sterm = ", sterm, " ; going to ", new_url);
window.location.href = new_url;
}
if (!$('#site-search-box').length){
$("nav.courseware").find("ol").append(lb + "section style='float:right'" + rlb + "input size='20'"
+ " id='site-search-box'"
+ rlb + "img src='" + course_static_url
+ "/images_search_glass.png'/" + rlb + "/input" + rlb + "/section" + rb);
}
$("#site-search-box").keypress(function(event) {
if (event.which == 13) {
event.preventDefault();
go_to_search();
}
});
// $('#site-search-box').bind("enterKey", go_to_search);
var get = function(x){
return eval(x);
}
return {'course_static_url': course_static_url,
'jump_to_url': jump_to_url,
'go_to_search': go_to_search,
'get_search_module': get_search_module,
'get_search_module_ficus': get_search_module_ficus,
'get': get,
}
}
var the_site_search = add_site_search();
var add_fix_transcript = function(){
if ($('div.wrap-instructor-info').length==0){
return;
}
$('div.xblock-student_view-video').each(function(key, vblock_e){
var vblock = $(vblock_e);
var vuid = vblock.data('usage-id').split('@');
var vid;
if (vuid.length==1){
vuid = vblock.data('usage-id').split(';_')
vid = vuid[5];
}else{
vid = vuid[2];
}
var mfnpre = vid.split("_video",1)[0];
var mfnid = mfnpre; // no periods
mfnpre = mfnpre.replace('8_370', '8.370'); // periods in gh filename
var lb = String.fromCharCode(60);
var rb = String.fromCharCode(62);
var mke = function(x){ return lb + x + rb; }
var ftid = "fix_transcript_" + mfnid;
if (!$('#' + ftid).length){
var html = lb + "span id='" + ftid + "' style='float:right'" + rb + lb + "a href='#'" + rb;
html += "contribute transcript fix" + mke("/a") + mke("/span");
console.log("html = ", html);
vblock.after(html)
}
$('#' + ftid).click(function(){
var cst = $('ol.subtitles').find('li.current');
var cindex = Number(cst.data('index'));
var gurl;
if (mfnpre.endsWith('_cq_sol')){
gurl = "https://github.com/mitocw/content-mit-8370x-cq-sol-subtitles/blob/master/";
}else{
gurl = "https://github.com/mitocw/content-mit-8370x-subtitles/blob/master/";
}
gurl += mfnpre + ".txt#L" + String(cindex + 10 + 1);
console.log("going to ", gurl);
window.open(gurl, "MITx 8.370x subtitle source");
});
});
}
try{
add_fix_transcript();
}
catch(err){
console.log(err);
}
try{
var rb = String.fromCharCode(62);
setTimeout(function(){ $('.math' + rb + 'span').css("border-left-color","transparent"); }, 3000);
setTimeout(function(){ $('.math' + rb + 'span').css("border-left-color","transparent"); }, 8000);
}
catch(err){
console.log(err);
}
</script></span>
</div>
</div>
</div>
</div>
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