I've been struggling for a day with the following problem on Quantum coherent states, so I was wondering if you could tell me if I'm going in the right direction (I've checked the books of Sakurai and Weinberg but can't seem to find a satisfactory answer)     Suppose a Schrödinger cat state given by $|\alpha_+\rangle :=N(|\alpha\rangle+|-\alpha\rangle)$, where $|\alpha\rangle$ are the coherent states of the harmonic oscillator and $N$ is a constant of proportionality. Verify the relationship of Robertson-Schrödinger for $X$ and $P$ operators.   The solution is simple as we only have to calculate $\langle x \rangle$, $\langle p \rangle$, $\langle x^2 \rangle$, $\langle p^2 \rangle$, and $rel(x,p)$.   From the definition of coherent states, $a|\alpha\rangle=\alpha|\alpha\rangle$ and with $x=x_0(a+a^\dagger)$, I began by calculating $\langle x \rangle$ in the following way: \begin{align} \langle x \rangle  &amp; = \langle\alpha_+|x|\alpha_+\rangle \\ &amp; = (\langle\alpha|+\langle{-\alpha}|)x(|\alpha\rangle+|{-\alpha}\rangle) \\ \langle x \rangle &amp; =\langle\alpha|x|\alpha\rangle+\langle\alpha|x| {-\alpha}\rangle+\langle-\alpha|x|\alpha\rangle+\langle-\alpha|x|-\alpha\rangle \\ &amp; = \frac{x_0}{\sqrt{2}}(\langle\alpha|a+a^\dagger|\alpha\rangle+\langle\alpha|a+a^\dagger|-\alpha\rangle+\langle-\alpha|a+a^\dagger|\alpha\rangle+\langle-\alpha|a+a^\dagger|-\alpha\rangle) \end{align}  From the definition $a|\alpha\rangle=\alpha|\alpha\rangle$, we deduce $\langle\alpha| a^\dagger=\langle\alpha| \alpha^{*}$.   Moreover (I'm not entirely sure of this step), by replacing $\alpha'\rightarrow-\alpha$ we deduce that $a|{-\alpha}\rangle=-\alpha|{-\alpha}\rangle$ and $\langle{-\alpha}| a^\dagger=\langle-\alpha| {-\alpha^{*}}$.  Applying the operators to the above expression, we finally get:  $$\langle x \rangle=x_0 ((a+a^{*})+(-a+a^{*})+(a-a^{*})+(-a-a^{*}))=0$$  In an analogous manner, $\langle p \rangle=0$.   So far, is this correct? My main doubt comes with the eigenvalues of the coherent states for $|{-\alpha}\rangle$.
121597
I've a brief question about coherent states in quantum mechanics.  As everyone knows, a coherent state is just the proper state of the anhilitation operator $a$, thus they're defined with the eigenvalue equation $a|\alpha\rangle=\alpha|\alpha\rangle$ (or inverting, $\langle\alpha|a^\dagger=\langle\alpha|a^{*}$).  However, sometimes I've seen coherent states of the form $|{-\alpha}\rangle$, and I was wondering about their physical meaning and if they satisfy the same eigenvalue equation as before, i.e., $a|{-\alpha}\rangle=-\alpha|{-\alpha}\rangle$ (and thus, $\langle{-\alpha}|a^\dagger=\langle{-\alpha}|{-\alpha}^{*}$).  Precisely, we can find these states, for example, in the case of the Schrödinger cat state.
7511
I confused about how to calculate the probabilities and getting a certain result of measuring Bell's states with Pauli matrices as the operator. When you measure something, the state involved would be projected onto an eigenstate of the observable. given $|\psi\rangle = \frac{1}{\sqrt2} (|01⟩ + |10⟩)$ as the state and $\sigma_x = \left[\begin{matrix}0&amp;1\\1&amp;0\\\end{matrix}\right]$ as the observable. if the probability is 1/2, how to calculate them actually? What is the state after measurement?
