My question doesn't go much beyond the title:  Why does $$\left | \psi \left ( x,t \right ) \right |^{2}$$ give us the probability density of something appearing at a certain location?  I understand that $$\left | \psi \left ( x,t \right ) \right |^{2} =  \psi \left ( x,t \right )^{*}\psi \left ( x,t \right )  $$ where $\psi \left ( x,t \right )^{*}$ is the complex conjugate, but I still don't understand how multiplying these two variants of the wave equation gives us a probability of a location.
60340
Is the  a fundamental , or can it be inferred from unitary evolution?
24398
1) Up until now, during practical work sessions, I always used these formulas for uncertainty propagation:  if $C = A+B$  or  $C = A-B$ $$\Delta C = \Delta A + \Delta B$$ if $C = AB$ or $C = \frac{A}{B}$ $$\frac{\Delta C}{C} = \frac{\Delta A}{A}  + \frac{\Delta B}{B}$$ if $C = A^m$ $$\frac{\Delta C}{C} = |m|\frac{\Delta A}{A}$$   These formulas are derived from the expression for the differential of a function of multiple variables:  if $C = f(A,B)$ $$ dC = \frac{\partial C}{\partial A}dA + \frac{\partial C}{\partial B}dB \Rightarrow \Delta C = \frac{\partial C}{\partial A}\Delta A + \frac{\partial C}{\partial B}\Delta B$$  and I think that makes sense because the value C is just a function of two variables that happen to be measurements.  2) But this morning, I was told that this is wrong and I should actually use these instead:  if $C = A+B$  or  $C = A-B$ $$(\Delta C)^2 = (\Delta A)^2 + (\Delta B)^2$$ if $C = AB$ or $C = \frac{A}{B}$ $$\left(\frac{\Delta C}{C}\right)^2 = \left(\frac{\Delta A}{A}\right)^2  + \left(\frac{\Delta B}{B}\right)^2$$  which are apparently derived from a general formula:  $$(\Delta C)^2  = \left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 +   \hspace{0.3cm}...$$  or   $$\Delta C  = \sqrt{\left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 +   \hspace{0.3cm}...}$$  3) So:   Why is this formula better?  Where does it come from? What does it actually represent? (Do I recognize the shape of a norm in that last formula?)  What is wrong with the formula of the differential?   I'm asking a lot, but the propagation of uncertainty always confused the ship out of me.
116108
The general formulation of quantum mechanics is done by describing quantum mechanical states by vectors $|\psi_t(x)\rangle$ in some Hilbert space $\mathcal{H}$ and describes their time evolution by the Schrödinger equation $$i\hbar\frac{\partial}{\partial t}|\psi_t\rangle = H|\psi_t\rangle$$ where $H$ is the Hamilton operator (for the free particle we have $H=-\frac{\hbar^2}{2m}\Delta$).  Now I have often seen used spaces like $\mathcal{H}=L^2(\mathbb{R}^3)$ (in the case of a single particle), but I was wondering whether this is correct or not.  In fact shouldn't we require to be able to derivate $\left|\psi_t\right&gt;$ twice in $x$ and thus choose something like $\mathcal{H} = H^2(\mathbb{R}^3)$?   If we treat directly $\psi(t,x) := \psi_t(x)$, shouldn't we require them to be in something like $H^1(\mathbb{R};H^2(\mathbb{R}^3))$? i.e., functions in $H^1(\mathbb{R})$ with values in $H^2(\mathbb{R}^3)$, e.g. the function $t\mapsto\psi_t$.
55596
I'm looking to calculate the expected values of a coherent state (of a harmonic oscillator) evolving in time. I know that the $x$ and $p$ expectation values are as in classical motion, but I'm wondering about $x^2$ and $p^2$.  Let's say I'm starting with the coherent state $| b \rangle$, with $b \in \mathbb{R}$, so the wavefunction is the ground state displaced by $bx_0\sqrt{2}$:  $$\psi_b (x) = \psi_0(x-bx_0\sqrt{2})$$  Or similarly the Wigner function will be   $$W_b(x,p) = W_0(x-bx_0\sqrt{2},p)$$  Now I know the expected values of $x$ and $p$ are classical:  $$\langle x(t) \rangle = bx_0\sqrt{2}\cos(-\omega t)$$ $$\langle p(t) \rangle = bp_0\sqrt{2}\sin(-\omega t)$$  But what about $\langle x^2(t) \rangle$ and $\langle p^2(t) \rangle$ and ?
