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The ladder operators satisfy: $\bf{a^{\dagger}}$$|n>=\sqrt{n+1}|n+1> \bf{a}$$|n>=\sqrt{n}|n-1>$ Taking into account $<n|m>=\delta_{n,m}$ , you get the answer.

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In the context of quantum mechanics, Hilbert spaces usually refer to the infinite-dimensional space of solutions to the time-dependent Schrodinger equation $$i\frac{d}{dt} \left|\psi (t)\right\rangle = H(t) \left|\psi(t)\right\rangle$$ for the state vector $\left|\psi (t)\right\rangle$. This space is completely determined by the (in general) time-dependent ...

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Your idea of what $\psi(\vec x,t)$ is supposed to be is essentially correct. Given a space of states $\mathcal{H}$, the "Schrödinger state" is a map $$\psi : \mathbb{R}\to\mathcal{H}, t\mapsto\lvert\psi(t)\rangle$$ where $\lvert \psi(t)\rangle\in\mathcal{H}$ for every instant $t$. If $\mathcal{H}$ is a space of functions in a variable $\vec x$, then $\lvert ... 0 The inner product on a finite-dimensional "bra-ket space" is not "Euclidean", as that denotes the standard inner product on Euclidean space$\mathbb{R}^n$. It is, however, in the case of$\mathbb{C}^n$given by $$x^\dagger y = \sum (x^i)^\ast y^i,$$ the Wikipedia article is just wrong about calling this Euclidean. Every complex inner product space ... 0 I would rather answer your question in a different way. I request you to be patient as the start diverges a bit from your question. There is a recent upsurge due to finding real eigenvalues for Non-Hermitian matrices and this has led to the idea of generalising hermiticity of operators in QM to something else, known as Pseudo-Hermiticity. If you study them ... 1 The situation is impossible and therefore doesn't happen. When you claim to have two states$\psi_1$and$\psi_2$then I'll assume they are linearly independent, otherwise they aren't really two different states. Then you claim you have an Operator$O$such that$POP^{-1} = \epsilon_3 P$where$P$is the parity operator and further that$P \psi_1 = ...

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Under conformal mapping z=>w(z) and $\bar{z}$=>$\bar{w}(\bar{z})$ a field of conformal dimension(h,$\bar{h}$) transforms as $\tilde{\phi}(w,\bar{w})=(\frac{\partial w}{\partial{z}})^{-h}(\frac{\partial \bar{w}}{\partial\bar{z}})^{-\bar{h}}\phi(z,\bar{z})$..

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Well, I think that first of all, you should understand, that the reason why you can decompose free field into plain waves is that equations of motions are linear. In case of interaction, equations are motions are nonlinear, so any linear combination of its solutions is no longer a solution. In the second place, about your second question: it depends on ...

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Your last integral is $$J = \int{d^3r \;\psi_0(r) \left[-2\pi\delta(\vec{r})Res(\psi_0(0)) + \sqrt{2}\pi a\delta(\vec{r})\frac{\partial}{\partial r}(r\psi_0(r))\right]} =\\ = 4\pi \int_0^\infty{dr\; r^2 \psi_0(r) \left[-2\pi\frac{\delta(r)}{2\pi r^2}Res(\psi_0(0)) + \sqrt{2}\pi a\frac{\delta(r)}{2\pi r^2}\frac{\partial}{\partial r}(r\psi_0(r))\right]} \\ = ... 2 Your confusion stems from the fact that \lvert x \rangle is not inside the Hilbert space of states. It cannot be because \langle x \vert x \rangle = \delta(x-x) = \delta(0) is not an allowed value for an inner product in a Hilbert space to have. There are several things to say about \lvert x \rangle: If you want to make precise what kind of objects ... 2 Assuming everything is defined in the correct Hilbert spaces, project the decomposition$$ |\psi\rangle = \sum_n {c_n |n\rangle} $$onto the position kets ("states") |x\rangle and obtain$$ \psi(x) = \langle x |\psi\rangle = \sum_n {c_n \langle x |n\rangle} = \sum_n {c_n u_n(x)} $$where the u_n(x) = \langle x |n\rangle are the wavefunctions ... 3 Each separable infinite-dimensional Hilbert space carries both irreducible and reducible representations of any noncompact Lie groups you can name. But this information in itself is of little use. The Hilbert spaces in quantum mechanics always come with distinguished representations that give certain operators an interpretation as distinguished ... 2 I know this question was asked a long time ago, but since I thought very hard about the same question today and didn't find the other answer very helpful, I decided to write my own. The problem is that OP is not really asking about why the product form used in the BO approximation is valid, but how the given expansion (which is claimed to be exact, see e.g. ... 0 The mathematical formalism starts from a C*-algebra A, that of observables, and a set of physical states. For any observable a\in A, i.e. a self-adjoint element (a^*=a) and a state \omega\in A^*, the expectation value of a on the state \omega is simply$$\omega(a)$$If you consider the C*-algebra generated by a you can then apply the ... 0 Probabilities of independent outcomes of the same measurement are always additive, which means that the expression$$P(U) = \sum_{n\in J}P(a_n),$$is perfectly correct. If you want something that looks more formal, you can express P(U) in as the expectation value of an appropriate operator, the projector$$ \Pi_U=\sum_{n\in ...

