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So here is the abstract approach: $$\langle \psi | H | \psi \rangle = \frac{1}{5}\bigg( \langle \phi_1 | H | \phi_1 \rangle + 2\langle \phi_1 | H | \phi_2 \rangle + 2\langle \phi_2 | H | \phi_1 \rangle + 4 \langle \phi_2 | H | \phi_2 \rangle \bigg) \,.$$ Now you know that $H|\phi_1\rangle = E_1 |\phi_1 \rangle$ and $H|\phi_2\rangle = E_2 |\phi_2 ... 1 For valid probabilities, use the normalized wave function. If all you care about is the shape of the probability distribution (i.e., which values are more likely than others, and relatively speaking how much more likely are they?), you can save time by working with the un-normalized function. 0 You can think of spin abstractly as in PipperChip's answer. A system's fundamental physics is not changed if we choose to rotate the co-ordinate system we use to describe Nature with. What this means is that, for an isolated system, you can describe its evolution by the minimisation of a Lagrangian which is unchanged if you transform co-ordinates by ... 1 Just some links.. The electron is comprised of two spherical scalar waves, one inward and one outward. A phase shift of the inward wave occurs in the wave-center region near where$\tau=0$, and spin appears as a required rotation of the inward wave in order to become the outward wave. This requirement is a property of 3D space termed spherical rotation. ... 2 At the level of particles, most people don't think of "spin" in terms of actual movement, but as something else entirely. Spin is often approached as a mathematical property only. This is because some things do not make a ton of sense if you think of electrons spinning about some axis. If you interpret spin as an electron (or other particle) actually ... 1 If you suppose that$\psi$is a real function, then as suggested by @Danu: $$\int_{-\infty}^\infty \dfrac{\partial^2{\psi^*}}{\partial x^2}\dfrac{\partial{\psi}}{\partial x}\, \rm{d}x = -\int_{-\infty}^\infty \dfrac{\partial{\psi^*}}{\partial x}\dfrac{\partial^2{\psi}}{\partial x^2}\, \rm{d}x,$$ and this gives zero, since$\psi=\psi^*$and we always ... 0 Honestly, the argument you're making here is a mess - the question is based on bad premises. So let me show you how to do it properly, and hopefully that will resolve your confusion. You're right that in order for a wavefunction to be normalized, it must satisfy $$\int_\text{all space} P(x)\mathrm{d}x = \int_\text{all space} \psi^*(x)\psi(x)\mathrm{d}x = ... 0 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. 1 Short answer The reason why a physical quantity such as probability is given by |\Psi|^2 rather than some other function of \Psi is geometry, namely Pythagoras' theorem. If you have a vector which points from the origin to the (\hat x,\hat y,\hat z) coordinates (x,y,z), then the length \ell is given by \ell^2=x^2+y^2+z^2. Why is this the ... 3 Let us consider the famous double-slit experiment with photons. With the usual set-up, we denote the number of photons passing through by N and we will denote the number of photons which hit the film between y and y + \Delta y by n(y). The probability that a photon will be detected between y and y+ \Delta y at a time t is given by: ... 3 I think the answer is "because it works". Early in the development of QM, that interpretation was given to the wave function, and over the decades it has proven to be a useful interpretation. It works. Additionally, the fact that it is possible to define an associated current, and that there is a quantum mechanical expression that guarantees that ... 4 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 ... 1 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 ... 1 Note: ChocoPouce's answer is the same as mine but is more mathematical. You have a (spherically symmetric) probability density distribution \rho in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say r = 0.1\mathrm{nm}? In other words, how much of this ... 1 The link uses a spherical shell element which is 4\pi r^2 \mathrm{d}r and has the dimension of a volume (r^2 is a surface and \mathrm{d}r is a length. The wavefunction of the ground state is spherical, if it weren't the calculation should have been made using a spherical volume element such as r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi. There ... 