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11

It's just notation. $H_1$ acts on Hilbert space ${\mathcal H}_1$, $H_2$ acts on Hilbert space ${\mathcal H}_2$. By $|\psi_1,\psi_2\rangle$ they have an implicit tensor product $$|\psi_1,\psi_2\rangle\stackrel{\rm def}{=} |\psi_1\rangle\otimes |\psi_2\rangle\in {\mathcal H}_1\otimes {\mathcal H}_2.$$ Then physiscists, usually without saying so, extend $... 7 It's not #1, because it's easy to write down QFTs which are unstable against pair production. One simple example is $$\mathcal{L} = \frac12 (\partial_\mu \phi)^2 + \frac12 m^2 \phi^2$$ which corresponds to particles with negative$m^2$. Since$E^2 = p^2 + m^2$, it is energetically favorable to produce infinitely many particles, since the rest energy is ... 6 It would be helpful for you to define what$H_1$and$H_2$are in your exercise. My guess is that your source is being sloppy in their notation for how they are defining tensor products of operators. Probably, they are working on some tensored Hilbert space$\mathcal{H}_A \otimes \mathcal{H}_{B}$and they mean something like $$H_1 = h_1 \otimes I_{B} \\ H_2 ... 5 "Differential" may be a bit of a red herring, as the generic \hat T operator may have strings of x s and \partial_xs. There is no good reason to hyperfocus on derivatives, which, in the coordinate representation you are using, correspond to the momentum operator \hat p,$$ \hat p = \int dx |x\rangle \frac {\hbar} {i} \partial _x \langle x|, $$so$$ \... 5 No. Mathematically multiplying your state vector by an operator is not how you determine the outcome of a measurement. First, that would assume you can get a deterministic value, which we know isn't true of measurements of quantum systems. Second, if your system is (most likely) in a superposition of momentum states, then this operation will not even give ... 5 There is actually a Hilbert space where mixed states (pure and non-pure) are represented by unit vectors up to phases. This Hilbert space is not the same Hilbert space where pure states are represented by all unit vectors (up to phases). However it includes the set of pure states (the vectors of the initial Hilbert space up to multiplication with scalars) ... 5 Consider that a state ket$\,\boldsymbol{|}\psi_1\boldsymbol{\rangle}\,$in the 2-dimensional Hilbet space$\,\mathcal{H}_1\,$is represented by its components with respect to a basis of this space by \boldsymbol{|}\psi_1\boldsymbol{\rangle}\boldsymbol{=} \begin{bmatrix} \xi_1 \vphantom{\dfrac{a}{b}} \\ \xi_2 \vphantom{\dfrac{a}{b}} \end{... 5 It's better to do it this way. Start with the expression without reference to any basis $$\hat{H} \vert{\psi}\rangle = E\vert \psi\rangle$$ Then bring in$\langle x|$$$\langle x|\hat{H} \vert{\psi}\rangle =\langle x| E\vert \psi\rangle$$ You are correct in having$\langle x|\psi\rangle=\psi(x)$, so the right hand side with scalar$E$easily becomes$E\...

5

Susskind is trying to stress the difference between elements (points) in a phase space and elements (vectors) in a (pre-)Hilbert space, admittedly causing some confusion by making oversimplified claims along the way. Phase space is not necessarily a vector (affine) space, i.e. it does not necessarily have a linear (affine) structure, respectively. The main ...