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In the video Prof. G. Rangarajan is considering the expectation value $$\mathbb{R}~\ni~ \langle \psi |\underbrace{\hat{L}_z}_{\text{self-adj.}} |\psi \rangle ~=~ \int\! d^3r ~\underbrace{\overline{\psi({\bf r})}}_{\text{real}} \underbrace{ (-i\hbar) \frac{\partial}{\partial \varphi}}_{\text{imaginary}} \underbrace{\psi({\bf r})}_{\text{real}} ... 3 I think the fundamental misunderstanding of superposition has a lot to do with the popular interpretation of quantum mechanics. That is, how Schrödinger's cat is portrayed in popular science. When a quantum system is in a state of superposition, it means that the outcome of a measurement of some property of that system is uncertain. The wacky thing about ... 2 The process of "collapse" can almost entirely be handled just by including your measurement apparatus in a quantum description. It's the quantum interaction between observer and observed that causes collapse. For the mathematics of this process, you might want to look at my answer to Particle interactions which are NOT considered observations? for more on ... 2 Just regarding question number 3: 1) If your friend knows anything at all about quantum mechanics, he will not say "Both", because quantum mechanics does not allow anything to be in two states at once, ever. 2) If your friend knows a small amount about quantum mechanics, he might make the mistake of saying "Neither", because, after all, most coins ... 3 I would say that you are suffering from a little bit of pop-science fatigue. You don't need to be convinced of these things because the great majority of working physicists don't think about quantum mechanics in this way. What if a brain dead person looks and doesn't comprehend? This is called "begging the question", I think. Except for a handful ... -3 At the moment, one observe a state, not the state does collapse, but it's our knowledge about this state, that collapses. This is right for cats in boxes with poison watches. An other thing is, if you have to do with superposition of quantum states. In a Bose-Einstein condensate all involved atoms are in a superosition and a powerful enouth observation of ... 0 Yes it is wrong because multiplication of matrices, you know it, gives matrices and I don't think it makes sense to put a matrice inside a ket or even a bra vector. Actually even with a constant (complex number) if you have k |v\rangle it does not make any sense to put it inside the ket vector like |kv \rangle. However if you have a constant k in the ... 6 Dirac notation is ill-suited for non-self-adjoint operators. Here's why: Let (-,-) be the inner product on our Hilbert space. The expectation value of AB is then$$ \langle AB \rangle_\psi = (\psi,AB\psi)$$by definition, and Dirac notation writes \langle \psi \vert AB \vert \psi \rangle. for this. But, in this notation, it is no longer clear to which ... 0  \langle\psi|AB|\psi\rangle  is a complex number (as opposed to a matrix), so taking its transpose gives you back the same thing, i.e.$$ \langle\psi|AB|\psi\rangle^{\dagger} = \langle\psi|AB|\psi\rangle^*, $$and therefore$$ = \langle\psi|B^\dagger A^\dagger|\psi\rangle . $$EDIT I just realised that you then equated this to  \langle\psi|B^\dagger ... 0 I don't know the term "imaginary operator". I take this to be an antihermitian operator, which eigenvalues are purely imaginary. Then the statement is clearly not true. Take as a counter example any hermitian operator \hat{A} and real wavefunction \psi with \langle \psi | \hat{A} |\psi\rangle = A_\psi \neq 0. A_\psi is of course real. Take now ... 5 Comments to the question (v5): In this quantum case the overline/bar notation \bar{A}=\langle A\rangle is borrowed from statistics and it denotes a quantum expectation value of a quantity A. See also Ehrenfest theorem. The problem from Ref. 1 considers a harmonic oscillator with Hamiltonian operator$$\tag{A} H~=~\frac{p^2}{2m} ...

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$\langle n| n\rangle = \frac{1}{n!} \langle0|(a)^n (a^{\dagger})^n |0\rangle$ Because $(a^{\dagger})^n |0\rangle = \sqrt{n!} |n\rangle$ and $(a)^n |n\rangle =\sqrt{n!} |0\rangle$ So we have $\langle n| n\rangle = \langle 0| 0\rangle =1$ In fact, I think the reasonable process is to assume the normalized $\langle n| n\rangle =1$ first, then we get the ...

