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열 The wave function offree particle at timet=0 is given by: P (2,0)=0 inseby: Y(2,0) = 0 ne herThe function 0(n) is real and differs gnificantfrom zero Only for values of lyingpe in theinterval-scals.For which range of Values Alle will theWave function atdifferent from Zero?later time t be significantly​

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열 by the way WHATS this?Explanation:This is the equation for a (non-relativistic) particle of mass m moving along the x axis whileacted by the potential V (x,t) ∈ R. It is clear from this equation that the wavefunction mustbe complex: if it were real, the right-hand side of (1.2) would be real while the left-hand sidewould be imaginary, due to the explicit factor of i.Let us make two important remarks:1 1. The Schr¨odinger equation is a first order differential equation in time. This means that ifwe prescribe the wavefunction Ψ(x,t0) for all of space at an arbitrary initial time t0, thewavefunction is determined for all times.2. The Schr¨odinger equation is a linear equation for Ψ: if Ψ1 and Ψ2 are solutions so isa1Ψ1 + a2Ψ2 with a1 and a2 arbitrary complex numbers.Given a complex number z = a + IB, a, b ∈ R, its complex conjugate is z∗ = a − ib. Let|z| denote the NORM or length of the complex number z. The norm is a positive number (thus√ real!) and it is given by |z| = a2 + b2 . If the norm of a complex number is zero, the complexnumber is zero. You can quickly verify that∗ |z|2 = zz . (1.3)For a wavefunction Ψ(x,t) its complex conjugate (Ψ(x,t))∗ will be usually written as Ψ∗(x,t).We define the probability density P(x,t), also denoted as ρ(x,t), as the norm-squared ofthe wavefunction:P(x,t) = ρ(x,t) ≡ Ψ∗(x,t)Ψ(x,t) = |Ψ(x,t)|2 . (1.4)This probability density so defined is positive. The physical interpretation of the wavefunctionarises because we declare thatP(x,t) dx is the probability to find the particle in the interval [x, x + dx] at time t .(1.5)This interpretation requires a normalized wavefunction, namely, the wavefunction used abovemust satisfy, for all times, ∞dx |Ψ(x,t)|2 = 1 , ∀ t . (1.6) −∞By integrating over space, the left-hand adds up the probabilities that the particle be foundin all of the tiny intervals dx that comprise the real line. Since the particle must be foundsomewhere this sum must be equal to one.Suppose you are handed a wavefunction that is normalized at time t0: ∞dx |Ψ(x,t0)|2 = 1 , ∀ t . (1.7) −∞As mentioned above, knowledge of the wavefunction at one time implies, via the Schr¨odingerequation, knowledge for all times. The Schr¨odinger equation must guarantee that the wavefunction remains normalized for all times. Proving this is a good exercise:2 - - - - - - - - - - - - - - -Exercise 1. Show that the Schr¨odinger equation implies that the norm of the wavefunctiondoes not change in time:d ∞dx |Ψ(x,t)|2 = 0 . (1.8) dt −∞You will have to use both the Schr¨odinger equation and its complex-conjugate version. Moreoveryou will have to use Ψ(x,t) → 0 as |x| → ∞, which is true, as no normalizable wavefunctioncan take a non-zero value as |x| → ∞. While generally the derivative ∂ Ψ also goes to zero as ∂x|x| → ∞ you only need to assume that it remains bounded.Associated to the probability density ρ(x,t) = Ψ∗Ψ there is a probability current J(x,t)that characterizes the flow of probability and is given by∂ΨJ(x,t) = Im Ψ∗ . (1.9) m ∂xThe analogy in electromagnetism is useful. There we have the current density vector Ji and thecharge density ρ. The statement of charge conservation is the differential relation∇ · Ji + ∂ρ = 0 . (1.10) ∂tThis equation applied to a fixed volume V implies that the rate of change of the enclosed chargeQV (t) is only due to the flux of Ji across the surface S that bounds the volume:i dQV i (t) = − J · dia . (1.11) dt SMake sure you know how to get this equation from (1.10)! While the probability current inmore than one spatial dimension is also a vector, in our present one-dimensional case, it hasjust one component. The conservation equation is the analog of (1.10):∂J ∂ρ + = 0 . (1.12) ∂x ∂tYou can check that this equation holds using the above formula for J(x,t), the formula forρ(x,t), and the Schr¨odinger equation. The integral version is formulated by first defining theprobability Pab(t) of finding the particle in the interval x ∈ [a, b]b bPab(t) ≡ dx|Ψ(x,t)|2 = dx ρ(x,t). (1.13) a aYou can then quickly show thatdPab (t) = J(a,t) − J(b,t). (1.14) dt3Z~ Z ZHere J(a,t) denotes the rate at which probability flows in (in units of one over time) at the leftboundary of the interval, while J(b,t) denotes the rate at which probability flows out at theright boundary of the interval.It is SOMETIMES easier to work with wavefunctions that are not normalized. The normalization can be perfomed if needed. We will thus refer to wavefunctions in general without assumingnormalization, otherwise we will call them normalized wavefunction. In this spirit, two wavefunctions Ψ1 and Ψ2 solving the Schr¨odinger equation are DECLARED to be physically equivalentif they differ by multiplication by a complex number. Using the symbol ∼ for equivalence, wewriteΨ1 ∼ Ψ2 ←→ Ψ1(x,t) = α Ψ2(x,t), α ∈ C . (1.15)If the wavefunctions Ψ1 and Ψ2 are normalized they are equivalent if they differ by an overallconstant phase:Normalized wavefunctions: Ψ1 ∼ Ψ2 ←→ Ψ1(x,t) = eiθ Ψ2(x,t), θ ∈ R .



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