# Tag Info

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When dealing with a single quantum mechanical particle, both the wavefunction and the electric field appear to belong to the familiar class of "fields", both $\mathbf{E}(x)$ or $\psi(x)$. This analogy completely breaks down when you consider multiple particles, in which case the wavefunction depends on all of the particle coordinates, i.e. ...

10

You can, in principle, measure the electric and magnetic field strength at every point in space and time. Thus, the EM field is real in the the sense that its value can be determined uniquely by measurements, and thus, also excitations of it - the EM waves - are real. You cannot, in principle, measure the wavefunction at any point. The true quantum state ...

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I am assumming you're solving the "Chemist's" Schrödinger equation, i.e. expressing the quantum state in position co-ordinates to find the shape of orbitals and probability density to find a particle at a particular position. In this case, the reason for the first derivative's continuity is the conservation of probability: we can define a probability flux ...

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If $A$ is self-adjoint, you can define $f(A)$ as a complex-valued observable, where $f: \mathbb R \to \mathbb C$ is a measurable complex-valued function: $$f(A) := \int_{\sigma(A)} f(x) dP^{(A)}(x)\:,$$ $P^{(A)}$ being the spectral measure (projector-valued) of $A$. $N= f(A)$ is a closed normal operator and admits a spectral decomposition $P^{(N)}$ ...

4

We divide by the norm of $\left|\psi\right>$ in order to take into account the case where the vector is not normalised. If $\langle \psi|\psi\rangle=1$ it makes no difference and if it's not, you recover the correct result as well.

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If the wave-function is a real thing or not, that doesn't depend on how many particles you put in a single wave-packet. If the wave-packet is of very low intensity, you should produce many copies of it. If the wave-function is a real thing, so it is for a thousand particles in a wave-packet, or for a single one particle. If the wave-function is a real-thing ...

3

There is no correct criticism of treating the wave function as real if by that you mean treating the equations of motion of quantum systems as if they describe how those systems actually work. There are many criticisms of treating the wave function as real that are no good: some examples follow. If quantum mechanics is a correct description of how ...

3

The non-negative real probability distribution can't interfere like a complex wave function can. To produce interference phenomena it is necessary for quantum mechanics to deal with probability amplitudes, not just probabilities.

3

You say that we are only interested in the probability distribution on the screen, $\rho(x,t) = \lvert \psi(x,t) \rvert^2$, which is essentially correct. So, why do we have $\psi(x,t) = \lvert\psi(x,t)\rvert\mathrm{e}^{\frac{\mathrm{i}}{\hbar}S(x,t)}$? Well, looking at the time evolution equation for the probability density, the continuity equation of ...

3

The time evolution of wave function is dictated by the Schrödinger equation, as you surely well know. Let's take the simple free particle in $\mathbb{R}^d$ (with mass $1/2$ and $\hbar=1$): $$i\partial_t \psi(t,x)=-\Delta_x \psi(t,x)\; ;$$ where $\Delta$ is the Laplacian operator (i.e. $\partial_x^2$). Mathematically, this is a linear PDE (partial ...

2

In practice it would be very difficult to truly reconstruct "the" wave-function from the spectra. For one thing: Which wave function? Measured spectral emission/absorption are the result of transitions between states (ground and excited, e.g.) and these states all have their own wavefunctions. Furthermore, a typical measured spectrum is measured over a ...

2

No, you cannot hear the shape of the drum. Even if you have a list of the eigenvalues of the Hamiltonian, you cannot reconstruct the Hamiltonian. For example, these two Hamiltonians have the same eigenvalues: The spectral line is the same for both systems. Assuming that the transition is from energy 2 to energy 1, the first system goes from a state that ...

2

It amuses me how often the cart is set in front of the horse in answers at this site. It is the data that drive theoretical formulations, not theoretical formulations the data. It was from the spectral lines of the hydrogen atom that the Schrodinger equation was established and the whole construct of the theory of quantum mechanics took off. The series ...

