If there is:  $$ U = \sum_{x,y \in \mathbb{Z}_2} |x \dot{+} y\rangle |y\rangle \langle x| \langle y| $$ acting over $|\psi\rangle$ which can be $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, does it act on the first qubit or on both? Also if I want to take the partial trace: $$ Tr_{1}[U(|\psi\rangle\langle\psi|U^{\dagger}] $$ would it be correct to do: $$ Tr_{1}[U(|\psi\rangle\langle\psi|U^{\dagger}] $$ $$ = \sum_{ n,u,w,x,y,z \in \mathbb{Z}_2} \langle n| x \dot{+} y\rangle \langle x | \psi\rangle\langle\psi|w\rangle\langle w\dot{+} z|n\rangle  |x \dot{+} y\rangle \langle x | \psi\rangle\langle\psi|w\rangle\langle w\dot{+} z| $$
113256
I am doing my bachelors dissertation based on an article by David Deutsch. He defines the action of a quantum gate as: $$ U = \sum_{x, y \in \mathcal{Z}_{2}} |x \dot{+}y\rangle|y\rangle\langle x|\langle y|  $$ where $\dot{+}$ is the OR operator from boolean algebra. Then, according to me: He then takes the partial trace:  $$ \text{Tr}_{2}[U(|\psi\rangle\langle\psi|\otimes\hat{\rho})U^{\dagger}] $$  and it is equal to: $$ \frac{1}{2}\hat{I} + \text{Re}[(\langle 0|\psi\rangle\langle\psi|1\rangle)(|0\rangle\langle 1|+|1\rangle\langle 0|)] $$  EDIT: where does the $\frac{1}{2}\hat{I}$ comes from?
91797
Consider the Hilbert space  $$\ell^2=\{(x_1,x_2,...,x_n),x_i\in\mathbb C\text{ for all }i\text{ and }\sum_{i=1}^\infty |x_i|^2&lt;\infty\}$$ with the inner product $$\langle(x_1,x_2,\dots,x_n)(y_1,y_2,\dots,y_n)\rangle = \sum_{i=1}^\infty x_i \overline y_i.$$ Ddefine $T:\ell^2 \to \ell^2$ by  $$T((x_1,x_2,...,x_n))=(x_1,\frac{x_2}2,\frac{x_3}3,...).$$ Then $T$ is a)neither self-adjoint or unitary b)both self-adjoint and unitary c)unitary but not adjoint d)self-adjoint but unitary  $$\langle Tx,y\rangle=\langle T((x_1,x_2,...,x_n)),(y_1,y_2,...,y_n)\rangle= \langle (x_1,x_2/2,x_3/3,...),(y_1,y_2,..,y_n)\rangle= \sum (x_i\overline y_i/i)= \sum x_i(\overline y_i/i)= \langle (x_1,x_2,...,x_n),(y_i,y_2/2,y_3/3,..y_n/n)\rangle=\langle x,Ty\rangle$$ so T is self-adjoint
119474
Let $V$ be an inner-product space, finite-dimensional over $\mathbb{C}$. Let the operator $T:V\to V$ satisfy $$T^2 = \frac{1}{2}(T+T^*)\,.$$  I'd like to prove that $T$ is normal, i.e., $T^*T = TT^*$, and that $\,T^2 - T = 0\,$ holds.  So I've tried the obvious: $$ \langle T(v), T^*(v) \rangle = \langle T^2(v), v \rangle = \frac{1}{2} \langle (T+T^*)(v), v \rangle = \frac{1}{2} \left(\langle T(v),v \rangle + \langle T^*(v), v\rangle \right) $$ I don't see we can infer something smart from those equalities. Any ideas?
8279
If the expectation value of operator $A$ in any state is real, then $A$ is Hermitian.   there is an uncompleted proof:  $$ \int(c_1\psi_1+c_2\psi_2)^* A (c_1\psi_1+c_2\psi_2)dx$$ $$=|c_1|^2\int\psi_1^* A \psi_1dx + |c_2|^2\int\psi_2^* A \psi_2dx +c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx $$  My teacher said that because  $$c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx$$  is real, then we can conclude that $$\int\psi_1^* A \psi_2dx = (\int\psi_2^* A \psi_1dx)^*$$  but I don't understand this last step, can someone explain it, or give another proof?
94685
I would like to know how to design a quantum circuit that given a Hermitian matrix $\hat{H}$ and time $t$, maps $|\psi\rangle$ to $e^{i\hat{H}t} |\psi\rangle$. Thank you for your answer.
43436
Apparently the (d-dimensional) maximally entangled state, $|E \rangle = \sum_{i} |ii\rangle /\sqrt{d}$ is invariant under operations of the form $U \otimes U^\dagger$. I want to prove this result, which amounts to showing that  $$\left(\sum_i U | i \rangle \otimes V |i \rangle = \sum_i | ii \rangle \right)\Rightarrow (V = U^\dagger )$$  I have no idea how to even start. I suppose it's some simple linear algebra result, but I don't see it. A hint would be appreciated.
