• Support PF! Buy your school textbooks, materials and every day products Here!

Qbits normalization homework

  • Thread starter killinchy
  • Start date
  • #1
4
0

Homework Statement



I'm a chemist, so forgive me. I'm looking at Leonard Susskind's course on 'Quantum Entanglement', and we've just started on Qbits. Electron spin: we have two column vectors (1,0) and (0,1) to represent the two states, call them 'up' and 'down'.

The vector for the state for the electron being 'across' is given by a column vector (1/√2, 1/√2)

Here's my problem; why not (1/2, 1/2)? I know that vectors that represent physical states must be normalized, but where does the √2 come in?

Thanks

John

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,523
737

Homework Statement



I'm a chemist, so forgive me. I'm looking at Leonard Susskind's course on 'Quantum Entanglement', and we've just started on Qbits. Electron spin: we have two column vectors (1,0) and (0,1) to represent the two states, call them 'up' and 'down'.

The vector for the state for the electron being 'across' is given by a column vector (1/√2, 1/√2)

Here's my problem; why not (1/2, 1/2)? I know that vectors that represent physical states must be normalized, but where does the √2 come in?

Thanks

John

Homework Equations





The Attempt at a Solution


Normalized means unit length. With the ##\sqrt 2## in the denominator the vector has length 1.
 
  • #3
418
28
Well, normalization basically says that a state's absolute magnitude must be 1. In other words, if we have a column vector [itex]\mathbf{v}=\begin{pmatrix} a \\ b\end{pmatrix}[/itex], then [itex]\mathbf{v}[/itex]'s inner product with itself must equal 1.

For two vectors [itex]\mathbf{v}_1=\begin{pmatrix} a_1 \\ b_1\end{pmatrix}[/itex] and [itex]\mathbf{v}_2=\begin{pmatrix} a_2 \\ b_2\end{pmatrix}[/itex], the definition of the inner product of [itex]\mathbf{v}_1[/itex] and [itex]\mathbf{v}_2[/itex], denoted [itex]\left \langle \mathbf{v}_1 | \mathbf{v}_2 \right \rangle[/itex] can be written in matrix notation as:[tex]\left \langle \mathbf{v_1} | \mathbf{v_2} \right \rangle=\mathbf{v}_1^\dagger \mathbf{v}_2 = \begin{pmatrix} a_1^* & b_1^*\end{pmatrix}\begin{pmatrix} a_2 \\ b_2 \end{pmatrix}=a_1^*a_2+b_1^*b_2 [/tex]
where * denotes the complex conjugate: e.g. if a=1+2i, a*=1-2i. It should be clear in this that the symbol [itex]\dagger[/itex] denotes the adjoint operation, which is the same as taking the complex conjugate and then the transpose (with a column vector, the transpose simply changes a column vector into a row vector.)

Equivalently, you could write the definition of the inner product of [itex]\mathbf{v}_1[/itex] and [itex]\mathbf{v}_2[/itex] in vector notation as:
[tex]\left \langle \mathbf{v_1} | \mathbf{v_2} \right \rangle=\mathbf{v}_1^* \cdot \mathbf{v}_2 = \begin{pmatrix} a_1^* \\ b_1^*\end{pmatrix}\cdot\begin{pmatrix} a_2 \\ b_2 \end{pmatrix}=a_1^*a_2+b_1^*b_2 [/tex]

As I stated earlier, the definition of [itex]\mathbf{v}=\begin{pmatrix} a \\ b\end{pmatrix}[/itex] being normalized is that its inner product with itself is 1:
[tex]1=\left \langle \mathbf{v} | \mathbf{v} \right \rangle=\mathbf{v}^\dagger \mathbf{v} = \begin{pmatrix} a^* & b^*\end{pmatrix}\begin{pmatrix} a \\ b\end{pmatrix}=a^*a+b^*b [/tex]

So let's see what happens if we use your suggestion that [itex]\mathbf{v}=\begin{pmatrix} \frac{1}{2} \\ \frac{1}{2}\end{pmatrix}[/itex] These are real elements so the complex conjugation does not change them at all. If we form the inner product of [itex]\mathbf{v}[/itex] with itself we get:[tex]\left \langle \mathbf{v} | \mathbf{v} \right \rangle=\mathbf{v}^\dagger \mathbf{v} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}\frac{1}{2} \\ \frac{1}{2}\end{pmatrix}=\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}[/tex]
So this state is not normalized by definition.

