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6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...


4

In general no. If you think of $\left|a\right>$, $\left|b\right>$ and $\left|c\right>$ as vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$, you can say that $\vec{a} = \vec{b} \times \vec{c}$, so that $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$, but this doesn't imply that $\vec{b} \cdot \vec{c} = 0$.


4

If you want to experimentally create a qubit, you need some actualization. One example is the z-component of the spin of a spin-1/2 particle such as an electron. There are two independent states which can be denoted $\left(\matrix{ 1 \\ 0}\right)$ and $\left(\matrix{ 0 \\ 1}\right)$ and each of which can be produced by orienting a stern-gerlach device in ...


3

There is indeed a way to construct the projection-operators, when you only know the operator itself and its eigenvalues. The derivation can be found in Julian Schwingers "Quantum Mechanics: Symbolism of Atomic Measurement" and leads to $$ P_j = \prod_{i\neq j} \frac{A-a_j}{a_i - a_j}, $$ where the product goes over all distinct eigenvalues $a_i$ of $A$. ...


3

Let's take the canonical commutation relations (CCR), in their exponentiated form (Weyl's relations): $$V(\eta)T(q)=e^{-i\eta\cdot q}T(q)V(\eta)\; ,$$ where $\{V(\eta)\}_{\eta\in \mathbb{R}^d}$ and $\{T(q)\}_{q\in \mathbb{R}^d}$ are objects of a given normed algebra with involution. This is a very general notion, that is nowadays taken as the definition of ...


3

Assuming that your operator has a spectrum consisting of isolated points you can look for all the independent solutions of the eigenvalue equation $$(Q-\lambda I)\xi = 0$$ Let these solutions generate a vector space $V_\lambda$ and then compute the dimension of $V_\lambda$. If it is greater than 1 then the eigenvalue $\lambda$ is degenerate.


3

Vectors (which kets are) don't have adjoints, they have duals. Whether the dual of $\lvert n_1,\dots,n_n\rangle$ is denoted by $\langle n_1,\dots,n_n\rvert$ or $\langle n_n,\dots,n_1 \rvert$ is entirely conventional.


2

For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question. When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for ...


2

$\renewcommand{ket}[1]{|#1\rangle}$ As others have pointed out, you can go whole hog and solve the characteristic equation $\text{det}(\hat{Q} - \lambda \hat{I})=0$ and find repeated solutions for $\lambda$. However, there is a simpler, more physically intuitive way to hunt for degeneracy: look for symmetry. When an operator $\hat{Q}$ has a symmetry there ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\o}{\mathbf 1}$ Physicists are lazy people we all are! When you see something like $S_{1z}+S_{2z}$ you should really think of the following: $$S_{1z}+S_{2z} \equiv S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z}$$ Since you get tired of writing it over and over you just shorten it by an addition ...


2

A state $\psi$ corresponds to an energy $E$ if: $$H\psi = E\psi$$ Clearly, if there is a state $\psi = \sum_i c_i \psi_i$ where $H\psi_i = E_0\psi_i\ \forall i$, then $$H\psi = \sum_i c_i H\psi_i = \sum_i c_i E_0\psi_i = E_0\psi$$ A linear combination of states with the same energy value again has the same energy value. Now consider $\psi = c_1\psi_1 + ...


1

Since $V(x)$ is bounded from above we have three possibilities. Either it oscillates at infinity with an upper bound, or it asymptotes to a constant $<E$ or it diverges to $-\infty$. Since we are interested in $x\rightarrow\infty$ we may average the oscillation in the first case to the mean, and if it diverges then we concern our selves with the leading ...


1

The coefficients $a_k$ quantity the projection of the state $|\psi\rangle$ onto the $k^\mathrm{th}$ basis state. So if you measure in that basis you would expect $|a_k|^2$ of the time to measure state $|\phi_k\rangle$. If you change the basis you will need to recalculate the projection coefficients. If you are in a non-orthogonal basis the projections are ...


1

As Phoenix87 said, you need to solve the asociated eigenvalue problem. $\lambda$ is an eigenvalue if and only if $\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator $\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side: ...


1

Eigenvectors are undetermined up to a scalar multiple. So for instance if c=1 then the first equation is already 0=0 (no work needed) and the second requires that y=0 which tells us that x can be anything whatsoever. This is fine, and correct, as $x=re^{i\theta}, y=0$ is a fine solution when c=1. If $c\neq 1$ then none of the equations are already 0=0 so ...


1

Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...


1

It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...


1

For a normalised linear combination of (orthogonal) states like this, the probability of measuring one of them is the absolute square of the coefficient in the combination: If $$\Psi = a_1\psi_1+a_2\psi_2+...$$ where $|a_1|^2+|a_2|^2+... = 1$ then $$P(\psi_1) = \left|a_1\right|^2, P(\psi_2) = \left|a_2\right|^2$$etc. Slightly more generally, if you know ...


1

I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace ...


1

I can understand your confusion. Let me start by saying that this is the "correct" way of writing down what you want to have in physics notation: $$ \langle e_i|\otimes \mathbf{1} \sum_j |e_j\rangle\otimes |w_j\rangle = \sum_j \delta_{ij} |w_j\rangle $$ Note the absence of the tensor product sign. Otherwise, you will never have this reduction of dimension, ...


1

Note: There is a short summary at the bottom. This is actually also described in Nielsen&Chuang: You don't learn about general measurements, because they are completely equivalent to projective measurements + unitary time evolution + ancillary systems, all of which is described in your usual QM formalism. The Measurement Postulate Let's start from ...


1

SUMMARY OF EDITED VERSION: You cannot place any conditions on $V(x)$ and $E$ that guarantee that solutions to the time-independent Schrödinger equation are normalizable, for something of a silly reason. Initial, partial answer: If the potential is bounded below by some value $V_\text{min}$, then a solution to the time-independent Schrödinger equation ...



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