Tag Info

New answers tagged

1

This wavefunction is an idealization of a wave function that has very very steep, but not discontinuous, behavior at $0$ and $a/2$. You are right, the wavefunction as written is not a proper wave function. That's a good observation. Often in physics the math is much easier if a real situation is modeled by one that is close to it, but mathematically more ...


2

$$ <\hat{A}> = \int \psi^*(x)\hat{A}\psi(x) dx $$ now $\hat{A}\psi(x)=a(x)\psi(x)$ so, $$<\hat{A}>=\int a(x)|\psi(x)|^2dx$$ Let $|\psi(x)|^2dx = d \mu$ now $\int|\psi(x)|^2dx=\int d\mu = 1$ $$ <\hat{A}> = \int a(\mu)d\mu $$


0

Let us use bra-ket notation. Suppose that the operator $\hat A$ has discrete and bounded eigenvalues $\mathcal A =\{A_1, A_2, \ldots\}$ with eigenkets $|A_1\rangle, A_2\rangle,\ldots$. The eigenkets form a complete orthogonal set since $\hat A$ is symmetric. Then any ket $|\psi\rangle$ can be expanded, $$|\psi\rangle = \sum \psi_n |A_n\rangle.$$ Since the ...


0

Observables are associated with linear operators, not measureable functions, so how can we talk about the expectation of a linear operator? The "expectation of a linear operator" is a term from quantum theory. It is defined by the integral $$ \int \psi^*(x)\hat{A}\psi(x) dx $$ or similar. The meaning of this number is not necessarily the same as ...


1

There are two answers to this. One answer simply points out that the probability of the jth outcome specified by the Born rule $p_j = tr(\rho\hat{P}_j)$, where $\hat{P}_j$ is the projector onto the jth outcome, satisfy the axioms of probability: http://mathworld.wolfram.com/ProbabilityAxioms.html. Another answer is that the Born rule can be explained ...


2

Since you want a bit of mathematical rigor: A quantum state is a self-adjoint positive trace class operator on a Hilbert space with trace 1. This is called density matrix $\rho$. In its simplest form, given $\psi\in \mathscr{H}$, $\rho$ is the orthogonal projector on the subspace spanned by $\psi$. Let $E_\rho(\cdot):D_\rho\subset\mathcal{A}(\mathscr{H})\to ...


2

The quantum state and the state vector is exactly the same thing. The word "vector" is meant to emphasize that quantum states form a vector space, the so-called Hilbert space. Because they're synonyma, the question in the first paragraph is a meaningless talkative tautology, like "Is a car meant to denote just one vehicle or are many cars meant to be many ...


1

You are maybe making confusion between the action of an observable (operator), and the measurement process. In particular: $A\psi$ is simply a vector of the Hilbert space. In my opinion it has not much sense of talking about "initial" and "terminal" state because you are not looking at a dynamical situation. If you want to know the average value of an ...


1

Regarding the first part of your question,they have just inserted a complete set of basis because $|\phi>$ is a basis in some infinite dimensional Hilbert space (in your case), therefore sum (integral) of all such bases is identity on the Hilbert space. Note that in second part $\langle r|\phi\rangle$=$\phi(r)$.


1

So first of all, the first equation you gave is only correct, if the $|ø\rangle$ form a basis. It has nothing to do with "in which basis they are". The easiest way to understand this is probably with a 3D vector-analogy. So if $b_i$, $i=1\dots3$ form a basis, for any vector $v$ it is legitimate to write $$v=\sum_{i=1}^3 b_i (b_i\cdot v)$$ There, the ...


1

No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


3

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


0

I think in quantum mechanics we assign to each system a specific Hilbert space i.e. if systems are different then their Hilbert spaces are different. Is this true? If not why? For differernt system I mean their hamiltonians are different. I found really interesting the question and @joshphysics, however I desagree with his point of view. It is true ...


0

Answer to question one : The Principle of Quantum Mechanics by R. Shankar page 149 reads "Barring a few exceptions, the schrodinger equation is always solved in a particular basis. Although all basis are equal mathematically, some are more equal that others. First of all, since H = H(X,P) the X and P basis recommend themselves....The choice between the two ...