5

The answer is NO Given any complete basis, one can construct another by taking independent combinations. Assuming the original basis is complete and orthogonal, and contains $V_1$ and $V_2$, (which are thus orthogonal). Replacing $V_2$ by $V_2'=V_1+V_2$ does not change the completude of the basis. But now two vectors of the new basis, namelt $V_1$ and $V'... 5 Most likely here$Q$is the generator of infinitesimal transformations that is a symmetry of your Hamiltonian. A finite transformation is then given by $$g(q) = \exp(i q Q) = I + q Q + \mathcal O(q^2)$$ for some real number$q$. For example if$Q$is the momentum operator,$g(q)$is the translation operator (translation of distance$q$). If$Q$is angular ... 4 What you've written from your text/exercise is an abuse of notation, but it is standard. The composite state$|\psi \alpha\rangle$is the tensor product of the states$|\psi\rangle \in \mathcal H_1$and$|\alpha\rangle \in \mathcal H_2$, which is sometimes written$|\psi\alpha\rangle \equiv |\psi\rangle \otimes |\alpha\rangle$. A very typical example of ... 3 It is $$p_x=-i\hbar\frac{\partial}{\partial x}$$ $$p_y=-i\hbar\frac{\partial}{\partial y}$$ $$p_z=-i\hbar\frac{\partial}{\partial z}$$ and $${\bf p}^2=-\hbar^2\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right).$$ The case you are considering is 1-dimensional and you can limit everything to$x$.... 3 The momentum operator is a function:$\hat p (\psi) = -i\hbar \frac{\partial}{\partial x} \psi$. Applying this function twice: $$\hat p (\hat p (\psi)) = -i\hbar \frac{\partial}{\partial x} \hat p (\psi ) = -i\hbar \frac{\partial}{\partial x} (-i \hbar) \frac{\partial}{\partial x} \psi =- \hbar^2 \frac{\partial^2 \psi}{\partial x^2}$$ Since$i\hbar$is ... 3 Try this: $$e^{A} B \ e^{-A} = B+[A,B]+\frac{[A,[A,B]]}{2!}+\dots$$ With$ A = - \frac{1}{2}(\xi^{\ast}\hat{a} - \xi\hat{a}^{\dagger} )$,$B = (a^{\dagger}-\hat{a})$, the above formula yields, $$(\hat{a}^{\dagger}-\hat{a})\hat{S} = \hat{S}(\hat{a}^{\dagger}-\hat{a})+\frac{(\xi-\xi^*)}{2} \hat{S}.$$ Repetitive application will recursively produce the ... 3 If a Hamiltonian is hermitian then its eigenvalues are real (and we deduce they are observable) and its eigenstates are complete (form a basis in the relevant Hilbert space). In general, a non-hermitian Hamiltonian will not have real eigenvalues and as such we would not propose being able to observe / measure a "complex energy." However, as Jon points out, ... 3 Option 3 is the closest match, but it's a bit like saying "Nothing guarantees that spacetime has a Lorentzian signature." We normally only consider spacetimes that do, because so much else depends on it. It's a requirement, not a theorem. Similarly, for relativistic QFT in flat spacetime, we normally only consider QFTs whose total energy has a finite lower ... 3 First, yes, in the notation$|x\rangle$and$|p\rangle$the symbols$x$and$p$label the eigenvalue of the eigenstate represented by the ket. In that sense$|x+a\rangle$is the eigenstate of position with eigenvalue$x+a$. It is very common that when working with Hermitian operators one labels the eigenstates of the basis by the eigenvalue (beware, though, ... 2 Each pure State$|\psi_n\rangle$of the quantum system is a multi-particle state describing all particles of the system. So your pure states are all possible states of the system. Since it is not prepared in a state$|\psi\rangle$there are plenty of possibilites of states in which your systems could be. Let us say you prepared your system in$|\psi\rangle$... 2 I assume that in transformation you restrict yourself to unitary transformations, such that probability is maintained$|| \hat{T} |\psi\rangle || = 1$. I will further assume that you limit the question to forms of transformation that are position-based somehow, so the notion of "Kernel"$\hat{K}(x,x')$makes sense (so for example rotations in spin-space are ... 2 In addition to the physical explanation given above, I'd like to point out the mathematical difference. A Hilbert space is a complex vector space equipped with an inner product that is also complete. A Minkowski space is a real vector space equipped with a symmetric bilinear form. A symmetric bilinear form is not the same as an inner product, that's why the '... 2 No. Expectation values of observable operators are always real. 2 Because the orthogonality is supplied by the azimuthal direction when the two$z$-components of the angular momentum differ. This is why we usually bundle them together into the spherical harmonics instead of handling them separately. 2 This is very simple, and really a purely mathematical question, not so much a physics one. What you want is to try and find a function$\psi(x)$so that $$\int_{-\infty}^{\infty} |\psi(x)|^2\ dx$$ converges but $$\int_{-\infty}^{\infty} |x \psi(x)|^2\ dx$$ diverges. One way to intuitively guess at what such a function might look like is to imagine it as ... 2 You can understand the relationship between these two expressions when using a common trick performed in quantum mechanical calculations, which is multiplication by one. The unit matrix in quantum mechanics can be expressed as the sum of the outer product over a complete basis, i.e.$ 1 = \sum_n \vert n \rangle \langle n \vert$, where the states$ \vert n \...

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This last line makes use of the fact that $$\sum_{n=0}^N n=\frac{N(N+1)}2.$$ This expression is called a triangular number and there are many great proofs for it. It also splits the sum in two parts \begin{align}\sum_{n=0}^N (n+1/2)&=\sum_{n=0}^N(n)+\sum_{n=0}^N(1/2)\\&=\frac 1 2N(N+1)+\frac 1 2(N+1)\\&=\frac 1 2(N+1)^2\end{align} Obviously you ...

2

An operator being self adjoint or not depends greatly on the Hilbert space upon which it acts. The momentum operator is self adjoint on functions defined over $\mathbf R^3$ when acting upon functions that are square integrable (I.e $L^2$ functions). The problem begins when one looks for eigen states or the operator because it turns out that plane waves are ...

2

The basic idea behind the factorization is to replace a 2nd-order differential equation by a pair of first order ones. It was made popular in physics by the work of Hull and Infeld: Infeld, Leopold, and T. E. Hull. "The factorization method." Reviews of modern Physics 23.1 (1951): 21 although in fact earlier examples, such as the factorization of the ...

1

Quantization, i.e. the appearance of discrete eigenvalues, is a feature of bound states. It is this discreetness of the energy eigenvalues that was difficult to explain using classical physics. Setting up the problem of finding energy levels as an eigenvalue problem for an equation that did not exist in classical physics was the giant step. Bound states ...

1

Bound states are characterized by the energy being less than the potential at $-\infty$ and $+\infty$. We typically say $E>0$ is a scattering state and $E<0$ is a bound state with the implicit assumption that the potential energy approaches zero at infinity. One canonical model which indicates this is the quantum harmonic oscillator: we generally write ...

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