17610
Some websites and textbooks refer to $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ as the correct formula for the uncertainty principle whereas other sources use the formula $$\Delta x \Delta p \geq \hbar.$$ Question: Which one is correct and why?  The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.  This is the mentioned derivation:     $\sin \theta = \frac{\lambda}{\Delta y}$      assuming $\theta$ is small:      $\tan \theta = \frac{\lambda}{\Delta y}$      de Broglie equation:      $\lambda = \frac{h}{p_x}$      $\Rightarrow \tan \theta \approx \frac{h}{p_x \cdot \Delta y}$      but also:      $\tan \theta = \frac{\Delta p_y}{p_x}$      equalize $\tan \theta$:      $\frac{h}{p_x \cdot \Delta y} \approx \frac{\Delta p_y}{p_x}$      $\Rightarrow \frac{h}{\Delta y} \approx \Delta p_y \Rightarrow \Delta p_y \Delta y \approx h$      $\Rightarrow \Delta p_y \Delta y \geq \frac{h}{2 \pi}$      $\Rightarrow \Delta p \Delta x \geq \frac{h}{2 \pi}$   So: Which one is correct and why?
53792
If I have a 3D dispersion relation   $E=E(k_x, k_y, k_z)$ I have an equation for the density of states, which is   $D(E)=\frac{1}{\nabla_k E}\int\frac{dS}{(2\pi)^3}$  1) I am confused about the integral. What am I integrating over, exactly? It must have units of the $k$ vector, so it's a k-space integral, but what is the surface that is relevant?  3) Where does that equation come from, how is it derived?
77016
The Schrödinger equation in non-relativistic quantum mechanics yields the time-evolution of the so-called wavefunction corresponding to the system concerned under the action of the associated Hamiltonian. And this wavefunction is, in general, complex, and its modulus squared yields the probabilities observed experimentally. Though, perhaps, this question has been asked many times, I am wondering if there is a direct physical interpretation - something that physically corresponds to - the wavefunction. Or is it just an intermediate calculational tool to arrive at the appropriate predictions for experimental outcomes, and nothing more? Of course, things like superposition and interference effects follow from the complex nature of the probability amplitude. So there must be something physical about it. What is it? Or are we not supposed to ask that question? Is it because the probability amplitude is complex that we have difficulty in relating it to something physical? Can we do quantum mechanics without complex numbers?
51101
In non-relativistic QM, on one hand we have the following relations:  $$\langle x | P | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial x} \psi(x),$$  $$\langle p | X | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial p} \psi(p).$$  On the other hand, despite the similarities, the relations cannot be directly applied to energy and time:  $$\langle t | H | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial t} \psi(t),$$  $$\langle E | T | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial E} \psi(E).$$  Just wondering, how can one mathematically prove that the "classical time" (which means no QFT or relativity involved), unlike its close relative "position", is not a Hermitian operator?  I ask your pardon if you feel the question clumsy or scattered. But to be honest, if I can clearly point out where the core issue of the problem is, I may have already answered it by myself :/
16985
We know that for a position variable $x$ and momentum $p$, the uncertainties of the two quantities are bounded by  $$\Delta x \Delta p \gtrsim \hbar$$  Now, this is usually first explained with $x$ being a simple linearly measured position and $p$ being linear momentum.  But it should apply to any good coordinate and its conjugate momentum.  It should, for instance, apply to angle $\phi$ about the $z$ axis, and angular momentum $L_z$:  $$\Delta \phi \Delta L_z \gtrsim \hbar$$  The thing is, $\Delta \phi$ can never be greater than $2\pi$.  I mean, you have to have some value of $\phi$ and $\phi$ only runs from 0 to $2\pi$.  Therefore  $$\Delta L_z \gtrsim \hbar/\Delta \phi \geq \hbar/2\pi$$  But, uh-oh!  This means it is impossible for $\Delta L_z$ to be zero, and we should never be able to have angular momentum states with definite $L_z$ values.  Of course, it doesn't mean that.  But I have never figured out how this is not in contradiction with the Schroedinger eqn. calculations that give us states with definite values of $L_z$.  Can anyone help me out?  One answer I anticipate is that $\phi$ is sort of "abstract" in that if you chose your origin at some other point you will get completely different values of $\phi$ and $L_z$, and ipso facto, usual considerations don't apply.  I don't think this will work, though.  Consider a "quantum bead" sliding around on a rigid circular ring and you get the exact same problem with no ambiguity in $\phi$ or $L_z$.  (Well, there will be some limited ambiguity in $\phi$, but still, there won't be in $L_z$.)
