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5

But how can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? I interpret that your question basically asks how do we ...


0

So I see your whole question as this: How can we guarantee that two solutions $\boldsymbol {\psi_1}$ and $\boldsymbol {\psi_2}$ to the time-dependent equation don't have $\boldsymbol {\psi_1(x,0)} = \boldsymbol {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffith's method is unique? This is a really basic ...


3

The form of the solution shown by Griffiths is not unique. That means that there exist cases where a basis $\{\psi_n(x)\}$ will reproduce $\Psi$ as $$ \Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}, $$ but there exists a second, different basis $\{\varphi_n(x)\}$ which (with different coefficients) also reconstructs $\Psi$: $$ ...


1

The derivative of $x f(x)$ is $x f'(x) + f(x)$, and here you're integrating $k \int_{-\infty}^\infty dx ~ x ~ f'(x)$ for some constant $k$, and some complicated function $f$. When you integrate this by parts, you raise $f' dx$ and lower $x$ to find:$$k \int_{-\infty}^\infty dx ~ x ~ f'(x) = k \left[x ~f(x)\right]_{-\infty}^{~\infty} - k \int_{-\infty}^\infty ...


4

Perhaps its a little clearer if you shorten the contents of the brackets (and lets drop the constants too): $$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$ $$ = \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} ...


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 ...


0

Looking at simple cases might help. You can have a vector $\vert\Psi\rangle$ in a Hilbert Space and represent it as a pure state such as $\vert\Psi\rangle\langle\Psi\vert$ and you could do the same for the vector $\vert\Phi\rangle$ in a Hilbert Space and represent it as a pure state such as $\vert\Phi\rangle\langle\Phi\vert.$ You could also imagine that ...


5

The inner product is always between two (ket) vectors. However, let $\mathcal H$ be the space in which they live. This is a complex vector space with a Hermitian inner product. The inner product defines a map $\mathcal H\to\mathcal H^\vee$, where $\mathcal H^\vee$ is the dual of $\mathcal H$, and consists of linear functionals on $\mathcal H$. The map is ...


-3

You can take inner product of any 2 vectors if they have same dimension. However, we consider the physics problem in quantum mechanics, where inner product of bra and ket vector (of course is complex conjugate of bra vector) of wave function means probability density of finding particle. Inner product of 2 arbitrary vectors (with same dimension) is no ...


3

The visualization method you choose is directly and completely determined by the information you need to see regarding your state. For the states of a single bosonic mode, there are multiple different visualization methods, and they all have their pros and cons. In particular, there is a direct trade-off between the amount of information you can display on a ...


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 ...


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).


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 ...


0

If you read the next paragraph, it seems that Dirac means that the eigenvalues are non-degenerate, i.e. for a set of simultaneous eigenvalues $\xi_1^\prime, \xi_2^\prime\dots$ there is exactly one corresponding eigenbra. There are no two distinct bras which are eigenbras of all the operators $\xi_1, \xi_2\dots$ and have the same eigenvalues for all of them. ...


0

After thinking about it, as long as the original eigenvalues are non-degenerate it should be possible to have the new Hamiltonian be represented by a differential equation of arbitrarily high order. The key is that the projection operators $P_n$ onto the eigenfunctions exist in the algebra generated by the original Hamiltonian $\hat{H_0}$. For instance say ...


4

http://arxiv.org/abs/quant-ph/9607007 discusses necessary conditions on $T$ (more precisely, on its singular values) for $\rho$ to be positive. They don't seem to derive sufficient conditions, however. The basic idea is that one can perform a rotation $U_A$ and $U_B$ on the two qubits, respectively, which correspondingly transforms $r\mapsto O_Ar$, ...


0

Here taking $|0\rangle$ and $|1\rangle$ as orthonormal basis for 2 dimensional hilbert space. Now $|00\rangle ,|01\rangle,|10\rangle,|11\rangle$ are orthogonal to each other ( take the inner product of any two it will be zero, eg. $\langle 00|01 \rangle= \langle0|0\rangle \langle 0|1\rangle =0 $ ). Thus any vector of a 4 dimensional dimensional hilbert space ...


0

$\newcommand{\ket}[1]{\left| #1 \right>}$Note that you can write a tensor product as a matrix in the following way: $$A\otimes B = \begin{pmatrix} A_{11}B & \ldots & A_{1m}B\\ \vdots & \ddots & \vdots\\ A_{m1}B & \ldots & A_{mm} B \end{pmatrix}$$ where $A$ is a $m\times m$ matrix and $B$ is a $n\times n$ matrix. Notice that ...


