Derive the Schrödinger's Time independent wave equation using kinetic energy and momentum.

Answers

Answer 1

Consider,

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =m\langle v(t)\rangle=m\int_{-\infty}^{\infty}x\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial t}\Psi (x,t)+\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial t}\bigg\}dx}[/tex]

Multiply both sides by ih and simplification will yield

[tex]{:\implies \quad \displaystyle \sf ih\langle p\rangle =m\int_{-\infty}^{\infty}x\bigg[\Psi (x,t)\bigg\{\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi^{*}(x,t)}{\partial x^2}-V(x)\Psi^{*}(x,t)\bigg\}+\Psi^{*}(x,t)\bigg\{V(x)\Psi (x,t)-\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}\bigg\}\bigg]dx}[/tex]

Some simplification, Then Integrate by parts and then knowing the fact that the wave function vanishes for [tex]{\bf x\to \pm \infty}[/tex] will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =\dfrac{ih}{2}\int_{-\infty}^{\infty}\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial x}\Psi (x,t)-\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}\bigg\}dx}[/tex]

Integrating by parts and knowing the same fact by some simplification will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =-ih\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}dx}[/tex]

The momentum is thus contained within the wave function, so we can then deduce that:

[tex]{:\implies \quad \sf p\rightarrow -ih\dfrac{\partial}{\partial x}}[/tex]

[tex]{:\implies \quad \sf p^{n}\rightarrow \bigg(-ih\dfrac{\partial}{\partial x}\bigg)^{n}}[/tex]

[tex]{:\implies \therefore \quad \displaystyle \sf \langle p^{2}\rangle =-h^{2}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

Now the kinetic energy

[tex]{:\implies \quad \displaystyle \sf \langle K\rangle =\dfrac{\langle p^{2}\rangle}{2m}=\dfrac{-h^2}{2m}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

The classical formula for the total energy

[tex]{:\implies \quad \sf \dfrac{p^2}{2m}+V(x)=E}[/tex]

Multiplying this equation by [tex]{\sf \Psi (x,t)=\psi (x)exp\bigg(\dfrac{-iEt}{h}\bigg)}[/tex] and use the above equations and simplify it we will be having

[tex]{:\implies \quad \boxed{\bf{\dfrac{-h^2}{2m}\dfrac{d^{2}\psi (x)}{dx^{2}}+V(x)\psi (x)=E\psi (x)}}}[/tex]

This is the Famous Time-Independent Schrödinger wave equation

Note:- If I write all the explanation then the Answer box willn't allow me to submit the answer


Related Questions

10. As a pharmacy technician, you will receive prescription orders with many different abbreviations, which will require conversions to calculate the dose. Now that you have learned about conversions and abbreviations, what would you do to fill a prescription with abbreviations? Also, what would you do with conversions you were not sure how to convert to the required dose. Explain

Answers

When you receive prescriptions that contain some abbreviations, you have to translate it into plain language that the patient can understand.

Who is a pharmacy technician?

A pharmacy technician is a skilled personnel who assists a pharmacist in a health facility. The pharmacy technician usually writes out prescriptions for patients.

When you receive prescriptions that contain some abbreviations, you have to translate it into plain language that the patient can understand. If you are not clear about the abbreviation you can ask the pharmacist for help or the doctor.

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an ldeal gas with internal energy U at 202°C is heated to 410°C.its internal energy then will be?

Answers

Internal energy is directly proportional to temperature

U1T2=U2T1U(273+410)=U2(273+202)683U=475U_2U_2=1.44U

Find the period of small oscillations of a barbell of length l with balls of mass m, located along a homogeneous electric field of intensity E. The charge of the barbell balls is ±q.

Answers

The answer to the question above would be

a) First write down what you see in the picture.

Answers

Answer:

which picture please,I don't get you

Ray X strikes a flat mirror at an angle of 0 = 26 degrees. What is the incidence?

