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yes, good example will be solution of free particle.Where solution is like a plane wave solution hence such sols do not represent physically accepted states.this is the reason why any problem related to free particle should have a initial wave function which can be normalized other wise we cannot proceed further.


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In Mahan may-particle physics P15: (for bilinear Hamiltonian)It is only necessary to find the eigenvalues of the Hamiltonian matrix. Usually the matrix is of infinite dimensionality. But one may often diagonalize it exactly for many problems. Computers allow very accurate solutions for any case of interest. If all Hamiltonians had only bilinear ...


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"projection operators commute → they're the same" Are you sure he said this predicate ? or it is your own consequence? However, it is not true ! Consider two dimensional X-Projector And Y-Projector , they commute but they are not the same!


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A complete set of eigenstates spans the whole space, not just the subspace the projection operators project on. In this set of eigenstates you also have a basis of the subspace belonging to the eigenvalue 0.


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A wavefunction can be nowhere continuous. It is enough that it belongs to $L^2(\mathbb R)$, so, in general, no regularity conditions are imposed on values attained at every given point of $\mathbb R$. It is only required that $\int_{\mathbb R} |\psi(x)|^2 dx < +\infty$. (Regularity conditions can be imposed when the wavefunction is required to belong to ...


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The wavefunction must be either normalizable or the limit of a sequence of normalizable functions which in general are known as distributions (generalizations of functions). A well known example of a distribution is the Dirac delta "function," $\delta(x)$. If the spatial wavefunction is $\psi=\delta(x_0)$, then the momentum wavefunction will be of the form ...


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From my limited knowledge of this subject, I would say that a non-normalizable wave-function wouldn't really make any physical sense. Remember, the wave-function is a function whose value squared evaluated between two points represents the probability that the particle will be found between those two points. So, the restriction that wave functions be ...


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This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? Your puzzlement arises because you are putting the cart in-front of the horse. The cart is the theoretical model of quantum mechanics and the horse is the data. As ...


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This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate. Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. ...


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To have emission (or absorption) of photons you must have a Hamiltonian that includes those degrees of freedom also. If your system consists of (a) the electromagnetic field and (b) a hydrogen atom, you can specify the state with (a) for each frequency, the number of photons with that frequency and (b) the state of the hydrogen atom, in your favorite way, ...


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Conservation of energy. If we measure the energy of an atom, we will always report an eigenvalue, because we are forcing it into an eigenstate (this is something like the quantum mechanical definition of measurement). Now suppose that we measure the energy of an atom twice, before and after it emits a photon. For conservation of energy to hold, the energy ...


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The idea here is increasingly complex depending on how deep into modern physics you want to delve, but also key to understanding quantum mechanics. So, I'll give a bit deeper explanation than it seems you've seen, but there's plenty more. It's understood that a photon acts both as a particle and a wave. As a particle it has an amount of energy associated ...


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For the raising operator case: We know that $\hat{a}_+^\dagger$ = $\hat{a}_-$ (Don't forget that you operator acting on your conjugate is daggered) Therefore $<\psi_n$|$\hat{a}_- \hat{a}_+$|$\psi_n$> = $|c_n|^2$ $<\psi_{n+1}|\psi_{n+1}>$ = $|c_n|^2$ But [$\hat{a}_-$, $\hat{a}_+$] = 1 and when you expand the commutator out you get $\hat{a}_- ...


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I will give a try at the answer. First I will say what can be said without knowing what $\psi_0$ and $\psi_1$ are. First I will normalize $\Psi_0$. $\Psi_0$ becomes $\frac{1}{\sqrt{5}}(\psi_0 + 2\psi_1)$. As innisfree says, we have $$\hat{P}\Psi_0 =\frac{1}{\sqrt{5}}(\hat{P}\psi_0 + 2\hat{P}\psi_1) $$. To get $\langle{\hat{P}}\rangle$, we just calculate ...


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You say that you know the results of $$ \hat P \psi_0\\ \hat P \psi_1 $$ in which case simply write $$ \Psi = \psi_0 + 2\psi_1\\ \hat P \Psi = \hat P \psi_0 + 2 \hat P \psi_1 $$ though you might need to check your normalisation. PS. I'm not sure that I understood your question. You write that $\hat P$ is a matrix but $\psi$, $\Psi$ are functions; you seem ...


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In finite dimensions, $$\langle\psi_n|U^\dagger U|\psi_m\rangle$$ extracts the $(n,m)$ component of $U^\dagger U$ in basis $\{|\psi_n\rangle\}$. Since the result is $\delta_{nm}$, so $U^\dagger U = I$. In infinite dimensions, you can apply double-integrals $$\int_n \int_m |\psi_n\rangle(\cdot)\langle\psi_m|\; dm\; dn = I$$ on both sides, separate the ...


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Starting with the first two identities you can actually show that the last expression is satisfied not just for the basis vectors but for any two vectors that are elements of the space spanned by the basis vectors, i.e. $\langle\eta|U^{\dagger}U|\theta\rangle = \langle\eta|\theta\rangle$. Therefore $U^{\dagger}U = I$.


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This problem could be done more simply through the application of linear algebra. You want to prove that $$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$ The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that $$\begin{aligned} \langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= ...


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It is very common to abuse the notation here, so I'll try to clarify a bit. The state of a physical system can be described by an abstract vector $\left|\psi\right\rangle$, which is an element of a Hilbert space. The wavefunction, $\psi(x)$ is the representation of that vector in the position basis, $\psi(x)\equiv\left\langle x | \psi\right \rangle \equiv ...


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You need to apply the operator first and then evaluate the integral: $⟨P⟩_ψ = i\hbar\int{\psi^*(x)\frac{d\psi(x)}{dx}dx}$


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You are right that $\langle A\rangle_{\psi}$ is a number and not an operator. However, people often write just a number when they actually mean the identity operator times that number. So in the right hand side, $\langle A\rangle_{\psi}$ should actually be $\langle A\rangle_{\psi} \mathbf{1}_H$, where $\mathbf{1}_H$ is the identity operator on your hilbert ...


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Short Answer. Different quantum systems often share the same Hilbert space. In particular, as we'll see in a moment, whether or not their Hilbert spaces are the same (isomorphic) depends only on the dimensions of these spaces. Details. There are two key theorems that together, give a decisive answer to this question. The first is from finite-dimensional ...


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They are not normalisable because they either come from, or extend to infinity. This essentially means that the probability density blows off and gives non-physical results.


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For Hermitian matrices eigenvectors corresponding to different eigenvalues are orthogonal. This guarantees that not only are the eigenvalues real, expectation values are too.


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There are many physically fundamental properties that would not hold if the completeness requirement were dropped. First of all the spectral theorem for self-adjoint operators would not hold. So, an observable (let us assume to deal with observables with pure point spectrum) would not have a complete set of eigenstates. There is a fundamental idea in quantum ...



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