when describing a two spin system with kets represented as column vectors, is it just convention which coordinates one attributes to +x and +y? (spin up in x and spin up in y)
i mean coordinates in the linear algebra sense, i.e., you linearly combine basis vectors to get some arbitrary vector. the scalars you use in the linear combination are the coordinates
for example, in townsend modern quantum, he representes $|+x \rangle$ by the column vector $\frac{1}{\sqrt{2}}(1 \ 1)$. Can I instead represent $|+x \rangle$ with column vector $\frac{1}{\sqrt{2}}(i \ 1)$ and change the other vectors representing other spin states accordingly? Also, the basis kets are $|+z\rangle$ and $|-z\rangle$.
@SillyGoose I think the more important insight is that wave functions are not all that physical to begin with. They are ensemble descriptions, like probabilities. They are not nearly as concrete as the momentum that your car sheds when it hits the concrete pillar of the road bridge....
Strange things can happen when we talk about an infinite set of similar things that are completely independent of each other... and strange things do happen in both probability theory and in quantum mechanics.
Coleman says that an acceleration field is geodesic if it has the form $\Gamma \dot{\gamma} \dot{\gamma}$ while apparently a spray gives rise to a linear connection if it's homogeneous of degree 2, so that $S(x, ay) = a^2 S(x,y)$
That does sound like the same thing
Also I think the spray condition is basically how to make sure you bump your $TTM$ vector down to $T^2M$
Bell made assumptions that need not be valid at all. One can dispute, for instance, his assumption of "statistical independence". That's, roughly, saying that if you don't know all the data, it is perfectly sensible to assume that all variables that you did not measure will come out with equal pr...
I was reading about Matter waves, and I came across the following paragraph :" Frequencies of solutions of the non-relativistic Schrödinger equation differ from de Broglie waves by the Compton frequency since the energy corresponding to the rest mass of a particle is not part of the non-relativistic Schrödinger equation."
If we consider the non relativistic SE, why do we ignore the $mc^2$^2 part?
Because I find it odd that in the case, where you'd observe a matter particle that moves extremely slowly, you'd ignore the rest mass energy and consider only it's momentum.
you don't "consider only its momentum", you expand the energy in powers of $v/c$ and ignore a constant because a constant shift in energy is irrelevant
Because the Hamilton operator is a linear one, then, the solution of the TISE, can be a superposition of eigenstates of H, and not only an eigenstate of H? Is it safe to say so?
the "TISE" is just the name for the equation that define what an eigenstate of the Hamiltonian is, it's silly to give it a special name, really
note that it's not really "the" TISE, but that $H\psi = E\psi$ is a different equation for each different value of $E\in\mathbb{R}$, and it makes no sense to try to apply the superposition principle to two different equations (i.e. two different values of $E$)
But if that's the case, then how should I interpret the following 2cases: $H(\lambda_1 \psi(\vec r)_1 + \lambda_2 \psi(\vec r)_2)= \lambda_1 E_1 \psi(\vec r)_1 + \lambda_2 E_2 \psi(\vec r)_2$
Now, above, I was told that the solution of the TISE, is an eigenstate of the Hamiltonian, and cannot be a superposition. Here, while it is said that $\Psi(\vec r,t=0)$ is a solution of the TISE, in the equations above, in the 2 last ones, this solution is expressed as a linear combination of the energy eigentates of H. Which is in contrdiction to what was said above, regarding the solutions of TISE.
I hope it's somehow clear the confusion, and where it arises from
@imbAF there are three different things we here need to distinguish: 1. A general solution to the TDSE. 2. A general solution to the TISE. 3. The ansatz $\psi(x,t) = \phi(x)\chi(t)$ for a particular set of solutions of the TDSE
you have two symbols $\psi$ and $\Psi$ for these three different things and even then you're not using them consistently :P
So let's write it like this: We want to find a general solution $\Psi(x,t)$ to the TDSE. In order to do so, we make the separation of variables ansatz $\psi(x,t) =\phi(x)\chi(t)$ for a particular solution $\psi$. Plugging that ansatz in, we see that $\chi(t)$ is a phase and $\phi(x)$ is a solution to the TISE
Now, we know that the solutions $\phi(x)$ to the various TISEs are labeled by their eigenvalues $\lambda_i$ of $H$, so we get $H\phi_i(x) = \lambda_i \phi_i(x)$, and consequently a bunch of particular solutions $\psi_i(x,t) = \phi_i(x)\chi_i(t)$
Every of these $\psi_i$ is a solution of the TDSE, and since the TDSE is a linear differential equation, so $\Psi = \sum_i c_i\psi_i$ is still a solution of the TDSE
if your Hamiltonian has discrete eigenstates and is time-independent, this is how it works, doesn't matter what sort of system that Hamiltonian describes
In other words, if the solution to the TDSE can have the form of the ansatz, then it's a superposition, or can possibly be, a superposition of solutions of TDSE?