2

Plane wave solutions to the Schrödinger equation are not normalizable because they extend to infinity with a constant amplitude. Any physical particle will be constrained to a finite space, though (at least to the visible universe), so you need to look at superpositions of plane waves. This means that your starting point is $$\psi(x, 0) = \int \mathrm dk ... 2 Schroedinger's equations may have both normalizable and non-normalizable solutions. The function$$ \psi_k(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2} $$is a solution of the free-particle Schroedinger equation for any real k and \omega = |k|/c. As a rule, if the equation has a class of solutions whose members are functions ... 2 In the position basis. \Psi_n(x) = \langle \phi _n, \Psi \rangle \phi_n(x), and if the set of \phi_n is a complete orthonormal set of functions, then:$$\Psi(x) =\sum_n \langle \phi _n, \Psi \rangle \phi_n(x)$$where \langle \cdot , \cdot \rangle is an inner product, and is not really Dirac notation. \langle \cdot | \cdot \rangle is the same thing ... 2 Think of your functions just like vectors. You can add two (square integrable) functions and get another (square integrable) function. You can multiply a (square integrable) function by a scalar (pointwise) and get a (square integrable) function. So they add like vectors, they can be scaled like vectors. You even have a scalar product. So they are just ... 2 To me, the whole subject feels somewhat like a comedy of errors, and your question reveals some misconceptions. Hopefully, I'll be able to clear up some of them without just spreading my own misconceptions ;) Let's start off with de Broglie, who wanted to represent matter as physical waves. But when quantum mechanics emerged, we did not end up with matter ... 2 This is not a "real" answer to your question but rather points you should consider while making a mental picture of the wavefunction. "as every respectable physics professor I've ever encountered has treated the wavefunction as an indisputably non-real mathematical tool" Quantum Mechanics, as taught in undergrad and early graduate courses, deals ... 2 The view that wave function is only a mathematical object rather than a real object is, unfortunately, only a (probably) majority view but it is not agreed upon universally. This is in part because not all people agree that mathematical object is not a real object. There are people who prefer to think mathematical objects can be real objects. This is partly ... 2 When it's written with an extra division of <\psi|\psi>, it is just that |\psi> is not normalized. 2 Let's drop the scaling constants and say that the wavefunction in momentum co-ordinates \psi_p(\vec{p}) is the FT of that \psi_x(\vec{x}) in position co-ordinates , i.e.$$\psi_p(\vec{p}) = \mathfrak{F}_{\vec{p}}(\psi_x(\vec{x}))$$where \mathfrak{F} is the Fourier transform. Now you propose to take the inverse Fourier transform of \psi_x. The ... 2 \newcommand{\ket}[1]{\lvert #1 \rangle}There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse. Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the z-direction ... 1 My question is, why am I doing this? Becase, by convention, we want the probability account for all possible outcomes to sum to unity. The fact that we will get some outcome at a measurement, is guaranteed, and with this normalization it reflects a total probability of 1. 1 I think what you are asking whether the relationship$$ \mathrm{normalizable} \iff \mathrm{continuous}$$holds, which is utterly wrong! The wave function has to be continuous*. Notwithstanding take \psi(x)=H(x-1/2)-H(x+1/2), where H(x) is the Heaviside step function.$$ \implies \int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1 $$(Area ... 1 The functions you write down are solutions of Maxwell's equations (if you think of them as lone, Cartesian components) and, as such, have an exact relationship the one-photon quantum state of the quantum photon field. Now, whether this is a photon wave function depends on your definitions. If you want to write down the quantum state of a one-photon, so ... 1 There is nothing wrong with this, as the solution to this equation doesn't live in the Hilbert space. That is to say that there is no eigenvector solution to the free particle equation since the Hamiltonian has only a continuous part in its spectrum. The best you can do if you want to stick with the Hilbert space is to find a sequence of vectors that ... 1 An observable that has a definite value is an eigenvalue of the operator. If A is a hermitian matrix of which |\psi\rangle is an eigenstate, we have$$\tag{1}A|\psi,a\rangle=a|\psi,a\rangle$$You asked about the wave function, not the state vector, though. We can still get all the information we need from the wave function \psi(x)=\langle ... 1 Let D be the subspace of Schwartz functions \psi of rapid decrease \mathscr{S} such that their primitive \Psi is in \mathscr{S}. Then A:D\to D (easy to see calculating the primitve with integration by parts); A defined on \mathscr{S} and B:D\to \mathscr{S}. Hence on D both AB and BA are well-defined, and$$[A,B]\psi=B\psi\;,\; ...

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I will expand on my comment, answering the title: “Reality” of EM waves vs. wavefunction of individual photons - why not treat the wave function as equally “Real”? What does Real mean in physics It is instructive to look at the definition of fields for physics: "A field is a physical quantity that has a value for each point in space and time." ...

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