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10

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 ...


6

I like the previous answer, but I thought you might want something less abstract. To simulate the Schrodinger equation, replace space by a grid of points, and the wavefunction values with a wavefunction value on each point of the grid. Then replace: $$ \nabla^2 \psi $$ in the Hamiltonian with $$ (\sum_e \psi(x+e) ) - N \psi(x) $$ Where e runs over the ...


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

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 }) ~=~ ...


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) ...


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

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

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 ...


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} ...


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} ...


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

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 ...


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

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.


1

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 ...


1

Now that there's an answer in traditional index notation, here's an alternative perspective from geometric algebra. You have some bivector $S = u \wedge v$ and its dual $\bar S = i (u \wedge v)$, where $i$ is the spacetime pseudoscalar. $\bar S$ is an oriented plane that is entirely orthogonal to $S$. In GA, the quantity you describe would just be denoted ...


1

The question formulation (v2) seems to mix the notions of invariant and covariant, which essentially is also the main point of user1504's answer (v1). Let's say we have a group $G$. The group $G$ could e.g. be a finite group or a Lie group. When we say that a theory is invariant under $G$, it normally implies at least two things. The group $G$ acts on ...


1

So far I can see only 12 different nonzero matrices. Perhaps you missed $\gamma^5\gamma^\mu$. EDIT: There is a proof of linear independence of the 16 matrices, e.g., in Bogoliubov and Shirkov's Introduction to theory of quantized fields. They prove (using trace invariance under cyclic permutations and anticommutation relations) that each of the matrices, ...


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 ...


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

A finite-dimensional matrix always has a discrete spectrum. For continuous quantities one needs operators with a continuous spectrum, which therefore must act on an infinite-dimensional space. Examples are the multiplication operator or the differentiation operator in the space of sufficiently nice square integrable functions of a real variable. (This ...


1

Applying $\hat p$ to a Hamiltonian eigenfunction gives another state indeed, but this state can be decomposed in the eigenstates of the Hamiltonian. Numerically it is good everywhere except, maybe, the boundary points, so for integration purposes such a decomposition (superposition) is good. You did not obtain a diagonal matrix exclusively because you took a ...


1

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 ...



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