On the other hand if we use [itex]\mathbf{v}=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix}[/itex], then we get[tex]\left \langle \mathbf{v} | \mathbf{v} \right \rangle=\mathbf{v}^\dagger \mathbf{v} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix}= \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}=\frac{1}{2}+\frac{1}{2}=1[/tex]
so this state is normalized by definition.
 
Last edited:
  • #4
33,314
5,006
This is easy enough to think about geometrically. If both vectors start from the origin, they extend to the points (1, 0) and (0, 1) on the unit circle. The point on the unit circle exactly between the two vectors makes an angle of 45° with the horizontal axis, and is the point whose coordinates are (1/√2, 1/√2). The vector that extends from the origin to this point on the unit circle has a magnitude of 1, so is normalized.
 
  • #5
418
28
The problem with the geometric interpretation of a state vector is that unlike a geometrical vector, in general, quantum mechanical state vectors can have complex entries. For example, if you insist on using geometrical ideas, you would either be baffled or get the wrong answer if you used geometry to figure out whether the state [itex]\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{pmatrix}[/itex] is normalized or not (e.g. you'd get that it has length 0 using the pythagorean theorem). Using my formulation, you can easily handle quantum mechanical states like this, which are actually quite common.
 
Last edited:
  • #6
33,314
5,006
The problem with the geometric interpretation of a state vector is that unlike a geometrical vector, in general, quantum mechanical state vectors can have complex entries.
I was responding to the OP's question, in which all the vectors involved had real entries. I was also trying to gauge my response to the level of mathematical experience I perceived in the poster, something that you apparently didn't consider.
For example, if you insist on using geometrical ideas, you would either be baffled or get the wrong answer if you used geometry to figure out whether the state [itex]\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{pmatrix}[/itex] is normalized or not (e.g. you'd get that it has length 0 using the pythagorean theorem). Using my formulation, you can easily handle quantum mechanical states like this, which are actually quite common.
 
  • #7
418
28
The OP's question was about a Leonard Susskind lecture course on quantum entanglement and qbits. I think he's probably referring to these online notes:
http://www.lecture-notes.co.uk/susskind/quantum-entanglements/lecture-2/quantum-states/
I don't think you'd get very far understanding qbits if you're stuck with the assumption of all real components and some sort of pythagorean concept of length. Susskind introduces complex components at the beginning of the linked lecture--which is only the 2nd lecture out of 9.
 
Last edited:
  • #8
4
0
Wow..... thanks everybody. I am in the middle of "Leonardo's" second lecture. If there is one thing I've learned so far in my adventure (OK I 'did' Schrodinger's take on QM as part of Chemistry but never really understood what was going on), it is this: Mathematicians and Physicists use a lot of weird and wonderful quirky things for no other reason than they work. Kolb, I should have tried the inner product. And thanks, Kolb for the link to the notes.

Totally off-topic. I am one handshake away from a physicist by the name of Albert Einstein. In the 70's I became friendly with a Chemist (Princeton) who did part of his PhD in Sweden with Arrhenius. Then he went off to Berlin to the Kaiser Wilhelm where he met (among others) AE. And where did AE end up, but Princeton. The point is this: it's not that long ago. BYW, Einstein, Wigner etc etc all thought Planck was the smartest of them all.
 

Related Threads on Qbits normalization homework

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
720
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
11
Views
2K
Replies
1
Views
2K
Top