Asked by qwerty
Consider a particle whose wavefunction at some fixed time t is represented by
-A if -2≤x≤0
ψ(x)= A if 0≤x≤3
0 otherwise
(a) What is the normalization constant A?
(b) What is the probability of finding the particle at a position x≤0?
(c) What is the expected value of position x?
(d) What is the expected value of momentum p?
-A if -2≤x≤0
ψ(x)= A if 0≤x≤3
0 otherwise
(a) What is the normalization constant A?
(b) What is the probability of finding the particle at a position x≤0?
(c) What is the expected value of position x?
(d) What is the expected value of momentum p?
Answers
Answered by
Count Iblis
|psi(x)|^2 integrated from minus to plus infinity is 5 |A|^2, this has to be equal to 1, therefore you can choose A as 1/sqrt(5) (you are free to multiply this by any phase factor
exp(i theta)).
The probability of the aprticle being at negative x is the integral from minus infinity to zero of |psi(x)|^2 dx, which is 2/5.
Expacted position: Integrate
x |psi(x)|^2 dx from minus to plus infinity, you get <x> = 1/2.
exp(i theta)).
The probability of the aprticle being at negative x is the integral from minus infinity to zero of |psi(x)|^2 dx, which is 2/5.
Expacted position: Integrate
x |psi(x)|^2 dx from minus to plus infinity, you get <x> = 1/2.
Answered by
Count Iblis
And, of course <p> = 0, because the fact that psi(x) is (or can be chosen to be a) real function of x, implies that the modulus squared of the momentum space wavefunction
psi(p) is an even function of p. To see this, note that:
psi(x) = 1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi(p) exp(i p x/hbar)
Take the complex conjugate of this:
psi*(x) = 1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi*(p) exp(-i p x/hbar) =
1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi*(-p) exp(i p x/hbar)
Since psi*(x) = psi(x), this means that
psi*(-p) = psi(p)
therefore
|psi(-p)|^2 = psi*(-p)psi(-p) =
psi(p)psi*(p) = |psi(p)|^2
psi(p) is an even function of p. To see this, note that:
psi(x) = 1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi(p) exp(i p x/hbar)
Take the complex conjugate of this:
psi*(x) = 1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi*(p) exp(-i p x/hbar) =
1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi*(-p) exp(i p x/hbar)
Since psi*(x) = psi(x), this means that
psi*(-p) = psi(p)
therefore
|psi(-p)|^2 = psi*(-p)psi(-p) =
psi(p)psi*(p) = |psi(p)|^2
Answered by
Gyanno
Answers to q6 and 8 ?
Answered by
hikikomori
Please q6 and q8 !
Answered by
help
prob 3 C
Answered by
Im Anononous
3C is 3/2
6A is pi/2 and pi/4
6B is 1/2 and 1/2
Looking for 6C and 6D
8A is 1/2, 1/2, 0, 0
8B is 0, 0, 1/2, 1/2
8C is 1, 1, 0, 0
Looking for 8D and 8E
8F is No
6A is pi/2 and pi/4
6B is 1/2 and 1/2
Looking for 6C and 6D
8A is 1/2, 1/2, 0, 0
8B is 0, 0, 1/2, 1/2
8C is 1, 1, 0, 0
Looking for 8D and 8E
8F is No
Answered by
s
Thank you.
And waiting for the rest answers
And waiting for the rest answers
Answered by
Help plz
5 plz....
Answered by
Help plz
7a 1/2 7b 1/2
Answered by
Anonymous
Problem 9 plzz
Answered by
BOB
8D ,8E,6C,6D and 9
Answered by
Anne
9 is 175.8
Anyone 8D ,8E,6C,6D?
Anyone 8D ,8E,6C,6D?
Answered by
Mary
6C,6D,8C,8D please..
Answered by
Bel
4a,b,c,d plz ?
Answered by
helper
for 4 some one already gave you so you have:
4a: 1/sqrt(5)
4b 2/5
4c 1/2
4d 0
4a: 1/sqrt(5)
4b 2/5
4c 1/2
4d 0
Answered by
Bel
3a and b please
Answered by
Bel
thx very much helper
Answered by
Gyano
5 plz?
Answered by
Mary
5A i 0 0 -i 5b -i 0 0 i 5C -1 -1 -1
3a is 2nd option 3b is 2nd option
6C,6D,8C,8D please
3a is 2nd option 3b is 2nd option
6C,6D,8C,8D please
Answered by
JLG
8c = 1, 1, 0, 0
8D= 1/n,1/n,0,0,1/n,1/n,
Please 6C and 6D
8D= 1/n,1/n,0,0,1/n,1/n,
Please 6C and 6D
Answered by
Im Anononous
JLG - Thanks for 8D and 8E. Didn't realize n was acceptable.
Anyone have 6C and 6D?
5 a) {{i,0}{0,-i}}
5 b) {-{i,0}{0,i}}
5 c) -1, -1, -1
Anyone have 6C and 6D?
5 a) {{i,0}{0,-i}}
5 b) {-{i,0}{0,i}}
5 c) -1, -1, -1
Answered by
Help
6C and 6D
Answered by
JLG
6c: 1/2+1/(2*sqrt(2)
1/2-1/(2*sqrt(2)
6D I continius lost..
Answered by
JLG
6D is the same that 6 C :)
Answered by
Gyano
thx all, 8e please
Answered by
See you in the Final ;)
8e: 0,1/n, 1/n
Answered by
Im Anononous
Thanks JLG
Answered by
Gyano
THANKS
Answered by
abdrrahim
THANKS ANNE
Answered by
Anonymous
6c second part ans. is not accepted,,,wat is its numerical value.
Answered by
See you in the Final ;)
Hey Guys... Ready for the final??
Answered by
Final ;)
Helloo Any answer for the Final?
Answered by
SS01
1] Given qubit 1 in the state 35|0>+45|1> and qubit 2 in the state 12√|0>+i2√|1>, what is the state of the composite system?
in form of a│00> + b│01> + c│10> + d│11>
find a b c d.
2] Which of the following states is entangled?
a) |++>
b) 45|01>−35|11>]
c) 1/√3|01> +1/√3|10>−1/3|11>
d) 1+i/√22|++> − 1+i/√22|+−>
+ 1−i/√22|−+> − 1−i/√22|−−>
in form of a│00> + b│01> + c│10> + d│11>
find a b c d.
2] Which of the following states is entangled?
a) |++>
b) 45|01>−35|11>]
c) 1/√3|01> +1/√3|10>−1/3|11>
d) 1+i/√22|++> − 1+i/√22|+−>
+ 1−i/√22|−+> − 1−i/√22|−−>
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