49222
In quantum mechanics, one makes the distinction between mixed states and pure states.  A classic example of a mixed state is a beam of photons in which 50% have spin in the positive $z$-direction and 50% have spin in the positive $x$-direction.  Note that this is not the same as a beam of photons, 100% of which are in the state $$ \frac{1}{\sqrt{2}}[\lvert z,+\rangle + \lvert x,+\rangle]. $$ It seems, however, that at least in principle, we could describe this beam of particles as a single pure state in a 'very large' Hilbert space, namely the Hilbert space that is the tensor product of all the $\sim 10^{23}$ particles (I think that's the proper order of magnitude at least).  So then, is the density operator a mathematical convenience, or are there other aspects of quantum mechanics that truly require the density operator to be a 'fundamental' object of the theory?  (If what I mean by this is at all unclear, please let me know in the comments, and I will do my best to clarify.)
100771
The wavefunction $\Psi(x,t)$ for a free particle is given by  $$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$  This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature?  Why then do we use free particles $\psi(\vec{x}) = e^{ikz}$, for example, in scattering theory?
43426
Most sources say that  acts like a joint phase-space distribution in quantum mechanics and this is justified by the formula   $$\int_{\mathbb{R}^6}w(x,p)a(x,p)dxdp= \langle \psi|\hat{a}\psi\rangle$$  as this allows us to compute the expectation value for the operator $\hat{a}$ corresponding to the physical quantity $a$. However, without the knowledge of this formula, how did Wigner come up with this definition:  $$w(x,p)=\frac{1}{(2\pi)^3}\int_{\mathbb{R}^3}\psi\left(x-\frac{v}{2}\right)\psi^*\left(x+\frac{v}{2}\right)e^{iv \cdot p}dv \quad ?$$  I would be greatly indebted for any mathematical motivation that anyone could provide. Similarly, I also wonder at the mathematical motivation for the Weyl quantization.
6953
Could someone please provide an explanation for the origin of ? (Ideally someone who I have been speaking with on the h-bar)  Any advanced maths beyond basic calculus will most probably leave me at a loss, though I do not mind a challenge! Please assume little prior knowledge, as over the past few days I have discovered that much of my understanding surrounding the process as virtual particle pairs is completely wrong.
55922
1.  Why is the wave function complex? I've collected some layman explanations but they are incomplete and unsatisfactory. However in the book by Merzbacher in the initial few pages he provides an explanation that I need some help with: that the de Broglie wavelength and the wavelength of an elastic wave do not show similar properties under a Galilean transformation. He basically says that both are equivalent under a gauge transform and also, separately by Lorentz transforms. This, accompanied with the observation that $\psi$ is not observable, so there is no "reason for it being real". Can someone give me an intuitive prelude by what is a gauge transform and why does it give the same result as a Lorentz tranformation in a non-relativistic setting? And eventually how in this "grand scheme" the complex nature of the wave function becomes evident.. in a way that a dummy like me can understand.  2.   A wavefunction can be thought of as a scalar field (has a scalar value in every point ($r,t$) given by $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{C}$ and also as a ray in Hilbert space (a vector). How are these two perspectives the same (this is possibly something elementary that I am missing out, or getting confused by definitions and terminology, if that is the case I am desperate for help ;)   3.   One way I have thought about the above question is that the wave function can be equivalently written in $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{R}^2 $ i.e, Since a wave function is complex, the Schroedinger equation could in principle be written equivalently as coupled differential equations in two real functions which staisfy the Cauchy-Riemann conditions. ie, if $$\psi(x,t) = u(x,t) + i v(x,t)$$ and $u_x=v_t$ ; $u_t = -v_x$ and we get $$\hbar \partial_t u  = -\frac{\hbar^2}{2m} \partial_x^2v + V v$$ $$\hbar \partial_t v  = \frac{\hbar^2}{2m} \partial_x^2u - V u$$ (..in 1-D) If this is correct what are the interpretations of the $u,v$.. and why isn't it useful. (I am assuming that physical problems always have an analytic $\psi(r,t)$).
