My exercise for a quantum optics course tells me to "find a normal ordering" for an Operator $\hat{O}(\hat{a},\hat{a}^\dagger)$, which is given as a (rather complicated) string of $\hat{a}$'s and $\hat{a}^\dagger$'s in random order.   To me, it is not clear, whether the tasks wants me to apply the normal ordering operator $:\ :$, which acting on an operator simply pulls all daggers to the right while neglecting the commutator rule or whether to use the proper commutation to create a normally ordered expression.  But that is not directly the question. When I open a textbook or my lecture note, when it is time to introduce the $P$, $Q$ and $W$ functions, they normally start by looking at some operator $\hat{O}$, then defining $:\hat{O}: = \hat{O}^{(N)}$. For example $:\hat{n}^2:= \left(\hat{a}^\dagger\right)^2\hat{a}^2$. Then an expectation value is calculated:  $$\left\langle:\hat{n}^2:\right\rangle = \int d^2\alpha\ P(\alpha)\left|\alpha\right|^4$$  My problem is, that I don't see why one would use the normal ordering operator, because clearly the normally ordered operator is different from the unordered operator and will have a different expectation value.  And to extend that question: Can I calculate the expectation value for any normally ordered operator using the $P$ function? If I wanted the expectation value for $\hat{n}^2$ instead of $\left\langle:\hat{n}^2:\right\rangle$, would I have to bring the operator into normal order using the commutator relation?
123591
According to the commutation relation of annihilation and creation operators,   $$[a,a^{\dagger}]=1. \tag{1}$$   I would like to calculate the vacuum expectation value of the normal order of this commutator. We claim that since this commutator is just a c-number, then there is no effect from the normal ordering. We therefore get   $$\langle 0|(:[a,a ^\dagger]:)|0 \rangle=1. \tag{2}$$  However, if we expand the commutator first and do normal order later, we will get something like   $$\langle0|(:[a,a^\dagger]:)|0\rangle=\langle0|(:aa^\dagger-a^\dagger a:)|0\rangle=\langle0|(a^\dagger a-a^\dagger a)|0\rangle=0.\tag{3}$$ Which is contradicting with itself.
13717
One of the exercises in Ravi Vakil's algebraic geometry notes, Ex. $5.4.$I(b), is to show that    $$ \operatorname{Spec}\left(k[x_1, \ldots, x_n]/(x_1^2 + \cdots + x_m^2)\right) $$   is normal, where $k$ is any field of $\operatorname{char}(k)\neq 2$, and $n \geq m \geq 3$.    I have absolutely no idea how to get started on this. Is there anyone that could give a hint as to how one would approach this problem?
89166
What are some elegant ways to calculate  $$[\hat{a}^{M},\hat{a}^{\dagger N}]\qquad\text{with} \qquad[\hat{a},\hat{a}^{\dagger}]=1,$$  other than brute force calculation?    (EDIT) I got the same result as Qmechanic. I think Prathyush's suggestion should be equivalent to mine of the correspondence up to a canonical transformation. Here is my calculation:  $\begin{array}{c} \mbox{representation of }\left(\hat{a},\hat{a}^{\dagger}\right)\mbox{ on polynomial space }span\left\{ \frac{x^{n}}{\sqrt{n!}}\right\} _{n\ge0}\\ \hat{a}\left[f\left(x\right)\right]=\frac{d}{dx}f\left(x\right)\;;\;\hat{a}^{\dagger}\left[f\left(x\right)\right]=xf\left(x\right)\;;\;\left[\hat{a},\hat{a}^{\dagger}\right]\left[f\left(x\right)\right]=id\left[f\left(x\right)\right]\\ \left|0\right\rangle \sim 1\;;\;\left|n\right\rangle \sim x^{n}/\sqrt{n!} \end{array}$  $\begin{array}{c} \mbox{calculate the normal ordering }\left[\hat{a}^{M},\hat{a}^{\dagger}{}^{N}\right]\mbox{:}\\ \sim\left[\frac{d^{M}}{dx^{M}},x^{N}\right]=\frac{d^{M}}{dx^{M}}\left(x^{N}\star\right)-x^{N}\frac{d^{M}}{dx^{M}}\left(\star\right)\\ \sim\left\{ \overset{min\left\{ M,N\right\} }{\underset{k=0}{\sum}}\frac{N!}{\left(N-k\right)!}C_{M}^{k}\left(\hat{a}^{\dagger}\right)^{N-k}\left(\hat{a}\right)^{M-k}\right\} -\left(\hat{a}^{\dagger}\right)^{N}\left(\hat{a}\right)^{M}\\ \end{array}$    One comment on 02-12-2012: The representation I was using is actually related to Bergmann representation with the inner product for Hilbert space (polynomials) being:  $$\left\langle f\left(x\right),g\left(x\right)\right\rangle :=\int dxe^{-x^{2}}\overline{f\left(x\right)}g\left(x\right)\,,x\in\mathbb{R}\,,\, f,g\in\mathbb{C}\left[x\right]$$
11306
