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8

The reason operators correspond to measured values has to do with what happens when you connect a measurement apparatus to the system under observation. Suppose the Hamiltonian of the system by itself is $H_S$ and the Hamiltonian of the measurement apparatus by itself is $H_M$. When $M$ is physically connected to $S$, we get a additional "interaction" term ...


7

Not equal, but equivalent, in the sense that they have the same effect on the wavefunction in question. More precisely, the book is using a slight abuse of terminology. Taking momentum as an example, it's not really the case that the dynamical quantity of momentum is equivalent to the operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$, because a number ...


4

Everything follows is valid at least when the various operators appearing below are defined on the dense subspace finitely spanned by the standard vectors $|n\rangle$, $n=0,1,2,\ldots.$ From $[a,a^\dagger]=I$ and $[A,BC]= B[A,C]+[A,B]C$, by induction, one easily finds that $$[a,(a^\dagger)^n] = n(a^\dagger)^{n-1}\:.$$ If $\alpha_k \in \mathbb C$ and ...


4

By definition, a value is in the continuous spectrum of $A$ if it not an eigenvalue, but the range of $A-\lambda I$ is a proper dense subset of the Hilbert space. There is nothing in this definition distinguishing separable spaces, or precluding operators in them from having continuous spectrum, and indeed some do. There is an equivalent definition in terms ...


2

A time reversal operator is an anti-unitary operator, which can be expressed as: $\mathcal{T}=UK$ where $K$ denotes complex conjugate and $U$ is a unitary operator. In case of spinless particles, $U$ is chosen to be Identity. Thus $\mathcal{T}=K$. If the system has $\mathcal{T}$-reversal symmetry: $$ KH\psi=HK\psi $$ which leads to: $$ ...


2

As argued by von Neumann, the measuring process has many properties that resemble those found in the theory of operator algebras. For instance, if you have an instrument, you can measure something, say the length of a table, to get a certain value $x$ within experimental errors. What you can now do is relabel the ticks of your instrument according to a ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$If your state is in an eigenstate of the energy operator then the answer is that you'll get the same value for the energy every time you measure the particle's energy. That is the reason why the energy eigenstates are also called stationary states. On the other hand you can also have a superposition of energy ...


2

Wick's theorem tells us that $$ \mathcal{T}(\phi_1\dots\phi_N) =\ :\phi_1\dots\phi_N: + :\text{pairwise contractions}:$$ where $:\ :$ is normal ordering. Immediately from the definition of normal ordering (all annihilators to the right, all creators to the left), the expectation value of anything that is normal-ordered and not a constant vanishes because the ...


1

If we consider local operators $M_A$ and $N_B$ acting on Alice's and Bob's part, respectively, then it holds that $[M_A,N_B]=0$, i.e., we can use this property in proofs involving local operations. Note that conversely, however, commutativity need not imply locality (to start with, there need not even be a tensor product structure).


1

As far as my experience goes (theoretical nuclear physics), they are pretty much equivalent. To make this clearer, I'll try to give an intuitive example for bosons: $$\langle cc{}^\dagger\rangle = cc{}^\dagger-:cc{}^\dagger:$$ which becomes $$\langle cc{}^\dagger\rangle = cc{}^\dagger-c{}^\dagger c = \left[c,c{}^\dagger\right]$$ so in this case $\langle ...


1

Hint: Think of way to write the Hamiltonian as $$\hat{H}~=~ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x)~=~\hat{B}^{\dagger}\hat{B} ~\geq~0 $$ for some first-order differential operator $$\hat{B}~=~a(x)\frac{d}{dx} +b(x) ,$$ with suitable functions $a(x)$ and $b(x)$. Here the potential $V(x)$ is given by formula (1). Can you see what the operator ...


1

Note that you only can identify the numbers with the operators when you hold a plane wave function. For an arbitrary wave functions, you can't identify the operator $(\hbar/i)(\partial / \partial x)$ with some number $p$, but you can hold the operator to have some significant meaning related to this numbers. If you apply $$ \langle P\rangle = \int dx \, ...


1

No, that is not quite correct. The Schrödinger equivalence says that if the system has a definite energy, then this energy can only be an eigenvalue of the system's hamiltonian $\hat H$. There is no requirement, however, for the system to have a definite energy; if the energy is undefined then an energy measurement may return different (eigen)values on ...


1

If you have an observable, that is like a symmetric matrix. So its eigenvectors span the whole space so you can form a basis for the whole space. So the span of the eigenvectors really is the whole space. What about eigenspaces? You did have to use all the eigenvectors to get the whole space, not just ones that share the same eigenvalue. If you only use ...


1

Equation (2.4.6): $T(z)X^\mu(0)\sim \frac{1}{z}\partial X^\mu(0)$ means that the RHS is the most singular term of the LHS. $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ So \begin{align*} T(z)X^{\mu}(0) & =-\frac{1}{\alpha'}:\partial X^{\nu}(z)\partial X_{\nu}(z):X^{\mu}(0)\\ & =-\frac{2:\partial ...


1

The question is really one of definition. In the math literature on self adjoint opertors the "discrete spectrum" is by definition that part of the spectrum which consists of normalizable states, while the "continuous spectrum" is that part where they are non-normalizable. It is possible to have a physical system (a random potential on the entrire real ...


1

The answer to the problem is that the operator used to change to the new frame, $U(t) = D(\alpha)$ is dependent on time through the time dependence of $\alpha(t)$ and for the purposes of the derivation this time dependence is kept quite general. This means that we cannot use the usual rewriting of the equation of motion, but eq. 81 in the lecture notes ...



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