New answers tagged

2

Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are not used because measurements must use real numbers, but rather because we almost always decide to use real numbers. As the OP mentions at one point, you might choose to use complex numbers to label a two-dimensional screen, and in that case you'll be able ...


3

When in doubt go back to the masters. From Dirac's Principles of QM When we make an Observation we measure some dynamical variable. It is obvious physically that the result of such a measurement must always be a real number, so we should expect that any dynamical. variable that we can measure must be a real dynamical variable. One might think ...


0

Use User Yuggib's hint for part (a); For part b), solve the equation of motion for $p$ first - it's really simple. For the equation of motion for $x$, as you realize, you need to find $$[x,\,H] \propto [x,\,p^2]=x\,p\,p-p\,p\,x\tag{1}$$ Before you plug anything in, try bringing one of the terms into the same order as the other with the help of the CCR; ...


0

Hint: OP's formula follows from a Wick-type theorem $$ \tag{1} T(f(\hat{A})) ~=~ \exp\left(\frac{1}{2}\hat{C}\frac{\partial}{\partial\hat{A}}\frac{\partial}{\partial\hat{A}} \right):f(\hat{A}): $$ between time-ordering $T$ and normal ordering $::$. Here $$\tag{2} \hat{C}~=~T(\hat{A}\hat{A})~-~:\hat{A}\hat{A}:$$ is a contraction. See e.g. this Phys.SE post ...


2

First, let's answer the questions precisely as you worded them: The point spectrum is always discrete in the sense that it consists of at most countably many points. This is true by proving the following results: a) the space spanned by all eigenvectors is a closed subspace of the Hilbert space, hence we have an orthonormal system of eigenvectors, b) two ...


4

(1) Yes, the point spectrum is countable in your hypotheses: otherwise the operator would have an uncountable set of pairwise orthogonal vectors since eigenvectors of a self-adjoint operator with different eigenvalues are orthogonal. This is impossible because, in every Hilbert space, every set of (normalized) orthogonal vectors can be completed into a ...


2

I assume you have not diagonalized the operator, else you would not be asking the question. The absolute value of the operator $\hat{O}$ is presumably $\sqrt{\hat{O}^\dagger \hat{O}}\equiv \hat{B} $, so the crucial question is what type of algorithm one would choose to evaluate the square root $\hat{B}$ of the operator $\hat{A}= \hat{O}^\dagger \hat{O}=\hat{...


4

You do the same thing you always do when applying a function to an observable - you diagonalize it, then apply the function to the eigenvalues, then undo the diagonalizing procedure. The formal underpinning of this procedure is given by Borel functional calculus.


2

I expect this to be a result of the Wick theorem. If you are considering an equilibrium situation with a quadratic Hamiltonian all odd moments will vanish. Thus you remain with even powers of your operator. If you now count the number of possible contractions you should get the correct result. P.S.: I just tried and it works indeed. Note the helpful ...


5

Whilst it is certainly true that Quantum Probability Theory (QPT) is an entirely different framework from Classical (Kolmogorovian) Probability Theory (CPT) (specifically because the event structure is non-Boolean and the random-variable structure is non-commutative), we can still identify enough formal similarity to borrow the classical terminology. In ...


0

Maybe you can be interested in another interpretation of Hermitian matrices. In a recent paper we have proposed to see them as gambles on a quantum experiment. We have then enforced rational behaviour in the way a subject accepts/rejects these gambles by introducing few simple rules. These rules yield, in the classical case, the Bayesian theory of ...


1

A quantum system can be described by a set of evolving quantum mechanical observables. This is not the same as describing a system in terms of a stochastic quantity described by a single number chosen at random. A quantum system really does have multiple values of any unsharp observable, see https://arxiv.org/abs/quant-ph/0104033. Those different versions ...


7

I'm going to try to explain why and how density operators in quantum mechanics correspond to random variables in classical probability theory, something none of the other answers have even tried to do. Let's work in a two-dimensional quantum space. We'll use standard physics bra-ket notation. A quantum state is a column vector in this space, and we'll ...


6

I believe it is misguided to think that classical probability makes sense any more than quantum mechanics, with its "peculiar" probability calculations, makes sense. I'm going to be slightly mischievous here and make a friendly attack your first paragraph: does really make sense? Of course it makes perfect sense as a measure-theoretic definition, but how ...


11

Quantum mechanics is indeed a probability theory, but it is a non-commutative probability theory. So it is not just a matter of having signed/complex measures, but really of having a non-commutative probabilistic framework. Quantum mechanics was developed, historically, before non-commutative probability theories and I think that people in probability ...