Then the solution of the particle in a potential well, which is in my notes must be wrong: $\Psi(\vec r,t)=A e^{-i\omega t}\phi (\vec r,t=0)$ $\Psi(\vec r,t)=A e^{-i\frac H \hbar t}\phi (\vec r,t=0)$ $\Psi(\vec r,t)=A e^{-i\frac H \hbar t}\sum_{n=1}c_n\phi_n(\vec r)$ $\Psi(\vec r,t)=\sum_{n=1}A e^{-i\frac {E_n}{\hbar} t}c_n\phi_n(\vec r)$. Because an expression that starts with the ansatz, ends being a superposition.
And the ansatz is just a solution of the TDSE, which as you said, if you take several solutions, you can have a new solution that is a superposition of these, and cannot be expressed as an ansatz
@ACuriousMind Then it ain't my fault for being confused for so long xD
There's the classic one, $\pi_{TM} : TTM \to TM$, built as a tangent bundle, and there's the pushforward one, which is $(\pi)_* : TTM \to TM$, with $\pi : TM \to M$
and the canonical flip is an isomorphism between the two
Basically if TTM has coordinates (x,y,v,w), it acts as $$j(x,y,v,w) = (x, v, y, w)$$
Basically to identify the fact that you do derivatives twice, I guess
So that y and v are redundant, somewhat
But apparently this identification breaks down if you're dealing with non-geometric forces
if the generator of rotations and the rotation operator (let's say both in the z-direction) both measure z-spin, is it useful to still deal with both concepts?
or like maybe i should say correspond to z-spin*
i guess i am wondering if two operators in quantum can correspond to the same observable and if so, why would we want two operators
oh wait sorry... Okay, so in Townsend QM, there is the $S_z$ operator which corresponds to z-spin. There is also the $J_z$ operator which is said to generate rotations about the z-axis and which turns out to produce the same eigenvalues that $S_z$ produces when acting on $|+z\rangle$ and $|-z\rangle$ in a 1/2-spin system
if we also write orbital angular momentum as $L$ you'll often see $J = L+S$, and these distinctions are relevant when you think e.g. about the angular momentum of electrons in higher shells of atoms
"We will see that a convincing case can be made that we should identify this operator $J_z$, the generator of rotations about the z-axis, with the z component of the intrinsic spin angular momentum of the particle..."
like, you can randomly postulate that all particles transform in some representation of an $\mathfrak{so}(3)$ and call the generators of that algebra $S$ and then you need to show that that $\mathfrak{so}(3)$ really is the rotation algebra and not some other random internal symmetry
okay wait to check my understanding, so you are saying that 1) we begin without knowing what specific operator would be related to spin. 2) let's now run through an argument to relate spin to rotations
also so in what you just said, a particle is represented by a state, then you can consider generators of rotational symmetries which live as representatives in so(3)?
and those generators spit out eigenvalues? or am i compeltely off base
i guess the state would be represented by a function defined on a sphere
and then you consider all elements of so(3) that are symmetries of that function, then condense into generators?