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?
82473
Let $$\langle x'|p'\rangle = N \exp(\frac{ip'x'}{\hbar})$$ be the overlap between position and momentum space, where $N$ is a normalization constant to be determined.  We can then compute $N$ by  $$ \langle x'|x''\rangle = \int \mathrm{d}p'\langle x'|p'\rangle \langle p'|x''\rangle\\ \rightarrow \delta(x'-x'')=|N|^2 \int \mathrm{d}p'\exp(\frac{ip'(x'-x'')}{\hbar}) \\ =2\pi\hbar |N|^2 \delta(x'-x'') $$  see Sakurai - "Modern Quantum Mechanics" 2nd, Pearson, p. 54.  How was the factor $2\pi\hbar$ obtained? The intermediate step should look something like this: $$ \rightarrow  \delta(x'-x'')=|N|^2 \int \mathrm{d}p'\exp(\frac{ip'(x'-x'')}{\hbar}) \\ =|N|^2 \int \mathrm{d}p'\delta(x'-x'') $$  But all I can think of is that one can say $\int \mathrm{d}p'=\hbar k$ where $k$ is the wavenumber.   What am I missing?
63304
Consider the time-dependent  (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm  dt^{\prime}\right]\Psi (0), \qquad t&gt;0, $$ where $\hat{\mathrm T}$ is the time-ordering operator.  It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake?
29657
I am reading Introduction to Quantum Computing by Kaye, Laflamme, and Mosca. As an exercise, they write  "Prove that if the operators $P_{i}$ satisfy $P_{i}^{*}=P_{i}$ and $P_{i}^{2}=P_{i}$ , then $P_{i}P_{j}=0$ for $i\ne j$.''   In the context of this problem, it has been assumed that $I=\sum_{i=1}^{n} P_{i}$, where I suppose that $n$ could be infinite. I have shown that this is true in the trivial case $n=2$, but the general case has been eluding me. How should I attack this?
86806
I have the following problem that I was unable to solve for class, but I had a couple first steps that I started with that I am unable to finish.  I know I can't get this since it's already been turned in, but I would like to see if this was even a viable option for doing this proof in the first place.  ``Given an orthogonal set of states, $\{\phi_{n}\}$, and a Hamiltonian, $\hat{H}$, show that the partition function, $Q_{\beta}$, satisfies the following $$ Q_\beta \geq \sum_{n}\exp\{-\beta \langle \phi_n|\hat{H}|\phi_n\rangle \}$$ with equality holding when the $\phi_n$ states are eigenstates of the Hamiltonian.''  I started by dropping in the identity in the exponential (as eigenstates of the Hamiltonian) $$ \sum_n \exp\{ -\beta\sum_k \langle \phi_n|\psi_k\rangle\langle \psi_k |\hat{H}|\phi_n\rangle\}=\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\}$$  Then I am left with showing that $$\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\} \leq \sum_k \exp\{ -\beta E_k\}$$ with equality showing up again the same way, with a Kronecker delta $\delta_{nk}$ collapsing the sum in the exponential.  I realize I didn't make it very far, so this might not be the best way to show this, but it seems manifestly true just by looking at it, but I can't actually show it.  Does anyone have any hints about how to continue with this?
78855
Can someone please explain me what is a ?  Also, how does putting $$X_a(\psi) ~=~ x_a\star\psi$$ realise $$[X_a,X_b]=i\theta_{ab}{\bf 1}?$$   Ref:
68342
Assuming we have two sets of $n$ qubits. The first set of $n$ qubits is in state $|a\rangle$ and second set in $|b\rangle$. Is there a fixed procedure that generates a superposed state of the two $|a\rangle + |b\rangle$ ?
