# Tag Info

7

If two states are orthogonal, this means that $\langle \psi | \phi \rangle = 0$. Physically this means that if a system is in state $|\psi\rangle$ then there is no possibility that we will find the system in state $|\phi\rangle$ on measurement, and vis versa. In other words the 2 states in some sense mutually exclusive. This is an important property for ...

6

Write an arbitrary state as $$|\Psi\rangle = \sum_{n=0}^{\infty} c_n |n\rangle \,.$$ Now apply the raising operator \begin{align} a^\dagger |\Psi\rangle &= a^\dagger \sum_{n=0}^{\infty} c_n |n\rangle \\ &= \sum_{n=0}^{\infty} c_n \sqrt{n+1} |n+1\rangle \\ &= \sum_{n=1}^{\infty} c_{n-1} \sqrt{n} |n\rangle \end{align} If $|\Psi\rangle$ is ...

5

Unitary matrices are (almost) never Hermitian, hence no finite version of the infinitesimal transformations described by the Hermitian observables will ever be Hermitian itself. The most prominent example is the time evolution $\mathcal{U}(t) = \mathrm{e}^{\mathrm{i}Ht}$, which is unitary, and not Hermitian.

3

Let $A$ be the C*-algebra of a quantum mechanical system, and suppose that $G$ acts on $A$ by symmetries through the group homomorphism $\alpha:G\to\text{Aut}(A)$. Let us further assume that $G$ is a simply connected Lie group with trivial $H^2(\mathfrak g,\mathbb R)$, $\mathfrak g$ being the Lie algebra of $G$. If $\pi$ is an $\alpha$-regular ...

3

If the eigenstates of your operator is not a orthogonal set, then your operator is not a hermitian operator, or in other words, is not an observable. Actually, non-hermitian operators "appears" all the time, but if you investigate decoherence mechanism you may note that this operators don't affect directly the classical realm. This is because you can't ...

3

Well, quantization of a classical system may not be a unique procedure. Classically, all variables commute, and, say, $\pi ~\partial_i \phi = \partial_i\phi~\pi.$ When we quantize, why should we choose $\hat{\pi}\hat{\phi}$ ordering, as OP does in his example? Why not instead the opposite $\hat{\phi}\hat{\pi}$ ordering? Or perhaps symmetrize? Or use normal ...

2

The two density matrices are expressed in a different basis of the same Hilbert space. If you compute the expectation values on any state you like you will find the same result with both $\rho_1$ and $\rho_2$. Hence by polarization you can conclude they are indeed the same operator, i.e. they're acting in the same way on the Hilbert space in question.

2

You have made an error in your calculation of ⟨p2⟩. You've evaluated two integrals, the second of which is off by a factor of two.

2

First you realize: $$\int\limits |q' \rangle dq' \langle q'|=1$$ Then using the definition of a linear operator on a bra: $$(\langle \phi |\frac{d}{dq})|\psi \rangle= \langle \phi |(\frac{d}{dq}|\psi \rangle)$$ Applying the first equation to both sides of the second one (at the right of the expression in parenthesis for the left side of the equation and ...

2

If you want to describe the way in which a mixed state transforms, this in generally expressed using non-hermitian operators, either the Lindbladian in the continous case, or e.g. the Kraus representation of trace preserving completely positive maps in the discrete case.

2

Algebraic Quantum Field Theory starts from the assumption that a quantum mechanical system can be described by a C*-algebra. But not every element in the C*-algebra has a physical interpretation, but just the self-adjoint part, which has the structure of a Jordan algebra through the product $$a\circ b = \frac{(a+b)^2 - a^2 - b^2}2.$$ The matrices you see ...

2

It's a matter of convention (basically the active v. passive distinction as mentioned in a comment by user jabirali). The punchline is that if the states transform as $|\psi\rangle \to T_g|\psi\rangle$, then operators should transform as $A\to T_gAT_g^{-1}$, and vice versa. To see why, suppose that states are taken to transform as $|\psi\rangle\to ... 1 In terms of the splitting$V=W\oplus W^{\perp}$, the projection operator$P_1=P=P_2$and unitary operator$U$are block diagonal $$P ~=~ \begin{bmatrix} {\bf 1} & {\bf 0} \\ {\bf 0} & {\bf 0}\end{bmatrix}$$ and $$U ~=~ \begin{bmatrix} U|_W & {\bf 0} \\ {\bf 0} & {\bf 1}\end{bmatrix},$$ respectively. Clearly, the two operators ... 1 Since$\lvert a_i\rangle$and$\lvert b_i\rangle$are both bases for the space$W$, there exists a unitary$U=\sum \lvert b_j\rangle \langle a_j\rvert$which maps$\lvert b_i\rangle=U\lvert a_i\rangle$for all$i$. This$U$can be naturally embedded in$V$, i.e., we can think of it as an operator$U:V\rightarrow V$. Then, $$P_2 = \sum \lvert ... 1 You are right. Think of the limit case of the Heritian identity operator I, it can be written in infinitely many ways, one for each Hilbertian basis of the Hilbert space. ADDENDUM. Referring to the EDIT, if you explicitly perform computations you see that U commutes with H and thus it is not effective as it must be. 1 When you give the form of an operator with respect to a certain basis, it would be good to explicitly state which basis you are using. The fact that you get H' = UHU^* is giving you a way to represent the same operator in a different basis. Let's say you have bases B and B' for a Hilbert space. What you have is the operator H in the basis B, and we ... 1 I) Let us for simplicity assume that the Hilbert space H is finite-dimensional. Let A:H\to H be a normal operator [and hence orthogonally diagonalizable] with eigenvalues$${\rm Spec}(A)~=~\{\lambda_1, \ldots, \lambda_n\}.$$We can uniquely decompose the Hilbert space$$H~=~\oplus_{i=1}^n H_i,$$in orthogonal eigenspaces$$H_i~:=~ {\rm ... 1 There is an error in my calculation. In fact:$\int ^{\infty}_{-\infty} e^{-kx^{2}}dx = \sqrt{\dfrac{\pi}{k}}$So $$\langle \widehat p^{2} \rangle=-2A^{2}ma\hbar [\frac{-1}{2}\sqrt{\frac{\pi}{k}}]=ma\hbar$$ given that$A^{2}=\sqrt{\frac{2am}{\pi\hbar}}$as calculated using the normalization condition. This, can be used to calculate ... 1 If you keep the order when expanding the inner product you get terms of the form$x^ip_i$which are not self-adjoint, due to the fact that$x^i$doesn't commute with$p_i$. Hence the spectrum of such operator won't lie in$\mathbb R$, nor there is any hope in finding simultaneous eigenstates for both$x^i$and$p_i\$. Expression like this are usually ...

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