i guess i am wondering about the comment you made. To me it seems like if something is an eigenoperator of a particular eigenstate can be related to symmetry because if an eigenoperator acts on a corresponding eigenstate, the state itself does not change, it is just scaled by some phase factor
like S_z acting on $|+z\rangle$, i am calling S_z an eigenoperator with respect to $|+z\rangle$ since it will act on the ket and produce an eigenvalue times that ket
but maybe this isn't a defined concept because it is not useful xD
@SillyGoose there are in general much more operators that powers of a single generator that have a fixed state as their eigenstate - consider that in a matrix representation where that state is a basis vector this just means the matrix of the operator has $n-1$ zeros and one $c$ in the column corresponding to that basis vector and there is no further restriction on the matrix
also, to my understanding spin is introduced via 5 stern-gerlach experiments with slightly increasing complexity. It's like silver atoms of this state go in, silver atoms of this state come out basically in different S-G set ups
so the thing here is this - you have, for some reason, considered "spin", i.e. an intrinsic bunch of operators $S_i$ that look exactly like generators of rotation but have nothing a priori to do with rotation in coordinate space
the question is: Are these operators meaningfully related to rotation in coordinate space? The answer is yes, in the S-G experiment spin couples to the magnetic field much like classical orbital motion of a charge would and the Einstein-de Haas effect demonstrates that total angular momentum is only conserved if you consider both orbital angular momentum and whatever spin is both as angular momentum
hm well now i'm confused again. In Sakurai a couple pages into chapter 3, he seems to replace J_k with S_k because they both satisfy the same commutation relations
well, you only have to do all these arguments if you want to be, uh..."philosophically careful"? I mean, we know spin works like angular momentum and not every intro text to QM wants to retread a lot of nitpicky arguments about what being a "rotation" really means? Just telling students it's a rotation and moving on is a defensible pedagogic strategy in my eyes :P
it'll be a while until they encounter another set of SO(3) operators that's not rotation (i.e. weak isospin) and if you're lucky they've forgotten about your parlor trick by then :P
when I say SO(3) I don't mean "the group of rotations" I mean "a group isomorphic to the group of rotations", it's a mathematical name, not a statement about the physical meaning of the group/algebra
but this is a distinction that requires a certain amount of care and abstraction to introduce and I'm saying intro QM sometimes doesn't want to bother with that so you'll just see them identifying spin with rotation without any explicit argument and moving on
or they'll make the argument but not carefully set it up (which sounds to me like what you got from Townsend) and just leave the reader confused why the argument was necessary at all :P
also an aside, since a hilbert space is a vector space, then you can create a group structure with the vectors of that hilbert space under vector addition (I think). are there meaningful quotient groups that you can make out of this group?
For the Space of the square integrable functions, one can say that it has the structure of a Hilber space, and that the functions must be regular enough. What does it mean for a function to be regular, in this case, or in general?
who says that? The space of square integrable functions is literally just the space of square integrable functions with no further regularity conditions
part of the requirements for a Hilbert space is that the zero vector is the only vector with zero norm
but the naive space of square-integrable functions does not obey that: Any function that is non-zero but only on a set of measure zero (e.g. the infinitely many functions $f_c(x)$ defined by $f_c(x) = 0$ for $x\neq 0$ and $f_c(0) = c$ for any $c\in\mathbb{C}$) has zero "norm" because its integral is just zero
in that context, it makes sense to ask whether there is something that gives you a statement like "every equivalence class contains at least one function that is continuous", and that leads to a highly technical line of questioning that results in Sobolev embeddings (See en.wikipedia.org/wiki/Sobolev_inequality).
@ACuriousMind ......lol...tf am I supposed to do now
In about 4 months, I have an oral exam, which includes quantum mechanics I, II, and statitical mechanics. So now I am reading QM I, and trying to understand every single thing, so if a question will come up, I'll know about it
And for just personal curiosity, how would you recommend me to start with all the above mentioned topics? Something I could do little by little in my free time
Also, my articulation might not be the best,so sorry in advance
If we consider the $n$ dimensional vector/state/hilbert space of a system. Now we consider a an subspace of dimension $k$ (k<n). Then the following expression (if I can say so): $P_q=\su_{i=1}^q |\phi_i \rangle \langle phi_i|$ is a projection operator. Then I can consider the projection of $\Psi$ in the subspace $H_q$.
But, if, for the sake of simplicity, we consider a discrete non-degenerate basis
Usually the nr. of basis vectors = nr. of component for a state vector (when we try to make a representation of the state,as a vector). A subspace has fewer basis vectors, and the states in the subspace, can be represented with vectors, which have less component then a vector of the vector space, in which the subspace belongs.