0 If E< V(x) everywhere, and if we assume that the kinetic energy operator T=\frac{p^{\dagger}p}{2m} is a (semi)positive operator, then the TISE implies$$ 0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0, $$which is impossible. Here H=T+V is the Hamiltonian operator. 0 This particle with have an unphysical wave function which blows up (as can be quite easily derived). Therefore, in quantum mechanics, we do not have any particles with E<V_\text{min}. 3 From my limited knowledge of this subject, I would say that a non-normalizable wave-function wouldn't really make any physical sense. Remember, the wave-function is a function whose value squared evaluated between two points represents the probability that the particle will be found between those two points. So, the restriction that wave functions be ... 0 Note that the problem in your link considers a particle with E>V_0. This means that k_2 is real, and thus e^{ik_2x} is bounded. On the other hand, if you speak in general, then for E<V_0 k_2 must be imaginary since kinetic energy is negative in the right side, and thus e^{ik_2x} is real and unbounded behaving like real exponential. In this ... 0 For the raising operator case: We know that \hat{a}_+^\dagger = \hat{a}_- (Don't forget that you operator acting on your conjugate is daggered) Therefore <\psi_n|\hat{a}_- \hat{a}_+|\psi_n> = |c_n|^2 <\psi_{n+1}|\psi_{n+1}> = |c_n|^2 But [\hat{a}_-, \hat{a}_+] = 1 and when you expand the commutator out you get \hat{a}_- ... 1 I believe you must have misunderstood something in the other textbooks but there is no way to know since I do not have a copy of the book mentioned. The wave comes from -\infty to x and the solution should be normalisable this way. 0 I will give a try at the answer. First I will say what can be said without knowing what \psi_0 and \psi_1 are. First I will normalize \Psi_0. \Psi_0 becomes \frac{1}{\sqrt{5}}(\psi_0 + 2\psi_1). As innisfree says, we have$$\hat{P}\Psi_0 =\frac{1}{\sqrt{5}}(\hat{P}\psi_0 + 2\hat{P}\psi_1) $$. To get \langle{\hat{P}}\rangle, we just calculate ... 0 You say that you know the results of$$ \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 ... 0 My initial equation was correct by I had neglected to include the modulus signs. Thanks to @jazzwhiz for pointing it out subtly. 1 The lecturer's answer assumes p can be positive or negative, and your answer assumes p=\sqrt{2mE} is positive. The factor of \exp{(-iEt/\hbar)} is just the time-dependent part of the separated solution. 1 The current density formula you consider is not right, because \frac{d(\psi\psi^*)}{dt} is always 0. The current density is defined as \bf{j}=(i\hbar/2m)(\psi\nabla\psi^*-\psi^*\nabla\psi). In your one dimensional case change \nabla to \partial/\partial x, you will get your answer. A more physical definition of current density operator is this: ... 4 You will not easily find the second one because in a sense it is trivial. Here is the single slit and double slit pattern from wikipedia . If you try to detect which slit the particle went through you get two single slits except for some experiments that are very careful not to disturb too much the wave functions of the setup. If you go to a lab that ... 8 The effect you are describing in your question is known as wave-particle duality and is a form of complementarity, it has been observed in various experiments. Realisations of Wheelers delayed choice thought experiment are what I find most interesting. In a delayed choice experiment the particles are not measured before they go through the slits but ... 0 Oh, I think I've figured it out... I sort of realised this just as I posted my question : From my earlier intros to quantum mechanics I know that p = mv = \hbar k. By simple substitution I can then obtain a final answer of : j = v\lvert{A}\rvert^{2} I'll wait for someone to confirm my line of thinking. 0 The speed only affects if there is an acceleration. If the atom is moving at a constant speed, you can do a galilean transformation, move with the atom, and it will be at rest. The two simplest reasons for an atom to accelerate are collision with other atoms and bombardment. The first one happens everytime you have a gas or a liquid. The atoms wiggle ... 2 From the Schroedinger equation,$$i\hbar\frac{\partial\psi}{\partial t}=\hat{H}\psi$$simply divide by i\hbar:$$\frac{\partial\psi}{\partial t}=\frac{1}{i\hbar}\hat{H}\psi=-\frac{i}{\hbar}\hat{H}\psi\\$$where we used$$ \frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i $$Now note that you've incorrectly applied the derivative. In quantum ... 