0

If the electron is confined to the $x-y$-plane, it's $z$-position is fixed, i.e. certain, and hence the $z$-momentum infinitely uncertain by the uncertainty relation. That paragraph is trying to say that, if the magnitude of $\vec L$ is larger than $L_z$, then $\vec L$ is not fixed, and if angular momentum is not fixed, i.e. conserved, then the motion does ...


0

If the particle is known to be in the $xy$-plane, then $\Delta z = 0$, and so (by the uncertainty principle) $\Delta p_z = \infty$. Roughly speaking, this means that there's a good chance that the particle's momentum would be greater than "escape momentum", and thus that it could escape from the hydrogen atom. (I'm kind of dubious about this argument ...


1

I think you could work it like this: $X_+ ={1 \over \sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) =a (\begin{matrix} 1 \\ 0 \end{matrix} ) +b(\begin{matrix} 0 \\ 1 \end{matrix} ) $.where $X_+$ is the eigenvector on the positive axon of $S_x$ Solve and find a,b and there you are. Note also that you can write a general spinor as $(\begin{matrix} cos\theta ...


1

Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation $U$. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we ...


1

Are you sure that's what the book is asking you to find? $\hbar/2$ is the eigenvalue of the $S_{x}$ operator corresponding to spin up, but it is not part of the state vector. If the question is really asking you to express the $\mid S_{x};+\rangle$ ket in the $S_{z}$ basis, then you're nearly correct, just a minor sign error: $$\mid S_{x};+\rangle = ...


0

Your second option is correct. It is a simultaneous eigenket of all three commuting operators. And that is how it should be interpreted. To be explicit, the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$ Is a triple of equations $\hat{x}\mid \mathbf x'\rangle = x'\mid \mathbf x'\rangle$ $\hat{y}\mid \mathbf ...


0

$\newcommand{\ket}[1]{\lvert #1 \rangle}$The Schrödinger equation does not say what you claim it does. The time-independent Schrödinger equation is an eigenvalue equation for the Hamiltonian operator $$ H \ket{\psi_E} = E \ket{\psi_E}$$ where solving for $\ket{\psi_E}$ for a concrete $H$ gives us the occuring eigenstates of $H$. It does not say that these ...


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 ...


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

The word separable (the property) has a precise and detailed meaning (about lack of factorizability) when discussing multiparticle states. But that meaning might not be the meaning in the context you consider. For instance the paper http://arxiv.org/abs/1302.7188 argues that separability (the principle) is not related to Bell's inequality. And in that paper ...


2

The notion of separability of a state has a precise and simple meaning: In natural language, a separable state of a system that has several subsystems is a state to which a unique state of every subsystem is associated. In classical mechanics, all states are separable in this sense - given two configuration spaces $Q_1,Q_2$, the configuration space of the ...


2

In general, the integral $$ V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$ is the integration of the identity over the space the measure $\mu$ is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.) Normalizing the measure means sending $\mu \mapsto ...


-1

you can make any state if you properly choose bases as, $|\Psi> = \alpha|0> + \beta |1>$ where $|0>$ and $|1>$ are assumed as the complete bases. In this case, $\alpha$ and $\beta$ are the probability amplitude to observe 0 or 1 respectively. Similarly if the system can be written by a continuous bases, like space $|x>$, any state can be ...


2

In the case of an infinite superposition of eigenstates it becomes more complicated but we can still write a general expression for it. If $$\psi(x) = \sum_{n=0}^\infty a_n \phi_n(x)$$ where the $\phi_n$ are the eigenstates of the Hamiltonian. The time dependent wavefunction will look like: $$\Psi(x,t) = \sum_{n=0}^\infty a_n \phi_n(x) T_n(t)$$ where $T_n = ...


-1

In that case, you have a sum of overlaps between a pair of functions having same eigenvalue index. The cross overlap will be zero because of the orthogonality of basis set which is very important.


1

I am not at all an expert of quantum gravity, but I think you have misunderstood the point. As I understand it, the point is not necessarily having distributions as quantum Hilbert space vectors, but having a distributional "configuration space", i.e. distributions as the domain of the function(al)s that are the quantum vectors. While in QM (i.e. for ...



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