When we do the projection of $\Psi$ that belongs to $H$, in the subspace $H_q$, are we effectively not considering some components?
is it kosher to say that $\int \hat{A^*} \hat{B^*} \Psi_m^* \psi_n dx = \int \hat{B^*} \psi _m^* \hat{A} \psi_n dx$? i do not think so but I see it in a calculation
a notion of subspaces is important in the context of entanglement of systems, but cruically the "projection" of a state to a subsystem is not a simple projection onto a subvectorspace, but a partial trace
don't write it as an integral but as the inner product it is and this is just $\langle A^\dagger B^\dagger \Psi_m,\psi_n\rangle = \langle B^\dagger\Psi_m,A\psi_n\rangle$, i.e. the definition of the adjoint
okay, so the thing i am confused about is in the LHS, $\hat{A}$ is acting on $\hat{B}$ and $\psi_m^*$, so normally if there was no $\hat{B}$, I see why we can move it, but why does the $\hat{B}$ not change anything?
@imbAF the partial trace simply isn't a projection - it is not a linear operator that sends state vectors the state vectors because it can turn pure states of the whole system into mixed states of a subsystem, see e.g. this answer of mine
@Relativisticcucumber why would the $B$ change anything?
$B^\dagger\psi_m$ is just a vector in the Hilbert space, and by the definition of the adjoint we have $\langle Av,w\rangle = \langle v,A^\dagger w\rangle$ for any two vectors $v,w$
@ACuriousMind Since it is mostly focused in pure/mixed states and these come later in my notes, and I don't won't to spread into to many things simultaneously, I'll save this link for when I reach lectures regarding this, read it, and probably ask you again xD.
so if i have a function that is an eigenfunction in hilbert space and I act on it using a hermitian operator, this resulting state/function will remain in hilbert space, right?
if you have something that somehow results in something that's not in the Hilbert space, what you have isn't really an operator (there are subtleties for unbounded operators but I'll only get into that if necessary)
@Relativisticcucumber the definition of Hermiticity is just that $A=A^\dagger$ for an operator $A$
If we consider a discrete set of square integrable functions {$$u_i(\vec r)}. If $(u_i,u_j)=\int u_i(\vec r)^*u_j(\vec r)d^3r=\delta_{ij}$ this would be the condition for orthogonality. One can describe this as the inner product of two different vectors, in the Hilbert space where every two functions of the set {$$u_i(\vec r)} are orthogonal. The condition for completeness is: $\sum_i u_i(\vec r)^* u_i(\vec r')=\delta(\vec r- \vec r')$. How do you describe what is going on here?
abstractly, the completeness condition is just that the sum of projectors is the identity operator, $\sum_i \lvert u_i\rangle\langle u_i\rvert = \mathbf{1}$
If this set {$$u_i(\vec r)} is in the end, because of orthonormality proven to be a basis of a vector space, what are these : $|\vec r\rangle$,$|\vec r'\rangle$
?
I am aware that $\langle \vec r|u_i\rangle=u_i(\vec r)$
behind that is even more terrible math about generalized eigenvectors and rigged Hilbert spaces :P
their usual usage in physics is just to say that they're eigenvectors of the position operator and form a "continuous basis" via $\psi = \int \psi(x)\lvert x\rangle\mathrm{d}x$ and not think all too hard about what that means or how it is compatible with the existence of countable bases :P
sure, because you're asking about all the things in QM that don't really make sense once you think about them!
and the reason they don't make sense is because the textbooks are lying to you in order to not have to do 2 years of math to even describe the free quantum particle
Anyway,so we establish a baseline here, as to how much I understand up to this point
We have this set {$u_i(\vec r)$}
orthogonality is easily proven
Now, in order to prove completeness, I jump to the Hilbert space of the position operator, and somehow use it's basis kets, which is a continuous basis?
Is this what is happening, in order to prove completeness ?
I doubt that. What was presented to you is that a basis is a set $u_i(r)$ that fulfills orthogonality and completeness
I really doubt anyone said that in practice you would get such sets and then have to prove orthogonality or completeness by some sort of computation
in practice, you usually use the spectral theorem to say "the eigenfunctions of this operator form an orthogonal basis" (where "this operator" is any operator for which you can use the spectral theorem, e.g. bounded self-adjoint) and then you solve the eigenvalue equation for that operator to get $u_i(r)$ and you then automatically know that they fulfill the orthgonality and completeness relations, no need to prove them explicitly
(because you either proved this in full generality when you proved the spectral theorem or, more likely, you believe in the spectral theorem after having proved it in the finite-dimensional case :P )
idk, maybe is should take some things on face value and not think to much, but, it's not that i am deliberately thinking about that. It just happens to cross my mind
Do you have to keep looking at the site to look at the comments and respond to them?