Answers

The incidence is maybe 26=9

A spherical convex mirror has a radius of 30 cm. An object with a height of 0.30 m is placed 20 cm from the mirror. Note that in +/- sign conventions, f is negative (–) if the mirror is a convex mirror. a. Calculate the image distance. b. Calculate the image height. c. Calculate the magnification. d. Summarize the properties of the image formed in terms of its LOST (location, orientation, size, and type). e. Draw the set-up using graphical methods (ray diagramming). Apply scale drawing. Make sure that your illustration matches well with what you have calculated and presented in a-d.​

Answers

a)A spherical convex mirror form the image at image distance is 8.57cm.

b) Mirror forms the image of height is 12.855 cm.

c)The magnification of the spherical convex lens is 0.4285.

What is convex mirror?

The mirror which is a part of sphere with radius R is painted from inside forms the convex mirror.

The focal length of convex lens,  f = R/2

Given , the radius of mirror is R =30cm, the focal length will be

f =30cm/2 = 15cm

a) Object distance u =20 cm, object height =0.30m, then the image distance will be 1/f = 1/u + 1/v where v is the image distance from the lens.

1/15 = 1/20 + 1/v

v= 8.57 cm

Thus, the image distance is 8.57cm.

b) Ratio of image distance to the object distance is equal to the ratio of image height to the object height.

Substitute the values into the expression, we get the image height as

h(image) = 8.57/20 x 30

h(image) = 12.855 cm

Thus, the image height is 12.855 cm.

c) The magnification is the ratio of image height to the object height.

Putting the values, we have

= 12.855/30

= 0.4285

Thus, the magnification is 0.4285.

d) Size of object is large as compared to the size of image. Object distance is small as compared to the image distance. The image formed is virtual.

e)  The setup drawing is attached.

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A triangular prism made of crown glass (n=1.52) with base angles of 30.0 is surrounded by air. If parallel rays are incident normally on its base as shown in fig., what is the angle of ϕ between the two emerging rays?

Answers

Answer:

es lo que es

Explanation:

es lo que es es que slep vaka vee

the angle ϕ= will be equal to angle of refraction and it is equal to 49.4°C.

What is refractive index ?

When a light is going from medium 1 to medium 2. The refractive index is defined as a ratio of velocity of light in medium 1 to velocity of light in medium 2. Refractive index is the factor which deals with the amount of bending of light. More refractive index means more it will bend in the medium 2. When it is 1 we can say that light has not been bent.

By Snell's law, Refractive index is given by,

sin (i) ÷ sin (r) = μ₂÷μ₁

where i is the the angle of incidence

r is the angle of refraction

μ₂ & μ₁ are refractive index of medium 2 and 1 resp.

Hence it is due to Refractive index.

In this problem the light is going from prism to air, the refractive index of air is 1. From figure it is clear that the angle of incidence is 30°C.

From snell's law,

sin (30) ÷ sin (r) = μ₂÷1.52

0.5 ÷ sin (r) = 1÷1.52

0.5×1.52 = 1×sin (r)

0.76 = sin (r)

The angle of refraction r = sin⁻¹(0.76) =49.4°C

Hence angle ∅ = 49.4°C.

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A 50.0-N bowling ball is held in a person’s hand with the forearm horizontal. The biceps muscle is attached 0.030 0 m from the joint, and the ball is 0.350 m from the joint. Find the upward force F exerted by the biceps on the forearm (the ulna) and the downward force R exerted by the humerus on the forearm, acting at the joint.

Answers

The upward force F exerted by the biceps on the forearm will be 583 N.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

The torque balancing equation is found as;

[tex]\rm \tau_i=\tau_R+\tau_F+\tau_{BB}=0\\\\ R(0)+F(0.0300)-(50.0)(0.350)\\\\ F=583 \ N[/tex]

Hence,the upward force F exerted by the biceps on the forearm will be 583 N.

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A battery of emf 20v and internal resistance 1 ohms connected in series with an external resistor4 ohms what is
a power dissipate due to ineternal resistor​

Answers

The power dissipate due to internal resistor​ is 400 W.

