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siniylev [52]
3 years ago
6

Problems that require solving the three-dimensional schrödinger equation can often be reduced to related one-dimensional problem

s. an example of this would be the particle in a cubical box. consider a cubical box with rigid walls (i.e., u(x,y,z)=∞ outside of the cube) and edges of length l. the general solution for this problem is
Mathematics
1 answer:
dusya [7]3 years ago
8 0
The wavefunction for a particle in a one-dimension box is a well-known problem, which has as a solution:

Ψ(x) = √(2/l) sin (nπx/l)

When the box has three dimensions, the general solution is simply the multiplication of the solutions for each dimension, therefore:

Ψ(x) = K · sin (n_x·π·x / l) · sin (n_y·π·y / l) · sin <span>(n_z·π·z / l)</span>
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Mr. Paulsen challenges Olivia and Angelica to move figure ABCDE onto figure A"B"C"D"E" using a series of two different transform
Jobisdone [24]

Problem 1

One method Olivia could have used is to reflect over the x axis and then translate 10 units to the right.

Let's apply these set of steps to point C.

C = (-2,-4)

C' = (-2,4) after reflecting over the x axis; y coordinate flips in sign

C'' = (8,4) after shifting to the right 10 units; add 10 to the x coordinate

Point C(-2,-4) moves to C''(8,4) which is shown in the diagram. I'll let you check the other four points.

-----------

Another method Olivia could have used is to shift to the right 10 units first, then reflect over the x axis. In this case, the order doesn't matter. Though for some other combinations of transformations, order does matter.

============================================================

Problem 2

Reflecting over the y axis will have the point (x,y) turn into (-x,y). Only the x coordinate changes, and we flip the sign here. The y coordinate stays the same.

A point like C(-2,-4) becomes C'(2,-4) after the y axis reflection. Shifting up 8 units means we add 8 to the y coordinate to arrive at C''(2,4) but this does not match what the diagram shows. Therefore, Angelica's method is incorrect.

You should find other inconsistencies for the other four points as well.

============================================================

Problem 3

Michael's method isn't correct either. He does not apply an x axis reflection and instead relies entirely on translations to move ABCDE to A''B''C''D''E''

At a quick glance, it looks like Michael may be correct. But upon a much closer look, you'll see that things don't line up perfectly.

Instead of point C, we'll pick on point D this time.

D is located at (-6,-2). If we shift up 8 units, then we add 8 to the y coordinate to get to D'(-6,6). Then add 10 to the x coordinate to shift 10 units to the right. This leads us to D''(4,6).

However, the diagram shows D'' is actually at (4,2)

I'll let you check the other points. You should find that only points C and E go to the correct locations, but the other points do not. Again, this is because Michael did not apply an x axis reflection. His figure of A''B''C''D''E'' is upside down compared to what it should be.

============================================================

To summarize, the methods described in problem 1 are the correct way to get ABCDE to move to A''B''C''D''E''. If Olivia followed either of those two methods, then she is correct. Both Angelica and Michael have incorrect methods.

5 0
3 years ago
7 + 9 + (-4) i have to wright extra smh
Oxana [17]

Answer:

12

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What is the average of 143, 98, and 128
asambeis [7]
Average\ A=\frac{S}{N}\\\\A=\frac{143+98+128}{3}=\frac{369}{3}=123\\\\Average\ of\ these\ numbers\ is\ 123.
8 0
3 years ago
Can someone please help me with this (-6) + (+5) =
qaws [65]

Answer:

-1

Step-by-step explanation:

This is is basically -6+5 which is -1

8 0
3 years ago
What is the equation for y=-2+3 and y=1/3x-1?
Crank
So y=-2+3 means y=3-2
y=1

so then subsitute 1 for y and solve
1=1/3x-1
add 1 to both sides
2=1/3x
multiply boht sides by 3
6=x
so x=6
y=1
4 0
3 years ago
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