# Tag Info

9

$\def\ket#1{\left|#1\right>}$Each of those kets represents the state of a system. In quantum mechanics the constant multiplying a wavefunction is not physically significant (multiplying a wavefunction by a number does not change the physical meaning of the wavefunction). So if $\ket{x}$ represents a wavefunction completely localized to the point $x$, then ...

9

In a nutshell: With a couple caveats: what is plotted here is really the wavefunction $-\langle x\lvert 2\rangle$ or $\langle x\lvert -2\rangle$ (so actually this graph is highly misleading in one way), and the spikes should be thought of as delta functions. But this is the basic idea.

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It means it's "the end of the line". The vacuum state is, as you correctly say, not the zero state. It has energy content, and physical meaning - it's the state with no particles. Annihilating the vacuum leaves...nothing. Trying to take a particle out of it is not possible - it gives you the zero vector, which does not represent a physical state, since it is ...

4

When dealing with the bra-ket notation, we must be clear that what we put inside the bras or kets is just a label and not something we make math with, although good labels makes visualization easier. When you write $|x\rangle$ you are considering the state of the particle in the $x$ position. (There is a problem here that this is not an acceptable state, ...

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You do factor out the 100, that is $$\langle\psi | 100 |\psi\rangle = 100\langle \psi |\psi\rangle.$$ In general, if $\psi$ is a correctly normalized state we should have $\langle \psi|\psi \rangle = 1$, hence $100\langle\psi|\psi\rangle = 100(1) = 100.$ Edit: Alternatively (more physically?) you can think of $100$ as a Hermitian operator in and of ...

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Although we can define the momentum as a self-adjoint operator in $L^2[0,1]$ as you proposed, I think it's rather artificial to think about it as having relation to momentum in the case of $L^2(\Bbb{R})$. Realize that the operator $p_1$ with domain $D(p_1)=\{\psi\in\mathcal{H}^1[0,1]\,|\,\psi(1)=\psi(0)\}$, is related to spatial translations via the unitary ...

