<p>There's a Version 2 with appendix and there's write...</p>

<ol start="3">
<li>Apply the Fourier transform to the register C. Denote the resulting basis
states with |ki, for k = 0, . . . T − 1. Define λ˜
k := 2πk/t0.</li>
</ol>

<p>if This does not answer properly, maybe take a look to this new version of this paper.</p>

CC BY-SA 4.0
2019-04-17T08:55:31.547Z
2019-04-17T08:55:31.547Z
Eduardo Vinicius Galle
<p>I remember struggling with this very same question! Ultimately I concluded that the $C$ is just a notational device (it doesn't represent any mathematical operation), just to indicate that, for a <em>specific</em> $\tau$, the Hamiltonian evolution operator $e^{iA\tau t_0/T}$ is performed on $\left|b\right&gt;$, but only when <em>conditioned</em> on $\left|b\right&gt;$ being tensored with $\left|\tau\right&gt;$. </p>

<p>If you ignore the $C$'s, I think a calculation will show that you just apply the operator $\sum_{\tau=0}^{T-1} \left|\tau\right&gt;\left&lt;\tau\right|\otimes e^{iA\tau t_{o}/T}$ "as-is". I hope this helps.</p>

CC BY-SA 3.0
2012-04-12T10:03:42.133Z
Ah thank you. By the way, I found a free source for the published version of the paper, in which the $C$ disappears. It lives here: http://dspace.mit.edu/handle/1721.1/51753
user14717
CC BY-SA 3.0
2012-04-11T20:51:37.317Z
2012-04-11T20:56:43.520Z
2012-04-11T20:56:43.520Z
Henry Yuen
<p>I am reading Harrow, Hassidim, and Lloyd's paper <a href="http://arxiv.org/pdf/0811.3171v3.pdf">Quantum algorithms for linear systems of equations</a>. On the third page of that paper, they write</p>

<blockquote>
  <p>Next we apply the conditional Hamiltonian evolution $\sum_{\tau=0}^{T-1} \left|\tau\right&gt;\left&lt;\tau\right|^{C}\otimes e^{iA\tau t_{o}/T}$ on $\left|\Psi_{0}\right&gt;^{C}\otimes\left|b\right&gt;\dots$</p>
</blockquote>

<p>For the life of me, I cannot figure out the meaning of the $C$. What is it doing there? How would $\sum_{\tau=0}^{T-1} \left|\tau\right&gt;\left&lt;\tau\right|^{C}\otimes e^{iA\tau t_{o}/T}$ act on (say) $\left|0\right&gt;\otimes \left|0\right&gt;$ or $\left|1\right&gt;\otimes \left|0\right&gt;$? </p>

<quantum-computing><notation>
Notation for a Conditional Hamiltonian Evolution Operator
user14717
# Notation for a Conditional Hamiltonian Evolution Operator

(Asked by: username_0 on 2012-04-11)

I am reading Harrow, Hassidim, and Lloyd's paper [Quantum algorithms for linear systems of equations](http://arxiv.org/pdf/0811.3171v3.pdf). On the third page of that paper, they write

> Next we apply the conditional Hamiltonian evolution $\sum\_{\tau=0}^{T-1} \left|\tau\right>\left<\tau\right|^{C}\otimes e^{iA\tau t\_{o}/T}$ on $\left|\Psi\_{0}\right>^{C}\otimes\left|b\right>\dots$

For the life of me, I cannot figure out the meaning of the $C$. What is it doing there? How would $\sum\_{\tau=0}^{T-1} \left|\tau\right>\left<\tau\right|^{C}\otimes e^{iA\tau t\_{o}/T}$ act on (say) $\left|0\right>\otimes \left|0\right>$ or $\left|1\right>\otimes \left|0\right>$?

**Tags:** quantum-computing, notation

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## 2 answers

### Answer by username_2 on 2012-04-11. Score: 6 | Accepted answer

I remember struggling with this very same question! Ultimately I concluded that the $C$ is just a notational device (it doesn't represent any mathematical operation), just to indicate that, for a *specific* $\tau$, the Hamiltonian evolution operator $e^{iA\tau t\_0/T}$ is performed on $\left|b\right>$, but only when *conditioned* on $\left|b\right>$ being tensored with $\left|\tau\right>$.

If you ignore the $C$'s, I think a calculation will show that you just apply the operator $\sum\_{\tau=0}^{T-1} \left|\tau\right>\left<\tau\right|\otimes e^{iA\tau t\_{o}/T}$ "as-is". I hope this helps.

> **Comments:**
> **username_0**: Ah thank you. By the way, I found a free source for the published version of the paper, in which the $C$ disappears. It lives here: http://dspace.mit.edu/handle/1721.1/51753  

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### Answer by username_1 on 2019-04-17. Score: 0

There's a Version 2 with appendix and there's write...

3. Apply the Fourier transform to the register C. Denote the resulting basis
   states with |ki, for k = 0, . . . T − 1. Define λ˜
   k := 2πk/t0.

if This does not answer properly, maybe take a look to this new version of this paper.


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