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In order to show that it is correctly normalised, you need to show that $| \left <n|n \right>|^2=1$ where in this case $\left <n \right|=\frac{1}{\sqrt{n!}} \left <0 \right| \hat a^n$. And here I suppose $\hat a$ is a ladder operator. Go take a look on this link to know more about ladder operators https://en.wikipedia.org/wiki/Ladder_operator. ...

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(1) Yes, take ${\cal H} = L^2(\mathbb R, dx)\oplus L^2(\mathbb R, dx)$ and thereon $\left(X (\psi, \phi)\right)(x,y) := (x\psi(x),y\phi(y))$. We have $\sigma(X)=\sigma_c(X)$ and the degeneracy is just $2$. (2) Yes, use the example (1) with a countably infinite copies of $L^2(\mathbb R, dx)$ and use the Hilbertian direct sum of Hilbert spaces. (There are ...

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Let $\{|e_i\rangle: 1\leq i\leq n\}$ be a basis. Since $|e_1\rangle$ is a vector, and $\hat{T}$ is a linear transformation then $\hat{T}|e_1\rangle$ is a vector. Since $\hat{T}|e_1\rangle$ is a vector and $\{|e_i\rangle: 1\leq i\leq n\}$ is a basis then $\hat{T}|e_1\rangle$ is uniquely expressed as a linear combination of the vectors in $\{|e_i\rangle: ... 1 First get the idea of kets as some component in$\mathbb{R}^n$out of your mind. kets are elements of a complex vector space, in this case a finite dimensional one. Yes the space will be isomorphic to$\mathbb{R}^n$, or rather$\mathbb{C}^n$, but by assuming they are$\mathbb{C}^n$, you may imbue them with properties of$\mathbb{C}^n$that are not true of a ... 2 Born calculated solution to Schroedinger's equation corresponding to electron scattering experiment and what he got was continuous function of scattering angles measured with respect to the original direction of propagation of electrons. However, in experiment electrons are always detected at definite points of a screen. Clearly, there is no direct match ... 0 There are already good answers. Here we merely use different words and stress different things. The$n$'th Fock space${\cal H}_n$is the vector space spanned by$n$particle states. In other words, if we omit the zero-vector, then${\cal H}_n\backslash\{0\}$is the space of$n$particle states. Be aware that we will often casually refer to${\cal H}_n$... 9 In field theory, there are two vacua. The non-perturbative vacuum$|\Omega\rangle$and the vacuum of the free theory$|0\rangle$. The wikipedia article makes reference to$|\Omega\rangle$in terms of$|0\rangle$and its excitations. The true vacuum is annihilated by the (dressed) annihilation operators, and can be thought of perturbatively in terms ... 6 The operator$X^{-2}$does exist and is self-adjoint as it follows from standard spectral theory. Its domain is $$D(X^{-2}) := \left\{\psi \in L^2(\mathbb R, dx) \:\left|\: \int_{\mathbb R} x^{-4} |\psi(x)|^2 dx \right.< +\infty\right\}$$ and thereon $$(X^{-2}\psi)(x) := x^{-2}\psi(x)\:.$$ The eigenfunctions of the Hamiltonian operator of the harmonic ... 0 Ali Moh's answer is nice. Here is a different perspective. Let me just completely forget about creation and annhilation operators and try to construct this matrix element using the wave function for the harmonic oscillator. I'll interpret$X$as the position operator (since that seems like what you have in mind). Then let's take a look at this expectation ... 1 when we define$X=a+ a^\dagger$,$\frac{1}{X}$and any power thereof do not exist. Proof: Consider the state$\left|x\right\rangle\propto e^{-a^{\dagger 2} + 2 x\, a^\dagger}\left| 0\right\rangle$, you can show that$X\left|x\right\rangle = x \left|x\right\rangle$, and in particular if you choose$x=0$, then$X\left|x=0\right\rangle = 0$. Now from ... 2 You can think of the spin state of an electron as represented by a vector$(\alpha,\beta)$. Depending on how you set things up, "Up" might be represented by$(1,0)$, "Down" by$(0,1)$, "Left" by$(1,1)$, and "Right" by$(-1,1)$. Up is orthogonal to Down, and Left is orthogonal to Right, but Up is not orthogonal to Left. 1 Angular momentum in quantum mechanics in general works like this: the total is measured by$L^2 = \hbar^2 \ell (\ell+1)$whereas the projection along any axis is measured by$L_z = \hbar~m$between$-\ell \le m \le \ell.$Both$\ell$and$m$are simultaneously measurable (i.e. the$L^2$and$L_z\$ operators commute), and they must be spaced by integers but ...

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