I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that $$\langle x|\alpha\rangle=f(x)$$ for some known function $f$. ($|\alpha\rangle$ is a Gaussian wave packet.) Sakurai says that this is given by:  $$\langle p\rangle = \int\limits_{-\infty}^{+\infty}\langle\alpha|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|\alpha\rangle dx.$$  I am wondering how we get to this expression. I know that we can express   $$|\alpha\rangle =\int dx|x\rangle\langle x|\alpha\rangle$$  and  $$\langle\alpha|=\int dx\langle\alpha|x\rangle\langle x|,$$   so my thinking is that we have:  $$\langle\alpha|\hat{p}|\alpha\rangle =\iint dx dx'\langle\alpha|x\rangle\langle x|\hat{p}|x'\rangle \langle x'|\alpha\rangle, $$  and if we can 'commute' $|x\rangle$ and $\hat{p}$ this would become: $$\iint dxdx'\langle\alpha |x\rangle\hat{p}\langle x|x'\rangle \langle x'|\alpha\rangle,$$  which is the desired result as $$\langle x|x'\rangle=\delta(x-x').$$ Is this approach valid?  I think my question boils down to: Does the operator $\hat{p}$ act on the basis kets $|x\rangle$ or on their coefficients? In the latter case, if we had some state $|\psi\rangle = |x_0\rangle$ for some position $x_0$, then would we say that for this state $$\langle p\rangle =\langle x_0|\left(-i\hbar\frac{\partial}{\partial x}\right)|x_0\rangle = 0$$ as the single coefficient is $1$ and the derivative of $1$ is $0$?
91320
In Assa Auerbach's book (Ref. 1), he gave an argument saying that in the normal process of path integral, we lose information about ordering of operators by ignoring the discontinuous path.   What did he want to say? I don't think there is any problem related to the ordering of operators.  References:   Assa Auerbach, Interacting Electrons and Quantum Magnetism, p.102, just below eq. (10.6).
113328
This relates a little bit to my previous question (), but it's different in a way and more historical.  One of the postulates of quantum mechanics is:     To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics ().   However, absent in the list of postulates is what these operators are.  So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms?  Specifically, where did  expressions come from?  Maybe the question can better be answered historically.  If I had to guess, I'd say it began with the wavefunction.  Some scientists thought perhaps the squared modulus of the wavefunction corresponded to charge distribution or something, but Born was the one who got it right.  He essentially made a guess that the squared modulus corresponded to the probability of finding a particle at a certain location.  What's funny is that he initially wrote down the wrong expression in his paper (), and his paper even got rejected by the first journal he submitted it to, but the right expression is in a corrective footnote.    But again, Born simply guessed what the squared modulus meant.  There was no procedure; he just thought of an idea and then experiment verified it to be correct.  Thus:  $$\langle\psi|\hat{X}|\psi\rangle = \langle\psi|x\rangle\langle x|\hat{X}|x\rangle\langle x|\psi\rangle = \int d\vec{x}\ \psi^*(\vec{x})\ \psi(\vec{x})$$  So basically in the position basis, the position operator is simply $\hat{X} = x$, which seems pretty intuitive.  I believe the momentum operator came about next; and it was "derived" by fiddling around with the de Broglie relation and just substituting classical expressions for momentum into the Schrodinger equation (can't remember where I read that).  The fact that plugging these classical ideas into a quantum equation just happens to match experiment for momentum measurements seems like a pure coincident to me (it's along the same lines as trying to make the Schrodinger equation relativistic; one way of shoehorning relativity into it gives you the Dirac equation and a different way gives you the Klein-Gordon equation).  I imagine the spin operator came about by simple observation.  An electron goes up or electron goes down in a magnetic field.  Not hard to guess what the expression for that operator might be.  So anyway, am I on the right track here?  Have all of the operators come about simply by guessing?  And if I'm wrong, how did they actually come about?  How would I come up with a new operator for a new type of measurement?