117672
There seems to be something very strange about the relationship between quantum field theory and quantum mechanics.  It is bothering me; perhaps somebody can help.  I'll consider a free Klein-Gordon field.  In standard treatments (e.g. Peskin &amp; Schroeder and Schwartz) the one-particle momentum eigenstates $| \vec{k} \rangle$ are normalized so that  $$ \langle \vec{p} | \vec{k} \rangle = 2 \omega_{\vec{p}} (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k}), \qquad 1 = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 \omega_{\vec{p}}}| \vec{p} \rangle \langle \vec{p} |. $$ Now, assuming $\langle \vec{x}' | \vec{x} \rangle = \delta^{(3)}(x-x')$ as usual, it follows that $$ \langle \vec{x} | \vec{p} \rangle = \sqrt{2 \omega_{\vec{p}}} e^{i \vec{p} \cdot \vec{x}}. $$ Now, one can compute (here in the Schrodinger picture; see Schwartz 2.76 or P&amp;S 2.42) that $$ \langle 0 | \phi(\vec{x}) | \vec{p} \rangle = e^{i \vec{p} \cdot \vec{x}}. $$ This is supposed to mean that $\phi$ creates a particle localized at position $\vec{x}$.  P&amp;S are a little cautious about the details, but Schwartz claims that the calculation implies  $$ \phi(\vec{x}) |0 \rangle = | \vec{x} \rangle. $$ But this is false because $\langle \vec{x} | \vec{p} \rangle \neq e^{i \vec{p} \cdot \vec{x}}$ with the normalization conventions used.  I suppose it could be true with some weird normalization of $| \vec{x} \rangle$, but I can't see what that might be (and at the very least this is not spelled out on the text).    Even if this works out, it seems extremely strange for there to be a relative normalization between the one-particle states of field theory and the states of one-particle relativistic quantum mechanics.  One ought to be able to redo the correspondence to make the normalization work out, but I don't see how.  (Note that the normalizations can be easily made to agree in the non-relativistic limit $\omega \approx m$, but that's besides the point.  Even if fully relativistic quantum mechanics is inconsistent [as some texts claim without reference], at the very least the perturbative corrections for $v \ll 1$ should be recoverable from field theory.)  [Edit: This seems to go beyond normalization.  We can get a feel for what kind of state $\phi(\vec{x})|0\rangle$ is by computing its wavefunction as a function of $\vec{x}'$, $$ \langle \vec{x}' | \phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec{p}}}} e^{i\vec{p} \cdot (\vec{x}-\vec{x}')}. $$ This wavefunction is peaked (I think divergent) at $\vec{x}'=\vec{x}$, so in some sense the particle is centered at $\vec{x}$, but it seems to be quite a stretch to say that it is at $\vec{x}$ (as the books do).  I would go so far as to say the claim is incorrect, since in quantum mechanics saying the particle is at particular position means the wave function is a delta-function there. I guess the same language is used in the Heisenberg picture, when two-point functions are called amplitudes for particles to propagate from one spacetime point to another.  This similarly seems false by the conventional meaning of amplitude as the overlap between two localized states.  Words of wisdom would be appreciated.]
84124
Suppose there're two systems $A$ and $B$, if we're in a pure state $|\psi\rangle\in\mathbb{H}_a\otimes\mathbb{H}_b$. Let $\hat A$ be an operator acting on $\mathbb{H}_a$ and $|\psi\rangle=\sum_{i,j}\psi_{i,j}|a_i\rangle|b_j\rangle$. The expectation value of A with respect to the state $|\psi\rangle$ can be calculated as $$ \langle A\rangle_{\psi}= Tr_{\mathbb{H}_a\otimes\mathbb{H}_b}(A|\psi\rangle\langle\psi|)=\sum_{i,j}\langle a_i|\langle b_j|A\otimes\mathbb{1}_B|\psi\rangle\langle\psi|a_i\rangle|b_j\rangle=... $$ I'm very confused about where these two equalities coming from? In particular, where does the term $|\psi\rangle\langle\psi|$ in the sum come from? Can I understand its function as generating a $\psi$ matrix with different constants $\psi_{i,j}$?