3

I don't find this proof a good one, since the notation is messy and not very clear (not to say wrong). One proof can be given in a similar way to the one you posted in the link. Suppose the spectrum of $H$ is discrete and the set of eigenstates $\{|\phi_n\rangle\}$ constitutes an orthonormal basis with eigenvalues $E_n$, such that $E_0\leq E_1\leq ... 3 The energy eigenstates can be expressed in the form of wavefunctions as well, e.g.$\psi_n(x) \equiv \langle x | E_n \rangle$. Then, you can compute the inner product of the two wavefunctions by integrating their product: $$\langle E_n | \psi \rangle = \int_{-\infty}^\infty \langle E_n|x\rangle \langle x|\psi \rangle \, dx = \int_{-\infty}^\infty ... 2 The notation \lvert \phi \rangle \lvert \psi \rangle is shorthand for \lvert \phi \rangle \otimes \lvert \psi \rangle. What you are doing is flipping a tensor product around. Though, in general, A \otimes B = B \otimes A, it is not the case that a \otimes b = b \otimes a for a \in A, b \in B (since the left and right hand side don't even live in ... 2 The integral without delta square would converge to 1, but squaring the function somehow breaks it? Yes. Recall the sifting property:$$\int_{-\infty}^{\infty}f(x)\delta(x - a)dx = \int_{-\infty}^{\infty}f(a)\delta(x - a)dx = f(a)$$Then, it follows formally that$$\int_{-\infty}^{\infty}\delta^2(x - a)dx = \int_{-\infty}^{\infty}\delta(x - ... 2 What happens when you integrate a function multiplied by a delta function? You get: $$\int f(x) \delta (x-y) dx=f(y)$$ (Because the delta function is zero everywhere, except at zero where its integral gives 1.) So, when integrating over$\delta^2 (x-y)$we get: $$\int \delta^2 (x-y) dx=\int \delta (x-y) \delta (x-y)dx = \delta (y-y) = \infty$$ 2 The Dirac delta is defined by the equation $$\int_a^b\mathrm{d}x \,\delta(x)f(x) = f(0)$$ for$a < 0 < b$. By direct application of this definition we get $$\int \mathrm{d}x\, \delta(x - y) \delta(x - y^\prime) = \delta(y - y^\prime)$$ If we let$a < c < 0$we can write ... 2 The$\delta$function has the following property: $$\int \text{d}x\; f(x)\delta(a-x)=f(a)$$ This actually answers both of your questions. First, the non-square-integrability: $$\int \text{d}x\;\delta(x-y)\delta(x-y)=\delta(y-y)=\delta(0)=\infty$$ according to the rule above if you choose one of the$\delta$'s to be the$f$. Your second question is the ... 2 Let $$|\psi'\rangle = e^{i\theta}|\psi\rangle.$$ The probability to be in the$n$:th eigenstate of the observable$\hat O$is given by the expectation value of$P_n$where$P_n$is the projector onto the$n$:th eigenstate. Since the expectation value in the state$|\psi'\rangle$is $$\langle \psi'|P_n|\psi'\rangle = \langle \psi|e^{-i\theta} P_n e^{i\theta} ... 1 The fact that D(p_0) is not big enough to define a self-adjoint operator does not mean it is not included into the domain of the self-adjoint momentum. The choice of boundary conditions for the function is a specification of the vector, not of the operator. Once you have fixed the self-adjoint extension you are considering (choosing the proper domain, ... 1 Let's run through the variational principle very quickly. The idea is that an arbitrary state \psi can be decomposed into a sum of orthogonal energy eigenstates: \left|\psi\right> = \sum c_n \left|\psi_n\right> where \sum |c_n|^2 = 1 and H\left|\psi_n\right> = E_n \left|\psi_n\right> Then the expectation of the energy ... 1 I see you have some great answers already, but here is a different way of looking at it which might help you with regard to your comment above. This is linear variational theory, meaning you have to keep all non-linear parameters constant throughout the calculation. We wish to find values of the linear parameters C_i that minimise the variational integral: ... 1 How about:$$ \left\langle\psi\ |100\,|\,\psi\right\rangle = 100\left\langle\psi\,|\,\psi\right\rangle $$I admit to guessing though as I have not seen such notation. But, my guess is 100 as a scaler can be factored out, or multiply the ket |\,\psi\rangle and then take the inner product of the Bra and Ket (and, the factor of 100 can of course be taken ... 1 As correctly pointed out by Daniel Sank in the comment section, the key to understanding the state space in quantum field theory is the realization that it contains information about the excitations of operator-valued functions (quantum fields) of spacetime. The latter consists of one time and three spatial coordinates (at least in the context of the ... 1 a^\dagger a\! \mid \! n \rangle = n\mid \! n \rangle. For vacuum you can write a^\dagger a\! \mid \! 0 \rangle = 0\mid \! 0 \rangle, if you like. No need to write 0 solely. 1 Your claim that the derivative is in the expansion of \langle x'|\hat{p}\hat{x}|\psi\rangle acts on everythin on the right is correct. Realize that$$\langle x|\hat{p}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\langle x|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\psi(x)$$So$$\langle x'|\hat{p}\hat{x}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx'}\langle ... 1 $$\sum _nf(n)\int \mathrm{d}p\, |\left< p|n\right> |^2=\sum _nf(n)\int \mathrm{d}p\int \mathrm{d}q\, \left< n|p\right> \left< p|n\right> =2\pi \sum _nf(n)\left< n|n\right> =2\pi \cdot \sum _nf(n),$$ where I used the fact that $$\int \mathrm{d}p\, \left| p\right> \left< p\right| =2\pi,$$ which follows from$$\left< ... 1 When you change the free field$A_\mu\$ by means of a gauge transformation, you can easily see that it affects longitudinal and timelike degrees of feedom. Since observables are gauge invariant, those degrees of freedom cannot be physical.

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