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16

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(\...


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 ...


4

Your $N^2-N^2$ calculation is naive, well, it is incorrect because not all $N^2$ generators of $U(N)$ are changing the Hermitian matrix $M$. If a generic Hermitian matrix $M$ is transformed to $$ M\to U M U^{-1} ,\quad U U^\dagger = 1,$$ then $N$ directions in $U$ i.e. in $U(N)$ don't change $M$ at all. This is easily seen in the basis in which $M$ is ...


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

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

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{...


3

Disclaimer at the bottom. I'm assuming we're working with the infinite (continuous position) case. A few things that may help you: In quantum mechanics, $\langle \hat{A} \rangle$ is usually defined to be $\langle \psi | \hat{A} | \psi \rangle=\langle \psi | \hat{A} \psi \rangle$. If you're not given $\psi$ you can't find $\langle \hat{A} \rangle$, which ...


3

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 ...


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

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 \...


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

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 ...


2

I'm not sure if the following is the main issue, but OP writes in the question formulation (v1): [...] $c_{ij}(t)$ is symmetric [...] The coefficient $c_{ij}$ is only $i\leftrightarrow j$ symmetric in case of bosons. In case of fermions, the coefficient $c_{ij}$ is $i\leftrightarrow j$ antisymmetric.


2

The matrices $\sigma_0\equiv \boldsymbol{1}_2$, $\sigma_x$, $\sigma_y$ and $\sigma_z$ form an orthonormal basis of your vector space w.r.t. the scalar product $$ (X,Y) \equiv \frac{1}{2}\operatorname{tr}(X\cdot Y), $$ where $X$ and $Y$ label any two complex $2\times 2$ matrices. The factor $1/2$ is just for convenience, you may as well normalise your ...


2

I like to put it this way: $$\left(\begin{array}{cc} w+z&x-iy\\ x+iy&w-z\end{array}\right)$$ So, for example: $$\left(\begin{array}{cc} 1&5\\1&2\end{array}\right) = \left(\begin{array}{cc} (1.5)+(-0.5)&(3)-i(2i)\\ (3)+i(2i)&(1.5)-(-0.5)\end{array}\right)$$ So $w=1.5, x=3, y=2i, z=-0.5$ and $$\left(\begin{array}{cc} 1&5\\1&...


2

In this answer we will basically expand on Lubos Motl's correct answer using some other words and introducing some terminology. In the Hermitian one-matrix model, the action $$\tag{1}S~=~ {\rm Tr} L(H)$$ is invariant under adjoint conjugation $$\tag{2} H\to UHU^{-1}$$ with unitary matrices $U$. Eq.(2) here play the role of the gauge transformations....


2

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. [Edit] In general, the Hamiltonian must be a densely defined operator on the physical Hilbert space. But a square well potential is too ...


2

$\lvert \psi_{n'm'l'} \rangle$ is the state you start out with. $A {\mid} \psi_{n'm'l'} \rangle$ is the new state you get when you apply $A$ to the original state. $\langle \psi_{nml} {\mid} A {\mid} \psi_{n'm'l'} \rangle$ is the projection of this new state onto $\lvert \psi_{nml} \rangle$; that is, it measures the overlap between the unprimed state and the ...


2

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 ...


2

You can solve the problem directly. Assuming a Schrodinger-like equation with a very simple "Hamiltonian", $$i \frac{d}{dt} \left[ \begin{array}{} \psi_1(t) \\ \psi_2(t) \end{array} \right] = \left[ \begin{array}{} 1 & 1 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{} \psi_1(t) \\ \psi_2(t) \end{array} \right], $$ it is straight-forward to show ...


2

There are five relevant quadrupole moment operators, and when labeled by the change $\Delta m$ in angular momentum projection they read $$ \begin{array}{c|ccccc} \Delta m & -2 & -1 & 0 & 1 & 2\\ \hat Q_{2m}& (x-iy)^2 & (x-iy)z & x^2+y^2-2z^2 & (x+iy)z & (x+iy)^2 \end{array} $$ These operators arise as (a basis for) all ...


1

The quantity $A|\psi\rangle$, for general Hermitian operators $A$, is mostly meaningless. The eigenvalues and eigenvectors of $A$ are meaningful, as is the quantity $\langle \psi | A | \psi \rangle$, but $A|\psi\rangle$ by itself is not. This is a sort of confusing point when first learning QM because it feels like the most important thing about operators ...


1

The solutions these equations give for the angles are different because they are for different rotation sequences. The first set of equations has the subscript xyz and the second has the subscript yxz, so these angles depend on the rotation sequence.


1

The matrix you show displays twelve degrees of freedom for two nodes, so six degrees of freedom per node. Transferring this matrix to global coordinates is as if you are rotating the beam (beam length is already included in the matrix). Rotation matrix $R(\vec{\theta})$ can be used to rotate a force $\vec{F}$. You already apply force in global coordinates, ...


1

this paper might help.$^1$ It's written pedagogically and hence is easier to read. It goes on to discuss a lot more than just color decomposition too. $^1$ Scattering Amplitudes, Henriette Elvang, Yu-tin Huang.


1

Note that in the sum $$=\sum_{S_1=\pm 1}...\sum_{S_N=\pm 1}\langle S_2| T_{_{NN}}^{\dagger}|S_1\rangle\langle S_1| T_{_{NNN}}|S_3\rangle\langle S_3| T_{_{NN}}^{\dagger}| S_2\rangle\langle S_2| T_{_{NNN}}|S_4\rangle...\langle S_1| T_{_{NN}}^{\dagger}|S_N\rangle\langle S_N| T_{_{NNN}}|S_2\rangle$$ every pair $|S_i\rangle\langle S_i|$ occurs twice with some ...


1

It turned out that some time away from my desk was what I needed to escape from this "Panic! I can no longer do any physics!" maze. Well, then for those who might be interested and for those who might tracked down some errors, here is my development. For the first quantization : \begin{equation} <a b ∣ V(r) ∣ cd > = \frac{1}{2} \Bigg[ \int d^3 r_1 \...


1

Focus on the integral $$ I_{ij}(k) = \int k_i k_j\ \mathrm{d}\Omega_k.$$ This is a rank 2 symmetric tensor which can only depend on $\vec{k}$ through its magnitude $k^2$, since the direction has been integrated over. So the only possibility is that $I_{ij}$ is proportional to the unit tensor (Kronecker delta): $$ I_{ij}(k) = f(k^2) \delta_{ij},$$ where $...



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