Power dissipated in a circuit

The power dissipate due to internal resistor​ is calculated as follows;

P = IV

P = (V/r)V

P = V²/r

where;

V is emf of the batteryr is internal resistor

P = (20²)/(1)

P = 400 W

Thus, the power dissipate due to internal resistor​ is 400 W.

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Need help asap!!science foundations

Answers

Warning

This may be wrong, but I am 75% sure on my answer. I sincerely apologize if if it wrong.
A seems like the most reasonable option. I believe it’s A because the age, density, and thickness of oceanic crust increases with distance.


It is often said the media can make or break a person especially, sports personalities. Explain how that happens. ​

Answers

I'm going to school tomorrow so I can get the best of it


If a sound wave has a wavelength of 7 m and travels at a speed of 330 m/s, what is the
frequency of the sound?

O 47.1 hz
O2.12 hz
2.31 hz
23.1 hz

Answers

Answer:

047.1

Explanation:

frequency equals velocity over lambda

A tree is at a distance of 50 meters from a place on earth. From this place, the Angle of the tree's peak is 30. Find out the height of the tree.

Answers

sine's law

h/sin 30 = 50/sin 60

h/0.5 = 50/(0.5√3)

h=28.9 m

A proton has a charge , what gives it a charge?

Answers

Answer:

The charge is believed to be from the charge of the quarks that make up the nucleons (protons and neutrons). A proton is made of two Up quarks, with 2/3 positive charge each and one Down Quark with a negative 1/3 charge (2/3 + 2/3 + -1/3 = 1).

Explanation:

Hope this helps :)

A sound wave travelling through water has a wavelength of 0.08m and a frequency of 18525 HZ. What is the speed of the wave?

a. 343 m/s
b. 1482 m/s
c. 5941 m/s

Answers

Answer:

1482m/s

Explanation:

velocity equals frequency x lamdha

A stationary sound waves has a series of nodes. The distance between the first and the 6th node is 30cm. What is the wavelength of the sound wave

Answers

Answer:

Assume a node at the left end and the right end

N-A-N-A-N-A-N-A-N-A-N    shows nodes and anti-nodes

10 quarter wavelengths are shown  (or 2.5 wavelengths)

30 / 2.5 = 12 cm wavelength      since there is 1/4 wavelength between node and anti-node

Nodes and anti-nodes are displayed in N-A-N-A-N-A-N-A-N.

The difference between the node and the anti-node exists 12 cm wavelength.

What is meant by stationary sound waves?

Standing waves are created when two identical waves move in opposition to one another along a line. Despite being composed of two waves that are moving in opposition to one another, standing waves do not travel through space or along a string.

A wave that is stationary is one that is not moving, is at a standstill, or is in a relaxed position. A standing wave is created whenever two waves with nearly identical frequencies, wavelengths, and amplitudes that are traveling in opposite directions collide.

A waveform signal that is carried in space or down a wire has a wavelength, This is the distance between two identical locations (contiguous crests) in the following cycles. This length is typically defined in wireless systems in meters (m), centimeters (cm), or millimeters (mm).

The distance over which a periodic wave's shape repeats is known as the wavelength in physics.

Assume that there exists a node at both the left and right ends.

Nodes and anti-nodes are displayed in N-A-N-A-N-A-N-A-N.

it displays 10 quarter wavelengths.

Since there exists a 1/4 wavelength difference between the node and the anti-node, 30 / 2.5 = 12 cm wavelength.

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73. Integrated Concepts
An electron microscope produces electrons with a 2.00-pm wavelength. If these are passed through a 1.00-nm single slit, at what angle will the first diffraction minimum be found?

Answers

The angle will the first diffraction minimum be found will be 0.034895°.

What is diffraction grating?

