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21

This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to ...


18

Physicists usually generously relax the condition that the norm should be finite and they sometimes say that $|\vec r\rangle,|\vec p\rangle$ belong to the "Hilbert space". It's exactly the same "generous" language that allows physicists to say that $\delta(x)$ is a "function", the delta-function, even though its values around $x=0$ are infinite or "ill-...


18

IF you know that your Hamiltonian is of the form $$ \hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1 $$ for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state $\Psi_0(\mathbf r)$ obeys $$ \hat H\Psi_0(\mathbf r) =...


6

Assume for simplicity that all the operators are bounded. If you know the wave function $\psi$ associated with the ground state of the unknown Hamiltonian $H$, then $H$ has the form $$H = E_0|\psi\rangle\langle\psi| \oplus K$$ where $K$ is another Hamiltonian defined on a subspace of the original Hilbert space of co-dimension 1, and $E_0$ is the energy of ...


5

General wave functions can be expressed in terms of any set of eigenfunctions. But for bound systems, the energy eigenfunctions have a couple of appealing properties that make them popular: The energy eigenfunctions are the solutions to the time-independent problem, so you can work on a steady-state system. This often makes the math a lot easier. The ...


4

Let me actually collect most of my comments in an answer attempting to be more coherent than they, or your labile question. In fact, you are piling up three different questions, logically distinct, but with strong and natural connections, so it might be worth splitting them apart, before bringing them back together in the final coda. First, there is plain ...


4

Because it doesn't have total spin $s=0$ - it has total spin $s=1$, with the spin component parallel to the $z$-axis being zero. If you looked at that state in a different basis (e.g. the $x-$ or $y-$ basis) it would very clearly not have spin 0.


4

There's no analytic proof, but numerical evidence suggests that if you know that the Hamiltonian is local, and it satisfies the Eigenstate Thermalization Hypothesis (which most local Hamiltonians do), then you can extract the entire Hamiltonian from a single excited eigenstate, though not from the ground state: https://arxiv.org/abs/1503.00729.


3

Use the Baker-Campbell-Haussdorf formula to commute the displacement operators, $$ e^{\alpha a^\dagger} e^{-\alpha^* a} = e^{|\alpha|^2} e^{-\alpha^* a} e^{\alpha a^\dagger} $$ and rearrange the matrix element as $$ \langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |m \rangle = \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \langle 0 | a^n e^{-\alpha^* a} e^{\alpha a^\...


3

If unknown part of Hamiltonian is potential $V({\bf{r}})$, then you can write a stationary Schrodinger equation and figure out what the potential should be.


3

Quantum states are rays in the Hilbert space. i.e. $e^{i\alpha}|\psi\rangle$ is the same as $|\psi\rangle$. The set of all $e^{i\alpha}|\psi\rangle$'s with $\alpha\in[0,2\pi)$ is said to form a ray in the Hilbert space. A point on the Bloch sphere represents a ray and not a state. If you want you can think of the point $(\theta,\phi)$ on the Bloch sphere as ...


3

The boundary conditions determine whether an operator is Hermitian or not. Once you know your operator is Hermitian, you have the results on the spectrum. Without boundary conditions the momentum operator need not be Hermitian, hence its spectrum can have non-real values (here I am assuming that by Hermitian you actually mean self-adjoint). As an example, ...


2

Here is what I understand: if you have a particle at state $|x \rangle$, active translating it by $a$ means moving the particle to state $ | x + a \rangle$. Passive transformation means you keep the particle in the same place, and change the coordinate by new variable $x = x' + a$ (note that the coordinate system is translated backwards $-a$). I am not very ...


2

As commentators have indicated Hilbert space is a vector space. A manifold is a space with an atlas-chart construction with maps on overlapping regions that define connection coefficients and ultimately curvature. It is certainly possible to think of a finite dimensional complex vector space that is a locally flat region in an otherwise curved space. This ...


2

Hilbert spaces are vectorspaces by definition. If you interpret a vector space as a manifold (which you can do) then it's a flat manifold.


2

The extra one simply reflects the commutation relationship $[a,\,a^\dagger] = \mathrm{id}$ of the quantum mechanical harmonic oscillator. If each annihilation operator in the expression for $E^+$ acts only on its corresponding mode, and likewise for the creation operators in $E^-$, then the action of $E^-\,E^+ -E^+\,E^-$ on a state involving only excitations ...


2

Spin operator of the total 2 electron system is tricky: the statistical requirement reduces the Hilbert space to a 3-d rather than 4-d version. Like a spin-1 system, the $S_z$, as represents in basis $|S_z=1,0,-1\rangle$ is (see http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html, for example) $$S_z=\hbar\left(\begin{matrix}1&0&0\\0&0&...


2

Your second equation isn't quite right. If you have a continuous complete set of states $\{|p⟩\}$, then the correct expansion of a given arbitrary state $|P⟩$ in that basis is of the form $$ |P⟩ = \int\mathrm dp \: f(p)|p⟩, \tag 1 $$ with a single arbitrary function $f(p)$ over the indexing variable $p$ as a (continuous) coefficient. Here the $dp$ denotes ...


1

Assuming you have it set up that $\langle n\mid m\rangle=\delta_{nm}$ Then taking the sum: $$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(n-l)!}\sqrt{(m-k)!}}\langle n-l|m-k\rangle$$ This simplifies to: $$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{...


1

Write $$ U = \frac{1}{2}(I_1 + Z_1) \otimes U'_{00} + \frac{1}{2}(I_1 - Z_1) \otimes U'_{11} + (X_1 + iY_1) \otimes U'_{01} + (X_1 - iY_1) \otimes U'_{10} = \left(\begin{array}{cc} U'_{00} &U'_{01} \\U'_{10}& U'_{11}\end{array}\right) $$ where $U'_{ij} \in G_n$. In terms of the latter, $$ U' |\Psi\rangle = \sqrt{2} \;\langle 0 | U | 0\otimes \Psi \...


1

I will answer with an example on how eigen functions of momentum do not exist in a Hilbert space in general. When they do not, we say the momentum operator is not self-adjoint. Consider a one dimensional infinite potential well. Lets us place the walls at $x=0$ and $x=L$. For $x\ge L$ and $x\le 0$, the potential $V(x)=\infty$ and therefore we put the ...


1

Sometimes a picture can help. Its not hugely rigorous (especially for a physics site) but using the vector model, pictures can be constructed to illustrate the states that two spins $\alpha$ and $\beta$ can form.



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