44655
Reproduced from Exercise 2.2 of Nielsen &amp; Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition):     Suppose $V$ is a vector space with basis vectors $|0\rangle$ and $|1\rangle$, and $A$ is a linear operator from $V$ to $V$ such that $A|0\rangle = |1\rangle$ and $A|1\rangle = |0\rangle$. Give a matrix representation for $A$, with respect to the input basis $|0\rangle, |1\rangle$, and the output basis $|0\rangle, |1\rangle$. Find input and output bases which give rise to a different matrix representation of $A$.   Note: This question is part of a series attempting to provide worked solutions to the exercises provided in the above book.
122916
I've been searching for a quantum algorithm to compute the the inner product between two $n$-qubit quantum states, namely $\langle\phi|\psi\rangle$, which is in general a complex number. One can get $|\langle\phi|\psi\rangle|^2$ through thte SWAP test for multiple qubits, but this is not really the inner product as the information of the real and imaginary parts are lost. I came across this , which claims that the task can be done by simply using CZ gates. I'm not sure if I am mistaken, but it seems to me that such a circuit works for computational basis states, but not for general quantum states. Could somebody help me confirm if this is actually the case? If the above algorithm doesn't work for general quantum states, what algorithm would you suggest?
99922
In Griffith's QM book, he introduces scattering matrices as an end-of-the-chapter Problem 2.52.  For a Dirac-Delta potential $V(x) = \alpha \delta (x - x_0)$, I've derived the scattering matrix and observed that it is unitary $S^{-1} = S^{\dagger}$.  I'm trying to explain why this is intuitively, but I don't really have an intuitive picture of what hermitian conjugation $S^{\dagger}$ is doing here. Thoughts?
117818
Please note that I do not have a background in physics, so if possible please refrain from a bunch of $ |x\rangle $ notations, unless clearly specifying what it symbolically means.  So I have been learning about representation theory lately in particular I have studied square integrable irreducible representations, and I'm interested in the applications of these. I have come to understand that given a square integrable irreducible representation $ U $ of a locally compact group $ G $ on a Hilbert space $ \mathcal{H} $ and an admissible vector $ g \in \mathcal{H} $, then the orbit $ \mathscr{O}_g := \{U(x)g\mid x\in G\} $ is a . Furthermore if the group $ G $ is the (Weyl-)Heisenberg-group, then these  are "classical coherent states"(?).  So I understand from this that coherent states can be described by these collections of vectors/functions in a Hilbert space and sometimes they constitute frames and possible wavelets(?). How exactly is such a collection of vectors understood in the context of coherent states? What does a coherent state describe? and why are they interesting?   If you can refer me to articles or literature explaining these questions in terms understandable by someone who has mostly had basic mechanics then it'd be much appreciated.
5833
I simply have no idea how this result is found $$I=\hbar  \sqrt{\ell(\ell+1)}.$$ The result seems to just be dumped in textbooks rather than explained. I can get the result that $I_z=\hbar m_j$. Please explain as clearly as possible how this is derived.  Also, any pointers towards any books/videos that may help me learn quantum mechanics would be appreciated.
25399
I'm studying for a test in quantum mechanics and I'm having a hard time understanding how to use . There are no examples in my text book, only definitions that I can't understand how to use, so I hope you can help me instead.  So, there's an assignment where you're supposed to calculate first order correction to the ground state using ladder operators. In the right answer they say that  $$E_0^1 = \langle 0|H'|0\rangle$$  where does the zeros come from? I understand that they're coming from the wave function $\psi$. But I don't really understand the theory behind this.  Then, after some calculations it's written in the right answer  $$ \langle 0|aa^\dagger a^\dagger|0\rangle +\langle0|aaa^\dagger a^\dagger|0\rangle =\langle0|aa^\dagger a|1\rangle +\langle0|aaa^\dagger|1\rangle$$  What are the mathematical rules of these ladder operators? How did these zeros become ones?  I would really appreciate if someone could help me understand this a little bit better.