112924
I'm trying to implement the main algorithm described in the . In order to do this, I have to create a quantum state $|x\rangle$ corresponding to a real vector stored in QRAM. I used an algorithm described more in detail in another paper, in which one can read:     For the runtime, there are $2^k$ rotations executed at the $k$-th level of the tree, apart from the last level where there are none. For a given level these rotations can be executed in parallel as they are all controlled operations on the same qubit, conditioned on different bit-string values of a shared register. To see this, let $U_x$ be a single qubit rotation conditioned on a bitstring $x \in \{0, 1\}^k$. Then the unitary $\bigoplus\limits_{x\in\{0,1\}^k} U_x$ applied to $|y\rangle\otimes|q\rangle$ achieves the desired parallel operation on the single qubit $|q\rangle$, where $|y\rangle = \sum\limits_{x\in\{0,1\}^k}\alpha_x\,|x\rangle$ is some superposition over bitstrings.   I'm not sure that I understand this notion of parallel execution. In the algorithm, at a given step $k$, one performs on one qubit $2^k$ controlled rotation conditionned on every qubit amongst the first $k-1$. Hence, we can summarize this as one big unitary matrix that acts on the first $k$ qubits, which will look-like $\begin{bmatrix}R_{\theta_1}&amp;0&amp;0\\\vdots&amp;\ddots&amp;\vdots\\0 &amp; 0 &amp; R_{\theta_{2^k}}\end{bmatrix}$, where for a given angle $\theta$, $R_\theta$ is the rotation of angle $\theta$ around the $Y$-axis of the Bloch sphere.  If we assume that this big operator is applied in $O(1)$, then I understand the notion of parallelism: you  don't have to successively apply the $2^k$ controlled gates, so you have a gain in complexity. But then, there is one problem I can't cope with. When you execute an algorithm on a quantum computer, I read that you are only allowed to use a specific set of gates. Hence, a compiler takes your code and transforms it to a compatible code. But am I sure that the compiler will always (or even sometimes) be able to decompose this big gate into a succession of smaller, allowed gates, without breaking the complexity?
104649
Let $H$ be a Hilbert space over $\mathbb{C}$. Let $\phi_{j} \in H$ and let $c_j$ be scalars. Prove that there exists a representation $$f=\sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j}, \forall f\in H$$ if and only if $$\|f\|^2=\sum_{j}c_{j}|\langle f,\phi_{j}\rangle|^2, \forall f\in H$$  My queries: For $\implies$: $\|f\|^2= \langle f,f \rangle = \sum_{i,j} c_{i} \overline{c_{j}} \langle f,\phi_{i} \rangle \overline{\langle f,\phi_{j} \rangle}\langle \phi_{i}, \phi_{j}\rangle$. It can not be simplified further due to $\phi_{j}$ are may not be orthogonal. Even if one assumes orthogonality it would become $\|f\|^2=\sum_{j}|c_{j}|^2|\langle f,\phi_{j}\rangle|^2\|\phi_{j}\|^2$ which is still not the desired form. Is there any way to apply the fact that the representation for $f$ is valid for  "all" $f$ in some way to deduce the desired equality?  For sufficiency I intend to apply the polarization identity to $\langle f,\phi_{j}\rangle$ then substitute it into the desired formula for $f$, but the polarization identity implied $\langle f,\phi_{j} \rangle =0$ which is not true in general, is there any point I have missed?
19516
For a real (or complex) inner product space $V$, the inner product can be expressed in terms of the norm as either $$ \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2) $$ or $$ \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2) $$ respectively.  These identities are not hard to verify directly once they are given. I would like to know, is there a systematic way one would derive these from scratch without knowing them beforehand, or are both polarization identities just clever observations?
34579
I'm reading Sakurai's Modern QM right now and in the first chapter he states a number of conditions required for a translation operator: unitarity, $$T(\mathrm{d}\mathbf{x})T(\mathrm{d}\mathbf{x^\prime}) = T(\mathrm{d}\mathbf{x}+\mathrm{d}\mathbf{x^\prime}),$$ $$T(-\mathrm{d}\mathbf{x}) = T^{-1}(\mathrm{d}\mathbf{x}),$$ and going to 1 for small translations.   He then makes the ansatz (1.6.20 in the new version): $$ T(\mathrm{d}\mathbf{x}) = 1-i \mathbf{K} \cdot \mathrm{d}\mathbf{x} $$ and demonstrates that this form of $T$ meets the requirements and is generally the correct choice. Here, $\mathbf K =(\mathrm K_x,\mathrm K_y,\mathrm K_z)$ with each component hermitian.  I was hoping somebody could provide some motivation for this particular choice. Is there any reason why it is to be expected or natural? How would you come up with this form of the operator by yourself?
70385
The reasoning my book uses is as follows:   Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.   I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$  is $10$. If someone could please explain this I would greatly appreciate it.   Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.
