열
The wave function of
free particle at time
t=0 is given by: P (2,0)=0 inse
by: Y(2,0) = 0 ne her
The function 0(n) is real and differs gnificant
from zero Only for values of lying
pe in the
interval
-scals.
For which range of Values Alle will the
Wave function at
different from Zero?
later time t be significantly
Answers
Answer:
열 by the way whats this?
Explanation:
This is the equation for a (non-relativistic) particle of mass m moving along the x axis while
acted by the potential V (x,t) ∈ R. It is clear from this equation that the wavefunction must
be complex: if it were real, the right-hand side of (1.2) would be real while the left-hand side
would be imaginary, due to the explicit factor of i.
Let us make two important remarks:
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1. The Schr¨odinger equation is a first order differential equation in time. This means that if
we prescribe the wavefunction Ψ(x,t0) for all of space at an arbitrary initial time t0, the
wavefunction is determined for all times.
2. The Schr¨odinger equation is a linear equation for Ψ: if Ψ1 and Ψ2 are solutions so is
a1Ψ1 + 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 complex
number 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 of
the 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 wavefunction
arises because we declare that
P(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 above
must 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 found
in all of the tiny intervals dx that comprise the real line. Since the particle must be found
somewhere 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¨odinger
equation, knowledge for all times. The Schr¨odinger equation must guarantee that the wavefunction remains normalized for all times. Proving this is a good exercise:
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- - - - - - - - - - - - - - -
Exercise 1. Show that the Schr¨odinger equation implies that the norm of the wavefunction
does 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. Moreover
you will have to use Ψ(x,t) → 0 as |x| → ∞, which is true, as no normalizable wavefunction
can 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
∂Ψ
J(x,t) = Im Ψ∗ . (1.9) m ∂x
The analogy in electromagnetism is useful. There we have the current density vector Ji and the
charge density ρ. The statement of charge conservation is the differential relation
∇ · Ji + ∂ρ = 0 . (1.10) ∂t
This equation applied to a fixed volume V implies that the rate of change of the enclosed charge
QV (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 S
Make sure you know how to get this equation from (1.10)! While the probability current in
more than one spatial dimension is also a vector, in our present one-dimensional case, it has
just one component. The conservation equation is the analog of (1.10):
∂J ∂ρ + = 0 . (1.12) ∂x ∂t
You can check that this equation holds using the above formula for J(x,t), the formula for
ρ(x,t), and the Schr¨odinger equation. The integral version is formulated by first defining the
probability Pab(t) of finding the particle in the interval x ∈ [a, b]
b b
Pab(t) ≡ dx|Ψ(x,t)|
2 = dx ρ(x,t). (1.13) a a
You can then quickly show that
dPab (t) = J(a,t) − J(b,t). (1.14) dt
3
Z
~
Z Z
Here J(a,t) denotes the rate at which probability flows in (in units of one over time) at the left
boundary of the interval, while J(b,t) denotes the rate at which probability flows out at the
right 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 assuming
normalization, 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 equivalent
if they differ by multiplication by a complex number. Using the symbol ∼ for equivalence, we
write
Ψ1 ∼ 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 overall
constant phase:
Normalized wavefunctions: Ψ1 ∼ Ψ2 ←→ Ψ1(x,t) = eiθ Ψ2(x,t), θ ∈ R .