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The equation of diffraction grating is given as

[tex]\rm d= \frac{m \lambda}{sin \theta }[/tex]

The angle between the incident ray and the point of the occurrence of the diffraction is found as;

[tex]\theta = sin^{-1}(\frac{m\lambda}{d} )\\\\\ \lambda= (2.0))(\frac{1.00 \times 10^{-12}}{1\ pm} )\\\\\ \lambda=2.00 \times 10^{-12}[/tex]

The angle will the first diffraction minimum be found is;

[tex]\theta = sin^{-1}(\frac{m\lambda}{d} )\\\\\ \theta = sin^{-1}(\frac{1 \times 2.000 \times 10^{-12}}{1\times 10^{-12}} )\\\\\ \theta= 0.0348995^0[/tex]

Hence,the angle will the first diffraction minimum be found will be 0.034895°.

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A sound travelling through some unknown medium has a wavelength of 0.05m and a frequency of 118,820 Hz. What is the speed of the wave?

a. 5941 m/s
b. 1482 m/s
c. 343 m/s

Answers

Answer:

5941m/s

Explanation:

velocity equals frequency x lamdha

Define the concept of heat. What is heat?

Answers

Answer:

In a simplest terms,

temperature is how hot or cold an object is, while heat is the energy that flows from a hotter object to cooler one

What is heat?
The transfer of energy between objects of different temperatures

How does the weight of an object relate to how fast it falls with a parachute?

Answers

Feathers fall slower than heavier objects. Parachutists fall slower when the parachute is open. Heavy objects push with greater force on your hand.


Two identical pucks of equal but unknown mass head straight towards each other with velocities
of 8.0 m/s and -6.0 m/s. They collide and stick together. What is their resulting velocity?

Answers

inelastic collision

m.8 - m.6 = 2mv'

2m = 2mv'

v' = 1 m/s

Answer:

1 m/s

Explanation:

According to the Law of Conservation of Momentum, (for inelastic collisions)

v = m₁v₁ + m₂v₂ / m₁ + m₂v = 8m - 6m / 2mv = 2m/2mv = 1 m/s

a simple generator has 200 loop square coil 10 cm on a side .how fast must it turn in a 0.25 T field to produce 24v peak out put​

Answers

Hi there!

Recall Faraday's Law:
[tex]\epsilon = N\frac{d\Phi_B}{dt}[/tex]

ε = Emf (V)
N = Number of loops

ΦB = Magnetic Flux (Wb)
t = time (s)

Since the magnetic field is constant, we can take this out of the time derivative:
[tex]\frac{d\Phi_B}{dt} = B * \frac{dA}{dt}[/tex]

Therefore:
[tex]\epsilon = N B \frac{dA}{dt}[/tex]

We can express 'A', the area in which the magnetic field passes as:

[tex]A = Acos(\omega t)[/tex]

Taking the time derivative:

[tex]\frac{dA}{dt} = A\omega sin(\omega t)[/tex]

ω = angular speed of coil (rad/sec)

Now, combine with the above expression:

[tex]\epsilon = NBA\omega sin(\omega t)[/tex]

The maximum output will occur when the loop's area vector is PERPENDICULAR to the field, so sin(ωt) = 1.

Therefore:
[tex]\epsilon = NBA\omega \\\\[/tex]

Rearrange to solve for ω:

[tex]\omega = \frac{\epsilon}{NBA}\\\\\omega = \frac{24}{(200 * (.10^2) * 0.25} = \boxed{48 \frac{rad}{sec}}[/tex]

A certain FM radio station broadcasts electromag-
netic waves at a frequency of 60,050,000 Hz. These
waves travel at a velocity of 30,000,000 m/s. What is
the wavelength of these radio waves?
v = λ x f

λ = v/f f=v/λ

Answers

Answer:

0.5 m

Explanation:

To find wavelength (λ) : apply the formula

λ = wave speed (v) / frequency (f)

Solving :

λ = 30,000,000 m/s / 60,050,000 Hzλ = 0.5 m (approximately)

How much work is done on the object as it moves from 6.0m to 9.0m

Answers

Answer:

W = 3F

Explanation:

Work is defined as the inner product between force and displacement. The inner product picks out the force components which point in the same direction as the displacement. The vertical components do not matter in the horizontal displacement. I am assuming you're dealing with 1-dimensional motion? Then the work would be:

[tex]W=Fd[/tex]

The displacement is 3 meters. Therefore, the work is:

[tex]W=3F[/tex]

You haven't gave the force being applied to the object, so I cannot give an explicit answer. Just multiply the force by 3 to get the work.