1 They have used the Schrödinger equation:$$i\hbar\frac{\partial \psi}{\partial t} = \hat{H} \psi$$And hence:$$\frac{\partial \psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\psi$$And since 1/i = -i:$$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$Similarly for \psi^* we take the complex conjugate:$$\frac{\partial \psi^*}{\partial t} = ... 1 To arrive at the 2nd term of the third line, they have simply employed the SE equation, $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$ Re-arranging the above equation gives: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Therefore by replacing$\frac{\partial \psi}{\partial t}\$ in the second line with ...
This problem could be done more simply through the application of linear algebra. You want to prove that $$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$ The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that \begin{aligned} \langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= ... 5 First, a note about the Hamiltonian and its time derivatives. I think that it is misleading to write that the Hamiltonian H = i\hbar\frac{d}{dt}, $$although the time-dependent Schrodinger equation is of course$$ H\psi = i\hbar\frac{d}{dt} \psi. $$To evaluate e.g. \frac{d}{dt}H you should consider H=H(p, q, t), rather than H = i\hbar\frac{d}{dt}. ... 1 Part 1: As described in the text book, \Psi is a linear combination of its eigenfunctions. a_n is just a scalar multiple of that state, which basically tells you how much goes into that state, its the probability amplitude of that state. Therefore the probability to be in state \Psi_n is equal to \|a_n\|^2 = a_n a^*_n. Part 2: It should be clear ... 1 One model is to say that the atom is in an impenetrable spherical box, and solve for the wavefunctions. See Y P Varshni Accurate wavefunctions for the confined hydrogen atom at high pressures J. Phys. B: At. Mol. Opt. Phys. 30 No 18 (28 September 1997) L589-L593. The Fermi Contact Term (electron density at the nucleus) greatly increases as the size of the ... 2 |\Psi|^2=\Psi^*\Psi=(c_1^*\Psi_1^* + c_2^*\Psi_2^*)(c_1\Psi_1 + c_2\Psi_2)=c_1^*\Psi_2^*c_1\Psi_1+c_2^*\Psi^*c_2\Psi_2 + c_1^*\Psi_1^*c_2\Psi_2 + c_1\Psi_1c_2^*\Psi_2^* = |c_1\Psi_1|^2+|c_2\Psi_2|^2 + c_1\Psi_1c_2^*\Psi_2^* + (c_1\Psi_1c_2^*\Psi_2^*)^* = |c_1\Psi_1|^2+|c_2\Psi_2|^2 + 2Re(c_1\Psi_1c_2^*\Psi_2^*) Than substituting 2.9, you get 2.10. 0 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 ... 1 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} 0 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. 3 The general solution is the first one you gave,$$\psi(x,t) =Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)}.$$To get to the second one, you need to impose appropriate boundary conditions, which essentially means having no incoming particle flux from +\infty. Alternatively, you can see the solution$$\psi_k(x,t) =Ae^{i(kx-\omega t)}$$as general enough to ... 0 The ostensibly sad reality is that any such \psi_{x_o} isn't really a position eigenfunction. Try acting on it with the position operator. You simply don't get x_o times the wavefunction. That said, there are important and meaningful ways in which it is nearly a position eigenfunction. Notice that such a Gaussian is extremely narrow for very small ... 6 There is a fundamental theorem already conjectured by von Neumann but proved just at the end of 1900 by Solèr (in addition to a partial result already obtained by by Piron in the sixties) which establishes (relying on the theory of orthomodular lattices and projective geometry) that the general phenomenology of Quantum Mechanics can be described only by ... 4 To add to @Luboš Motl's great answer, I just want to mention the connection to the electric current and electric charge density from electrodynamics, which you may be more familiar with. Note that probability is unitless, so probability density has units of 1/Volume and probability current has units of 1/area*time. These are the same units as electric ... 12 The total probability of all mutually excluding alternatives must always be 100%, so it is conserved. The conservation law in the spacetime tend to be "local", so just like for the charge conservation, we may derive the conservation of the probability in Schrödinger's equation from the local continuity equation$$ \frac{\partial \rho}{\partial t} + \mathbf ...