-
Notifications
You must be signed in to change notification settings - Fork 0
/
U1L3f.txt
158 lines (135 loc) · 5.06 KB
/
U1L3f.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
#
# File: content-mit-8370x-subtitles/U1L3f.txt
#
# Captions for course module
#
# This file has 149 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
The next thing I want to do is explain
how to rotate the Bloch sphere.
So let's write down these matrices, minus i 0 i
0 and sigma z equals 1 minus 1 0 0.
So these are all two-dimensional matrices,
which means they perform transformations
on a two-dimensional state space.
And these are the three Pauli matrices, sigma x, sigma y,
sigma z.
And let's try to figure out what transformations they perform.
So 1 over root 2--
no wait-- 1 0, what happens when we apply sigma x to 1 0?
We get 0 1.
What happens when we apply sigma 1 1, sigma x to 1 1?
We get 1 over root 2 1 1.
And if we have 1 over root 2 1 minus 1,
it goes to 1 over root 2 minus 1 1.
So what does that do to the Bloch sphere?
Well, 1 1, which happens to be the x-axis is left unchanged.
And if we-- it takes minus the z-axis to plus the z-axis,
and it took minus the y-axis--
1 over root 2 1 1--
1 minus 1 goes to 1 over root 2 minus 1 1.
1 over root 2 1 i goes to 1 over root 2 i 1, which happens just
happens to equal to 1 over root 2 1 minus i.
But this is just the negative of this.
So as I said before, this is essentially
the same vector as this.
So it takes-- it leaves this vector unchanged.
It leaves this vector unchanged.
It takes this guy to the top, and it takes this minus y--
or this minus y unit to the y unit and vice versa.
So it's a 180 degree rotation around the x-axis.
And I'm not going to go into the calculations,
but this and this are 180 degree rotations around the y-axis
and the z-axis.
So this is-- and you see that this transformation just
rotates the Bloch sphere.
And, in fact, all the quantum transformations
you can do, essentially, rotate the Bloch sphere.
So let's figure out another one.
H is equal to 1 over root 2 minus 1 over root 2--
no, 1 over root 2 1 over root 2 1 over root 2 minus 1
over root 2.
OK, what does that do?
Well, it takes a vertical, which is 0 to 1 over root 2 1
over root 2 which is 0 plus 1 over root 2 to the right.
And, in fact, if you square this, you get the identity.
Because you square this, you multiply this by that--
I mean, 1 over root 2 times 1 over root 2 plus 1 over root
2 times 1 over root 2 is 1.
And this times this plus this times this is also 1.
And the other two terms are 0, because they have
a plus and a minus in them.
So H squared equals I. And so it interchanges that and that.
Well, the down-- if you have 0 1, you get 1--
you get a 1 minus 1 vector.
So it interchanges that and that.
And what does it do to 0 plus i 1?
OK, maybe I need to--
so that's what 1 i is equal to--
well-- OK, this gets complicated.
1 plus i or 1 over root 2 1 plus i gives you 1 plus 1/2.
So the top thing is 1 plus i.
And the bottom thing is 1 minus i.
Well, you can multiply the top and bottom of this by 1
plus i over root--
or 1 minus i over root 2 1 minus i 1 minus i 1 over root 2.
So this is a--
so we're essentially multiplying the whole thing.
I mean, this is supposed to be a scalar, I'm sorry, 1 minus i.
So this is a scalar, and it's a unit scalar.
So it does not change what this vector really is.
And the top thing becomes 2.
And canceling all the powers of 2 out, you get a 1
over root 2 1.
And the bottom thing, 1 minus i times 1 minus i is 2 i,
you got minus i.
So what you've done is you've swapped the front
and the back of the sphere.
So it's not a reflection, because you've
taken this guy to this guy, this guy to this guy, and the back
to the front of the sphere.
It's a rotation around--
I should have brought colored chalk clearly.
It's a rotation around this axis.
OK, so there are a lot of--
yeah.
In terms of x, y, z what is the vector
that points in the direction on that axis?
OK, so this rotation around--
well, this is cosine pi 4 or rather 1 over root 2 1
over root 2 is cosine--
OK, cosine pi over 8 sine pi over 8 really.
And I forgot to put cosine pi over 8.
The y-axis is 0.
And the-- no, the z-axis is cosine pi over 8.
And the y-axis is 0.
And yes, so sine pi over 8 cosine pi over 8.
So you know, this 1 over root 2 0 1 over root 2
is cosine pi over root 4--
sine pi over root 4 0 cosine pi over root 4 and 1 0 or 0
0 1, which is this point in--
I guess, Euclidean coordinates is sine 0 0 cosine 0.
And this point is supposed to be halfway in between them.
So it's really sine pi over 8 0 cos pi over 8.
And that's good.
So I want to--
so we're going to be doing a lot of quantum gates.
Yeah, another question.
Yeah.
So H is basically [INAUDIBLE] sigma x plus sigma [INAUDIBLE]??
And the [INAUDIBLE] rotation is basically
like the x plus [INAUDIBLE].
[INAUDIBLE] that like if you decompose it,
then you can figure out what [INAUDIBLE]??
Yes.
Well, there's a very nice formula
for exactly how rotations correspond to axes,
which we won't get to yet.
But we should be doing it at some time
in the next few lectures.
But what do I want to say?
I want to say that you need to know--
that we need a little more background before we