9

You could certainly model any one quantum observable as a random variable. The problem comes in when you have multiple observables, which you might attempt to model as classical random variables with some joint distribution. From this joint distribution, you can compute various probabilities (like $\textrm{Prob}(Y\neq X)$, for example), according to the ...


1

This functional analysis textbook is quite good, it even has a chapter dedicated on some of the mathematical foundations of quantum mechanics without using this name. This book is quite compact though, the classic text by Walter Rudin is more detailed. Prerequisites are topology and measure theory.


4

Preliminaries: If you "limit" your description of quantum mechanics to $L_2$ Hilbert spaces, all your bases will be discrete, both bounded or unbounded. You can have Hilbert spaces of any cardinality, but the one in "standard" quantum mechanics is $L_2$, the space of square integrable functions, which has a countable cardinality, $\aleph_0$. In this case ...


2

The fields $\pi$ and $\phi$ are quantum fields which satisfy equations of motion, given classically by $$ d_t \pi=\{\pi,H\},\\ d_t q=\{q,H\}, $$ and these classical equations are established in any text on Hamiltonian mechanics. Heisenberg equations are just the quantization of those (replacing the Poisson bracket by the commutator). It is therefore not any ...


0

There are already good answers, here are a few more ways to help see it: If you already accept that vectors can have units (like in $\mathbf{F} = m \mathbf{a}$), then you already have accepted that algebraic constructs that aren't just numbers can have units. The operator case isn't any more complicated. You've already seen examples of operators with units ...


-1

I think this can be explained if one considers the Schrödinger picture to be more fundamental. There one has to define some configuration space, and this configuration space is something which does not change in time at all - only trajectories in the configuration space change in time. And the configuration of a field is described by all the field values in ...


0

A nonunitary operator can be thought of as the sum of unitary and antiunitary operators. So let me let $T = aU + bA$, a, b = normalization constants, and we evaluate $$ \langle T\phi|T\psi\rangle = \langle\phi|T^\dagger T|\psi\rangle. $$ The product is $T^\dagger T = U^\dagger U + U^\dagger A + A^\dagger U + A^\dagger A$. The unitary operator is easy $U^\...


0

I see two sides to your question, in addition to what was already pointed out in comments: For a transformation to preserve the scalar product for any pair of kets $|\psi\rangle$, $|\phi\rangle$ in the Hilbert space, it is sufficient that it be an orthogonal transformation: Say $U$ is such a transformation. If $\lbrace |\psi_n\rangle \rbrace_n$ is an ...


0

in the Schrodinger picture operator are time independent and states evolve with time.but in the Heisenberg picture operators are time dependent that are given by unitary transformation which you mentioned. when H is time dependent then integral in the exponent will come and time ordering also matters.Dyson formula work there. In the interaction picture, the ...


2

Two ways that might help you see that operators in general must have units: The quantum Hamiltonian must have units of energy, because $\exp\left(i\,\frac{H}{\hbar}\,t\right)$ is the time evolution operator, so that the exponent is dimensionless; otherwise put: Schrödinger's equation is $i\,\hbar\,\partial_t\,\psi = H\,\psi$, so that $H$ must have the same ...


3

The creation and annihilation operators are not observables. They are obviously not hermitean because $a \neq a^\dagger$. But, regarding your question, consider the number operator $( N_k = a_k a^\dagger_k )$. As the eigenvalue of the number operator is a dimensionless number, the creation and annihilation operators must be dimensionless as well. The full ...


3

The raising and lowering operators are dimensionless. The position and momentum operators are written according to $$ x = \sqrt{\frac{\hbar}{m\omega}}q,~\frac{\partial}{\partial x} = \sqrt{\frac{m\omega}{\hbar}}\frac{\partial}{\partial q} $$ with $p = -i\hbar\partial/\partial x$ we then write the raising and lowering operators according to these ...


2

$$ [P_i,L_j]=\varepsilon_{jkl} ([P_i, X_k]P_l+X_k[P_i,P_l])=-\mathrm{i}\hbar \varepsilon_{jkl}\delta_{ki}P_l = -\mathrm{i} \varepsilon_{jil} P_l = \mathrm{i}\varepsilon_{ijl} P_l$$ Which shows that as expected $P_k$ is a vector. The lesson here is that you should use not use the same letter twice as an external index and a dummy index. Here $i, j$ are ...


0

If the parity operator commuted with the Hamiltonian, you'd be able to simultaneously diagonalize the parity operator and the Hamiltonian. That would mean you could find a complete set of eigenstates of the Hamiltonian that were also eigenstates of the parity operator. The eigenstates of the parity operator are just even/odd functions, but odd potentials ...