22572
This is an exercise from "Contemporary Abstract Algebra" I'm not sure how to solve.     Exercise: Let $\langle a\rangle $ be a (cyclic) group of order $n$. Prove that the order of $a^k=\frac{n}{\gcd(n,k)}$.   Direction: (1) Let $d=\gcd(n,k)$, thus by the Euclidian algorithm we can find $X,Y\in\mathbb{Z}$ s.t. $d=Xn+Yk$, thus, $a^d=a^{Xn+Yk}=a^{Xn}a^{Yk}=(a^n)^X(a^k)^Y=(a^k)^Y$. What to do from here?  (2) We know that $d|n$, thus $\langle a^{n/d} \rangle$ is of order $d$ and $\langle a^d \rangle$ is of order $\frac{n}{d}=\frac{n}{\gcd(n,k)}$. Is it mean that $d=k$? Where is my mistake?
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 $\psi_{i,j}$?
45342
I wish to use the  to calculate the variance of a normal-distributed function. For this, I need the expected value of $X$ as well as the one of $X^2$. Intuitively, I would have assumed that $E(X^2)$ is always equal to $E(X)^2$. In fact, I cannot imagine how they could be different.  Could you explain how this is possible, e.g. with an example?
28591
I'm trying to evaluate $(\hat{a}+\hat{a}^\dagger)^k|n\rangle$ Where $\hat{a}$ and $\hat{a}^\dagger$ are ladder operators and $|n\rangle$ the $n$th Fock state. For this, I separated the problem in three parts: $k&lt;n$, $k=n$ and $k&gt;n$. In this part I'm a bit confused: Is it possible to evaluate such operations? Some tips on how to start would be really great.
23262
Let $E$ be  a complex Hilbert space. Let $T\in \mathcal{L}(E)$.  I have two questions:     Why it is not true that for an arbitrary operator $T\in \mathcal{L}(E)$, we have $\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0$? And is  this property true for normal operators?   I think it is true for self adjoint operators because the norm of a self adjoint operators is given by  $$\left\|T\right\|= \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}$$  Thank you.
30841
So I'm going through my notes and I think I've confused myself. We often imply  $$ |\psi\rangle \to \psi(x)\\ \langle\psi| \to \psi(x)^* $$ for instance when we talk about eigenvalue equations we interpret  $$ \hat{H}|\psi\rangle =E|\psi\rangle $$ as simply  $$ \hat{H}\psi(x)=E\psi(x) $$ but I don't understand why we say $|\psi\rangle \to \psi(x)$ because if that was the case then  $$ \langle\psi|\psi\rangle=|\psi|^2 $$ when clearly $$ \langle\psi|\psi\rangle=\int_{-\infty}^\infty |\psi|^2 dx = 1 $$ I'm obviously missing something simple, could anyone point out where I'm going wrong?
124060
I'm going through the first chapter in a text on real analysis, which contains preliminaries on ordered fields, the real numbers, etc. Supposedly I had learned about such things already, in calculus, but I thought it wouldn't hurt to go over it again.  Up until the paragraph which is the subject of my question, everything is thoroughly proven and examples are provided. However the following excerpt just goes through facts, whose proof I cannot conceive:      Although $\mathbb Q$ is an archimedean field, these properties cannot be used to define $\mathbb Q$ since $\\$ there are many Archimedean ordered fields. What distinguishes $\mathbb Q$ from the other $\\$Archimedean ordered fields is that $(\mathbb Q,&lt;)$ is the $smallest$ ordered field in the following sense: $\\$ if $(X,\prec)$ is an ordered field, then $X$ contains a sub-field which is (field) isomorphic to $\mathbb Q$.$\\$ Furthermore such an isomorphism preserves the order relation.   I've concatenated the paragraph a bit, so the text isn't idem from the book (if anyone wishes to know it's Phillips, An Introduction to Analysis and Integration Theory, Dover Publications). Also $\mathbb Q$ contains no proper subfields, and this can be verified by the apparently easy-to-prove fact that, if a field has characteristic zero, then it contains a subfield isomorphic to $\mathbb Q$. Then it remains to prove the preservation of order.  I'm guessing that some concepts from abstract algebra or field theory would easily suffice, but at the moment such topics are a bit over my head, so all I can think of doing is actually constructing the isomorphism $\phi:\mathbb Q\to\hat{\mathbb Q}$, where $\hat{\mathbb Q}$ is a certain subfield of  $X$ in such a way that field operations are preserved, but I really can't come up with anything. So the question is: how do I construct this isomorphism, or, if there's a better way of proving this, how is it done? Thanks for any help.  Edit: Assume the existence of $(\mathbb Q,&lt;)$ as an archimedean ordered field.