0

You have to clarify the term "basis" in infinite dimensions. By definition, finite-dimensional spaces have a finite basis (so that, for instance, $\mathbb{R}^n\approx\mathbb{R}^m$ iff $n=m$). But for infinite-dimensional spaces, there are various types of bases (assuming you can take infinite sums). A natural one (using finite sums) is the Hamel basis, but ...


5

First, the term "basis spaces" isn't standard in quantum physics but let us assume that we understand what the sentences approximately mean. Second, the momentum is continuous (not quantized) if the position space is noncompact (infinite). The momentum only becomes quantized if the position space is compact (or periodic), and indeed, it's been ...


2

An observable is a self-adjoint operator $\mathcal{O}$ on the Hilbert space of states $\mathcal{H}$. The spectral theorem tells us that such an operator has an orthonormal basis of eigenvectors in $\mathcal{H}$, if it is compact. If it is not compact, we have to "enlarge" the Hilbert space to something called rigged Hilbert space or Gelfand tripel. A ...


0

Unfortunately not every operator, which is used to get an observable gives you enough elements to make a base of the Hilbert space. As far as I know every eigenstate is orthogonal to each other. But the “length” could be arbitrary. Mostly used eigenstates in quantum mechanics are normalized for better usage.


4

The problem is that entities like the Dirac field are not observable. We cannot measure the value of the electron field at a particular point in space; it is impossible to experimentally find a specific configuration of an electron field in space. What we can measure are results of scattering experiments and cross sections. For this reason the QFT is ...


3

The Projection Operator $\delta_{\varepsilon\, a}$ is $1$ for some particular $|a\rangle$ and is $0$ on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by \begin{equation} \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = ...


4

The position operator in three dimensions is a vector operator, which, as in mpv's answer, acts as $$ \hat{\mathbf r}\psi(\mathbf r)=\mathbf r \psi(\mathbf r). $$ Note that the hat denotes an operator and not a unit vector. The definition of a vector operator is somewhat tricky, and it is indeed startling that an operator can have vector eigenvalues. The ...


3

The position operator in 3D is a vector in 3D: $$ \hat{\bf r} \psi({\bf r}) = {\bf r} \psi({\bf r}) $$ See here.


1

1) It's basically just a description of what a particle's doing. This involves its energy level, the probability of where to "find it", its spin, etc. 2) I'd say it's somewhat analogous to describing a ball rolling down a hill in terms of its energy, where you describe its potential energy, kinetic energy, and rotational energy. But there's no "equivalence" ...


1

The wavefunction, $\psi$, is the most complete possible description of a particle (or collection of particles). From the wavefunction, one can calculate a probability distribution for the the outcome of any measurement. Remember, quantum mechanics is probabilistic - one cannot make exact predictions for all measurable quantities, even if you know the ...


1

I will give you an example from classical wave theory. Take Melde's experiment with a rope where you can have different modes in this rope. Those modes are discrete, like say a photon in a box. And you have a dispersion relation that relates the wave number $k$ to the frequency $\omega$. Therefore, if we try to make an analogy with a photon in a box, the ...


6

Dirac being opaque and hard to follow? Well I never... In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates $|\varepsilon\rangle$ and some general state $|P\rangle$. Then I can write $|P\rangle$ as \begin{equation}|P\rangle = \sum ...


0

This is why I think Schrodinger introduced this thought experiment. You can explain why Schrodinger developed this thought experiment by first considering the double slit experiment. You fire a photon at a double slit, and the photon goes through the double slit device and then strikes a screen. If you don’t try to determine which slit the photon has gone ...


1

You are not wrong, the symmetries of a theory are essential to finding the right space of states. The space of states must carry a representation of all symmetries of the theory (though it might be the trivial one). For example, for a quantum system that is invariant under rotation (think of the hydrogen atom), the fact that we must represent the rotation ...


5

Isn't your Ex 3 such an example? In any event, you might have trouble producing a pure photon state $\left|\,001\right>$, but you should have no trouble creating a coherent state. Just turn on your laser pointer. The coherent state is a superposition of single photon states, summing over all single photon number states. You can't impose a cutoff. ...



Top 50 recent answers are included