2

Your expression for: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\vec \sigma_1 |+\rangle\otimes \vec \sigma_2 |+\rangle_2$$ Is wrong. It sould read: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\sigma_{1x}|+\rangle_1\otimes \sigma_{2x}|+\rangle_2+\sigma_{1y}|+\rangle_1\otimes \sigma_{2y}|+\rangle_2+$$ $$\...


3

Well, you know that the eigenfunction of $\hat p$ is $\exp(ipx)$, so let's try to find what the expectation value of $x$ is, to begin with: $$\langle p | x | p\rangle = \int \exp(-ipx) x \exp(ipx) \, dx = \int x\, dx$$ and this integral doesn't exist. Then it shouldn't come as a surprise that trying to take the time derivative gives nonsense. The underlying ...


0

Update: Comments by Robin pointed out my confusion. Consider the following: $$ [x, p^2] = x p p - p p x = x p p - p x p + p x p - p p x = [x,p] p + p [x,p] = 2 i h p $$ If you plug this in the initial expression, you will get exactly what you would expect from this observable: $p/m$. But formally correct manipulations that you performed provide a ...


4

Let us write the eigenvalues as follows: $$\lambda=a+ib$$ Where $a$ and $b$ are real. By definition we must therefore have: $$\lambda^*=a-ib$$ Equating these gives us: $$a+ib=a-ib$$ $$2ib=0$$ $$b=0$$ and therefore: $$\lambda=a$$ Which is a real number.


1

The solution on part (a) should be correct, and it is solved correctly to my knowledge. However, on part (c), there may be something wrong on your understanding. Here is what I think is the correct way to solve it. What you want to calculate is the expectation value of $\hat{P}_1$, which can be expressed as \begin{align} \langle \hat{P}_1\rangle &= \...


5

No, an arbitrary operator does not represent a change of basis. And even those that can be used to perform changes of basis should not always be interpreted as such. A "change of basis" in a Hilbert space is usually meant to be a change from one orthonormal basis to another. The operators that map orthonormal systems to orthonormal systems are precisely the ...


3

The operator $aa^\dagger$ is easily seen to be nothing else than $$ aa^\dagger = a^\dagger a + 1$$ which is equivalent to $[a,a^\dagger]=1$, a commutator that easily follows from $a\sim (x+ip)/\sqrt{2}$ with some extra coefficients and $[x,p]=i\hbar$, so $aa^\dagger$ is the number operator plus one, i.e. taking values $1,2,3,4,\dots $ instead of $0,1,2,3,\...


1

I think the proof can actually work, but it needs to be formulated a bit better and it needs a bit more of explaining. Take $|\psi_n \rangle$ as a non-degenerate Eigenstate of $\hat{A}$. Then for all other Eigenvectors $|\psi_m \rangle$: $$\langle \psi_m | [\hat{A},\hat{B}]| \psi_n \rangle = (a_m - a_n)\langle \psi_m | \hat{B} | \psi_n \rangle = 0$$ So: $\...


1

For future convenience denote $$ {\hat \phi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_j} $$ $$ {\hat \chi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_{j+1}} $$ The average you want to calculate reads then $$ \langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\langle 0 | \prod_n{{\hat \phi}_n} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\...


3

You may just complete the square: $$ K = (a^\dagger+\lambda^*)(a+\lambda) - \lambda\lambda^* $$ Expand the product and subtract the last term to see that you get the same three terms. One may define $b=a+\lambda$. Then $$ K = b^\dagger b - \lambda \lambda^* $$ and $[b,b^\dagger]={\bf 1}$, so these $b$ operators are isomorphic to $a$ and the spectrum of $K$ ...


1

Essentially, separation of variables in the time-independent Schroedinger equation amounts to diagonalizing the Hamiltonian. One can see this easily by considering the case where the Hilbert space is finite-dimensional, and the Hamiltonian is a Hermitian matrix. In case of a partially continuous spectrum one gets the same, except that the sum must be ...


2

Comments to the question (v3): The main fields assumption that goes into the proof of the Wick's theorem for fields $\hat{\phi}^i\in{\cal A}$ is that their (super)commutators $$[\hat{\phi}^i,\hat{\phi}^j]~\in~ Z({\cal A}) \tag{1}$$ are central elements of the operator algebra ${\cal A}$, cf. e.g. this Phys.SE post. For free fields $\hat{\phi}^i\in{\cal A}$,...



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