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24

All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation, $$-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)$$ and therefore for any solution you can construct a second solution by taking the complex conjugate $\psi(\... 13 The action (to the right) on the ket space naturally induces an action (to the left) on the bra space. The bra space is the dual space of the ket space (that is the space of linear functionals over the kets). We can simply define$\left< \psi \right| X$by its action on kets (or, since it's linear, on a basis of ket space and linear extension): $$\big(\... 13 This is what computer scientists would call (ad-hoc) polymorphic or “overloaded” functions: basically, an operator X on the Hilbert space \mathcal{H} is not just one function X: \mathcal{H}\to\mathcal{H}, but a family of two functions$$ X = \{ X_{\mathcal{H}}, \quad X_{\mathcal{H}^\ast} \} $$with$$ X_{\mathcal{H}}: \mathcal{H}\to\mathcal{H}, \... 6 Indirectly. To start with, set the pesky and useless constants$\hbar/m\omega \mapsto 1 $, to work in natural units. You then recognize the matrix, as Eric emphasizes by identifying to his (5), as written in the number basis of oscillator discrete energy eigenstates$|n\rangle$. Now for a celebrated basis change. Infinity should not bother you. Consider ... 5 I thoroughly approve of your approach. Putting things on a computer, if you're careful, is a good way to learn this stuff. You have to understand the basic machinery clearly in order to express it in code, and once you've got your code working, you can study examples that are too hard to solve by hand. About your question: You transform vectors in an ... 5 The Hilbert space for a particle in a box has no associated momentum operator (as a momentum operator implies that the state space is invariant under space translations). So your attempt do define one leads to artifacts.  In general, the Hamiltonian must be a densely defined operator on the physical Hilbert space. But a square well potential is too ... 5 That "matrix" form of the position operator, infinite-dimensional or not, is meaningless since neither you nor the page you link say what basis it is supposed to be with respect to. Since the$\delta$-distributions which one might take as the eigenvectors of the position operator are not part of the Hilbert space of square-integrable functions the position ... 5 This is an interesting question. It is reminicent of the popular (but fallacious) "proof" that $$\exp\{ia\hat p\}\psi(x) \equiv \exp\{a\partial_x\}\psi(x)=\psi(x+a)$$ that claims that applying the exponential of the derivative operator to$\psi$gives the Taylor expanion of$\psi(x+a)$about$x$. The problem is that if$\psi(x)$is$C^\infty$and of ... 4 I will write a vector given with respect to the initial basis$[x_1,x_2]_{B_1}$.$[x_1,x_2]_{B_1}$represents the configuration where the first object is at position$x_1$and the second object is at position$x_2$. We have found the equation giving the time derivatives of the coordinates in this basis has the form $$\left \lbrack \matrix{ \ddot{x}_1 \cr \... 4 First remark: The correct expression is$$\langle0|\Phi(0)|p\rangle =(2\pi)^{-3/2}\left(2\sqrt{\vec{p}^{2}+m^{2}}\right)^{-1/2}N=(2\pi)^{-3/2}\left(2p^0\right)^{-1/2}N,$$where \vec{p} represents a spatial vector. Second remark: Under a homogeneous Lorentz transformation \Lambda, the creation operator of a scalar transform as$$U_0(\Lambda) a(\vec{p})... 4 We start with the definition $$\tag{1} S^{\alpha \beta}~:=~u^\alpha v^\beta-u^\beta v^\alpha.$$ Indices are raised and lowered with the metric. Up to an overall factor, one has $$\tag{2} \bar{S}_{\alpha \beta}~\propto~ \epsilon_{\alpha \beta \gamma \delta} S^{\gamma \delta},$$ so that the matrix trace $$\mathrm {Tr} (\mathbf{\bar{S}\cdot S }) ~=~ \bar{... 4 In addition to Emilio's great answer, and in answer to your second question: Specifically in 1D potential problems (i.e. \hat H = \frac{1}{2m} \hat p^2 + V(\hat x)), all the bound states can simultaneously be made real. This is because of the theorem that bound states in 1D are non-degenerate; then, \psi and \psi^*, which are both solutions in any ... 4 It's not non-sensical at all, except that there shouldn't be a minus sign (as mentioned in the comments) and that you took an operator outside of an expectation value, which I think worked out OK in this case but in general should be avoided. More conservatively,$$ \hat x = i \hbar \frac{d}{d \hat p} $$it follows that$$ \langle q \mid \hat x \hat p \... 4 I found the answer, at least for some of the cases that I thought were intractable, including my examples. The main tool is the disentangling theorem for the relevant group. Example 1: the transformation is in$SU(2)$, so we need the$SU(2)$disentangling theorem: $$\exp(z J_+-z^* J_-) = \exp(\tau_+ J_+)\exp(\tau_ 0J_0)\exp(\tau_-J_-)$$ Here$\{J_0,J_\pm\}...

4

You, in fact, did not make any mistake. You just did not pursue the problem to the end, changing bases to reduce your reducible representation. In fact, regardless of your Lorentz embedding usage, this is just the Kronecker product composition of two spin 1/2 representations into Kronecker sum of a spin 1 and a spin 0 representation... undergraduate addition ...