108406
Suppose $\mathcal{B}=\{|0\rangle, |1\rangle, |2\rangle, ... \}$ is the energy eigen-basis of a quantum simple harmonic oscillator. I want to create the state \begin{equation} |\Psi\rangle = c_{0}|0\rangle + c_{1}|1\rangle + c_{2}|2\rangle + ... \end{equation} Besides the normalization condition \begin{equation} \langle\Psi|\Psi\rangle = |c_{0}|^2+ |c_{1}|^2+ |c_{2}|^2+...=1, \end{equation} are there any other mathematical constrains on the complex amplitudes $c_0,c_1,c_2,...$?  Also, is there a protocol, perhaps from quantum optics, for creating an arbitrary $|\Psi\rangle$ as defined above?   For example, the complex amplitudes which define the state can be taken as \begin{equation} \{c_n\}=\{ \frac{1}{\sqrt{\zeta\left(3\right)}}\frac{1}{\left(1+n\right)^{3/2}} \}, \end{equation} where $\zeta\left(3\right)\approx 1.202$ is given by the . In this case, the normalization condition holds. Also, we get a nice finite value for the mean energy \begin{equation} \langle\Psi|\hat{n}|\Psi\rangle \approx0.019, \end{equation} where $\hat{n}=\hat{a}^{\dagger}\hat{a}$ is the number operator (i.e. the Hermitian operator corresponding to the eigen-basis $\mathcal{B}$). However,the variance in the energy \begin{equation} \text{Var}[\hat{n}]=\langle\Psi|\hat{n}^2|\Psi\rangle-\langle\Psi|\hat{n}|\Psi\rangle^2  \end{equation} is infinite. Can such a state exist in nature?
95947
How is the    $$ i\hbar\frac {\partial }{\partial t}\psi=H{\psi }$$  derived?
96541
Using Minkowski signature $(+,-,-,-)$, for the Lagrangian density   $${\cal L}=\partial_{\mu}\phi\partial^{\mu}\phi^{\dagger}-m^2\phi \phi^{\dagger}$$  of the complex scalar field, we have the field  $$\phi(x)=\int{\frac{d^3 \vec{p}}{2(2\pi)^3\omega_\vec{p}}}(a(\vec{p})e^{-ipx}+b^{\dagger}(\vec{p})e^{+ipx}).$$  I'm trying to now find an equation for the $a(\vec{p})$ and $b(\vec{p})$ (with the final goal of finding an expression for $[a(\vec{p}),b^{\dagger}(\vec{q})]$ using $[\phi(x),\Pi(y)]$).   What I should note is that we are considering all of this in the Schrodinger picture (t=0) so I suppose the very first thing to do is change all the $x$'s to $\vec{x}$ right?  The strategy I'm struggling to implement and failing at many points along the way:   Find the momentum $\Pi^{\phi}(x)=\frac{\partial L}{\partial \dot{\phi}}=\dot{\phi^{\dagger}}$. Add some combination of $\phi(x)$ and $\Pi(x)$ to get rid of one of the creation/annihilation operators.  Do an inverse Fourier transform to find $a(\vec{p})$ in terms of $\phi(x)$, for example.    None of the major text books seem to actually carry this through, and instead write something like "and it's easy to show…". However I don't find it too easy, especially part 3. as I'm not a Fourier transform expert.   Could anyone either direct me somewhere where the above is computed explicitly (in more than 2/3 lines), or help me understand each of the 3 steps above?  (I realise it is easy to find a reference where this is done for the real scalar field, in which case we have $a(\vec{p})$ and $a^{\dagger}(\vec{p})$. Even still, I find it hard to follow parts.)