31063
In my quantum mechanics textbook it says that the relation between the basis $|x\rangle$ and $|p\rangle$ is given by:  $\langle p | x \rangle = \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \, .$  However, i'm not sure how to go about proving this relation. My idea was that the eigenstates of momentum can be written as below:  $\phi(x,p) = \Large \frac{1}{\sqrt{2\pi \hbar}} e^{-ipx/\hbar},$  which form an orthogonal basis.  But, we can expand any state, $|\text{state}\rangle$, in terms of the othorgonal basis right? So is my notation below correct for the expansion of the state $| x \rangle$ ?  $|x\rangle = \int_{-\infty}^{+\infty} \phi(p', x) |p'\rangle dp'$  If it were correct then the following would also be correct?  Using the fact that $\langle p | p' \rangle = \delta(p - p')$, $\langle p | x \rangle$ can be computed:  \begin{align} \langle p | x \rangle =&amp; \int_{-\infty}^{+\infty} \phi(p', x) \langle p | p' \rangle dp' \\ =&amp; \int_{-\infty}^{+\infty} \phi(p', x) \delta(p - p') dp' \\ =&amp; \phi(p, x) \\ =&amp; \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \end{align} as required.  Does the logic work here?
86751
I have the state vector $|p\rangle$ made up of 4 qubits. Say system A is made up of the first and second qubits while system B is made up of qubits 3 and 4. I want to find the reduced density matrix of system A.   I know I could separately extract qubits 1,2 and 3,4 into their own state vectors then find their density matrices and compute the reduced density matrix for system A.  I want to figure out how to do this without having to extract and separate the systems. First I would find the density matrix of $|p\rangle$ and then do a partial trace with respect to system B. I am not sure how to do the partial trace of system B since the system contains 2 qubits.  Can anyone help me figure this out? I am using Python and NumPy for reference.
109577
Basically what I'm trying to show is $\forall h_1, \ h_2 \in \mathscr{H}$ and $U: \mathscr{H} \rightarrow \mathscr{K}$ then $\langle Uh_1, \ Uh_2\rangle_\mathscr{K} = \langle h_1, \ h_2 \rangle_\mathscr{H}$  (So your standard definition). Now I figure the easiest way to do so is by polarization.  Since we have an isometry then   $\begin{eqnarray*} \langle h_1, \ h_2 \rangle_\mathscr{H} &amp;=&amp; \frac{1}{4} \bigg( \|h_1 + h_2\|_\mathscr{H}- \|h_1-h_2\|_\mathscr{H} +i\|h_1+ih_2\|_\mathscr{H} - i\|h_1 - ih_2\|_\mathscr{H} \bigg) \\&amp;=&amp;\frac{1}{4} \bigg( \|Uh_1 + Uh_2\|_\mathscr{K}- \|Uh_1-Uh_2\|_\mathscr{K} +i\|Uh_1+iUh_2\|_\mathscr{K} - i\|Uh_1 - Uih_2\|_\mathscr{K} \bigg) \\&amp;=&amp; \langle Uh_1, \ Uh_2\rangle_\mathscr{K} \end{eqnarray*} $  But what I'm wondering is where surjectivity comes into play?
2190
For a system has both continuous and discrete spectrum, is it possible that a physical states is something like:  $$\psi(x)=\sum_{n=n_1}^{n=n2}c_n\psi_n(x)+\int_{E_0-\Delta_E}^{E_0+\Delta_E}\psi_E(x)\mathrm{d}E$$  I am asking this because of some comments under my  to this Phys.SE question:
58823
Isn't it obvious for $|a|=|a^{-1}|$, since $\langle a\rangle = \langle a^{-1} \rangle$?  For $|ab|=|ba|$, I think we should go like this: $e=(ab)^n\Rightarrow e=(ab)(ab)^{n-1}\Rightarrow (ab)^{-1}=(ab)^{n-1}\Rightarrow b^{-1}a^{-1}=(ab)^{n-1}\Rightarrow a^{-1}=b(ab)^{n-1}$ But I get stuck here, since I don't know whether to get from here to $e=(ba)^n$.  For $|a|=|cac^{-1}|$, do I have to show that $\langle a\rangle = \langle cac^{-1}\rangle$?
61362
This problem is from Kunze Hoffman book. I think I go in the right direction to solve this but I miss some point to finish it. Can anyone help me?          Suppose $U$ is a self-adjoint unitary linear operator on a finite-dimensional inner product space $V$. Prove that we can find a subspace $W$ such that, if we express every $\alpha$ in $V\,$ in the form $\alpha = \beta + \gamma$, here $\beta$ in $W$ and $\gamma$ in $W^{\bot}$, then we have $U(\alpha) = \beta - \gamma$      If we denote $W = range(U+I)$, $W' = range(U - I)$,then must have $W \subset null(U-I)$ and $W \subset W^{'\bot}$. We can finish the proof if we can prove that $W' = null(U+I)$ or $W' = W^{\bot}$. But I can't prove this...  Thanks