4

If the Hilbert space is infinite-dimensional, representing operators in terms of finite matrices is not possible. You can however use cumbersome infinite matrices which are not very useful... In any case some of the properties of matrices survive the passage to the (separable) infinite-dimensional case, at least dealing with bounded operators. Sometimes ...

4

From the form you've given of the operator $S_x$ and the basis vector you've given, you can easily calculate the matrix representation of it. Since $$S_x = \frac{\hbar}{2}(|\uparrow\rangle\langle\downarrow|+|\downarrow\rangle\langle\uparrow|)$$ and $$|\uparrow\rangle = \left(\begin{matrix}1\\0\end{matrix}\right)\implies\langle\uparrow| = \left(1\;\;\;\... 4 Yes,it is possible. Your representation is correct. It is easily to check using explicit form of \hat{S}_x:$$ \begin{pmatrix} \langle\uparrow|\hat S_x|\uparrow\rangle & \langle\uparrow|\hat S_x|\downarrow\rangle\\ \langle\downarrow|\hat S_x|\uparrow\rangle\ & \langle\downarrow|\hat S_x|\downarrow\rangle\end{pmatrix} = \frac{\hbar}{2} \begin{...

3

Not quite. The matrix elements you are talking about are called "off-diagonal" for obvious reasons: If you'd write down the matrix, these elements would occur somewhere other than on the diagonal. A non-zero off-diagonal element of an operator $B$ does not necessarily mean that you cannot diagonalize $B$ at all. It just means that in the currently used ...

3

A slow construction would go...  $$\begin{pmatrix}a&b\\c&d\end{pmatrix} = a\begin{pmatrix}1&0\\0&0\end{pmatrix} +b\begin{pmatrix}0&1\\0&0\end{pmatrix} +c\begin{pmatrix}0&0\\1&0\end{pmatrix} +d\begin{pmatrix}0&0\\0&1\end{pmatrix}$$ $$\begin{pmatrix}1&0\\0&0\end{pmatrix} =\frac{1}{2} \begin{... 3 That is the transition dipole moment integral. It is basically the probability that an electric dipole (i.e. a photon) can cause a transition between the states \psi_{nml} and \psi_{n'm'l'}. 3 A geometric approach can be achieve via an orbifold construction. I) Start with a free 1D quantum particle \psi(x)=\psi(x+2\pi R) on a circle S^1 of circumference 2\pi R=2L. The momentum p=n\frac{\hbar }{R} is then quantized, n\in\mathbb{Z}. The Hilbert space H has momentum basis (| n\rangle)_{n\in\mathbb{Z}}, where$$\hat{p} | n\rangle~=~n\...

3

It works just like in classical mechanics: the Hamiltonian generates infinitesimal time translations. Take the Schrodinger equation, $$i \frac{d}{dt} | \psi \rangle = H |\psi \rangle$$ and expand it for small times. Then $$|\psi(t)\rangle \approx (1 - iHt) |\psi(0) \rangle.$$ That is, $H|\psi \rangle$ tells you what $|\psi \rangle$ will instantaneously ...

3

from your profile you seem to be an amateur (gifted?, teenager?, still in school?) self-studying GR. Great! So drop the Old Man's Book (the Meaning of Relativity) and get yourself a modern intro to GR or - my suggestion - the wonderful MTW's Gravitation. I did the same when I was 16. It was published in 1973 and Kip Thorne himself told me two weeks ago that ...

3

As a partial answer, I took the advice of @JonCuster and pulled the following equations from "Quantum Mechanics of One- & Two- Electron Atoms" by Bethe and Salpeter. All results are in C.G.S units. The sum of dipole transition intensities for initial state $(n,l,m)$ and final states $(n',l+1,m')$, where $m'$ runs over the three allowed values ($m'=m-1,m,... 3 Let$M_2(\mathbb{C})$denote the set of all$2\times2$complex matrices. We also note that dim$(M_2(\mathbb{C}))=4$, because if$M\in M_2(\mathbb{C})$and$M=\left( \begin{array}{cc} z_{11} & z_{12}\\ z_{21} & z_{22} \\ \end{array} \right)$, where$z_{ij}\in \mathbb{C}$, then$M=\left( \begin{array}{cc} ...

3

As a practical matter, this is something where you expect the exponentiation to converge as the size of the truncated input grows. So, without knowing anything about the problem in question, I would find the desired size of the answer matrix (say, $n\times n$), truncate the input matrix at twice that size ($2n \times 2n$) and calculate the exponential, ...

3

That's not how you take the inverse of a 2x2 matrix. $$\pmatrix{t_1 & 1 \\ t_2 & 1}^{-1} = \frac{1}{t_1-t_2}\pmatrix{1& -t_2 \\ -1 & t_1}^T = \frac{1}{t_1-t_2} \pmatrix{1 & -1 \\ -t_2 & t_1}$$ The inverse of a 2x2 matrix is one over the determinant times the transpose of the cofactor matrix. Most generally,  \pmatrix{a & b \...

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