An object is being shot from a horizontal ground at an incline angle of 30 degree with respect to the ground at a speed of 52 m/s. Find the duration in seconds that the object is above the height of 14 m. Give your answer with one decimal place.

Answers

When an object is thrown at an angle from the horizontal, the path followed by the body is called the projectile motion. The duration in seconds that the object is above the height of 14 m is 4.698 s.

What is speed?

Speed is the time rate at which velocity is changing.

Vertical speed component is

V₀y = 52sin 30°= 26 m/s

Given is the height h=14m

Using second equation of motion,

S=ut+ 1/2 at²

Substituting the values, we get

14 = 26t - 4.9t²

4.9t² - 26t +14 =0  

Solving the quadratic equation, we get the time as

taking the positive sign, t =4.698 s

taking the negative sign , t = 0.6082 s

Thus, the duration in seconds that the object is above the height of 14 m is 4.698s.

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Find the charge on the capacitor in an LRC-series circuit at t = 0.03 s when L = 0.05 h, R = 6 Ω, C = 0.005 f, E(t) = 0 V, q(0) = 4 C, and i(0) = 0 A. (Round your answer to four decimal places.)
Determine the first time at which the charge on the capacitor is equal to zero. (Round your answer to four decimal places.)

Answers

The charge on the capacitor in an LRC-series circuit is 2.8907 C.

The first time at which the charge on the capacitor is equal to zero is 0.1409s.

What is capacitor?

The capacitor consists of two parallel plates used to store the charge when current flows through it, and provides energy when shuts off.

By the Kirchoff's second  law,

Lq'' +Lq' +q/C =E(t)

Where, L =0.05h, R = 6 Ω, C = 0.005 f, E(t) = 0 V, q(0) = 4 C, and i(0) = 0 A.

Equations becomes q''+120q' +4000 =0

The auxiliary equation is m² +120m+4000 =0

Solving this quadratic equation, we have

m =-6± 20i

Then , q(t) = Ae^(-60t)cos 20t + Be^(-60t)sin 20t    where, A and B are constants.

At t=0, q(0) =7 =A

and q'(t) = A [ -20e^(-60t)sin 20t - 60e^(-60t)cos 20t ] +B  [ 20e^(-60t)cos 20t - 60e^(-60t)sin 20t ]

So, i(0)= q'(0) =0 =-60A +20 B

Putting the value of A=7 above, we have B =21.

So,  q(t) = 7e^(-60t)cos 20t + 21e^(-60t)sin 20t  

The charge on the capacitor at time t =0.015

q(0.015) = 2.8907 C

Thus the charge on capacitor is  2.8907 C.

q(t) = 7e^(-60t)cos 20t + 21e^(-60t)sin 20t  =0

Solve for t, we get

e^(-60t)cos 20t + 3e^(-60t)sin 20t\

-1/3 = sin 20t /cos20t

t =1/20 arctan(-1/3)

As, arctan is defined within -90 to +90,

So, 20t = -0.3218 +π =2.8197

t =0.1409s

Thus the first time at which the charge is zero is 0.1409s.

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Which of the objects below would experience lift?

Answers

Answer:

the first one, The object moving through air

What type of wave is illustrated below?

Longitudinal

Transverse

Surface wave

Seismic wave

Answers

Answer:

Longitudinal

Explanation:

I searched it up.

The expression of x=4t+2t^2 Where x is distance and t=time What is displecement ?​

Answers

Answer:

x = 4 t + 2 t^2      

the displacement is x which is zero at t = 0 and then increases indefinitely as t increases indefinitely

is it second or first law

Answers

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration.

Explanation:

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