69769
I have a question concerning the time ordering operator. Let's suppose we have a time evolution generated by some Hamiltonian $H(t)$ given by $$ U(t)=T_\leftarrow\exp\left(-\mathrm{i}\int_0^t\mathrm{d}s\,H(s)\right)\tag{1}. $$ In Breuer and Petruccione, it's said that if the commutator of the Hamiltonian at some time $t$ with itself at some other time $t^\prime$ is a $c$-number function/ a complex function, i.e. $$\left[H(t),\,H(t^\prime)\right]=f(t,t'),$$ then the time evolution is given by $$ U(t)=\exp\left(-\frac{1}{2}\int_0^t\mathrm{d}s\int_0^t\mathrm{d}s^\prime\,\left[H(s),\,H(s^\prime)\right]\Theta(s-s^\prime)\right)$$ $$\times \exp\left(-\mathrm{i}\int_0^t\mathrm{ds}\,H(s)\right),\tag{2} $$ where $\Theta(s-s^\prime)$ is the Heaviside function. No proof or reference is given and I could not find any explanation anywhere, this is why I ask here. Any help would be very appreciated.
104966
For the definition of the momentum operator $$\hat{P } = -i \hbar \nabla$$ in quantum mechanics, as I understand you can derive this by either considering a more general definition of momentum, i.e. 'canonical momentum' which is an operator and then apply this operator to wave functions. This is shown here in this . We can alternatlively start with translations and use the face that the momentum operator is a generator of translations as is done here in this .   What I am interested in is what is the more fundamental derivation for the position operator: $$\hat{X} = x.$$ To this point I have considered that the motivation for defining position operator is from the definition of the expectation value $$\langle x \rangle = \int dx x |\psi(x)|^2 = \langle \psi|x | \psi \rangle$$ where $| \psi \rangle$ is normalised. Is this the full extent of the motivation or the main point to consider? Is there another motivation for the position operator or is it taken from this motivation and confirmed by experiments to be acceptable?
90167
If $a, b \in G$ commute   , show that  $o(ab) \mid\operatorname{lcm}(o(a), o(b))  $.  (Hint: Let $L=\operatorname{lcm}(o(a), o(b))$ and prove that $(ab)^L =e$.    ??  So I know that  *Suppose a has order n ,and b has order m ; that is, a^n=e and b^m=e   Then, since G is abelian, we have $(ab)nm=^{nm}=(a^nma^{nm})(b^nmb^{nm})=((a^n)^m))(^m\cdot (b^m)^n))=e^n=e$.  Now, the order of ab may be smaller, but it must divide $     nm=\operatorname{ord}(a)\cdot\operatorname{ord}(b)$.*  how iHow I can answer this question?
36135
In the Quantum Operations section in Nielsen and Chuang, (page 358 in the 2002 edition), they have the following equation: $$\mathcal E(\rho) = \mathrm{Tr}_{env} [U(\rho \otimes \rho_{env})U^\dagger]$$ They show an example with $\rho_{env} = |0\rangle \langle0|$ and $U = \mathrm{CNOT}$, and claim that the final solution is: $$P_0\rho P_0 + P_1\rho P_1,$$ where $P_0=|0\rangle \langle0|$ and $P_1=|1\rangle \langle 1|$. These are my steps so far to get this, but I don't know how to trace out the environment after this: Let $\rho$ be $|\psi \rangle \langle \psi |$, so that $\rho \otimes \rho_{env} = |\psi, 0\rangle \langle \psi, 0|$. Applying the unitary $U$, we have $$ |00 \rangle \langle 00| \psi, 0 \rangle \langle \psi, 0 | 00 \rangle \langle 00 |  + |00 \rangle \langle 00| \psi 0 \rangle \langle \psi 0 | 10 \rangle \langle 11 | \\  + |11 \rangle \langle 10| \psi 0 \rangle \langle \psi 0 | 00 \rangle \langle 00 |  +  |11 \rangle \langle 10| \psi 0 \rangle \langle \psi 0 | 10 \rangle \langle 11 |. $$ I don't know how to trace out the environment in the above state. Also, I realize that I have considered only a pure state, if anyone can show it for a general state that would be great.