The Schmidt decomposition in QC

  • #1
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Summary:

How to do Schmidt decomposition in QC?
Hi, there. I have some problems when learning Schmidt decomposition in Nielsen's QC.

The statement of Schmidt decomposition is simple and clear, however, the book doesn't give a clear procedure to do the Schmidt decomposition. I don't know whether the proof under the theorem is the the one I should use or not. After all, it seems practicable to me.

Well, then I got confused again when solving the exercise that ask me to find the Schmidt decompositions of the states
##\frac {\left| 00 \right>+\left | 11 \right >} {\sqrt 2}## ; ##\frac {\left| 00 \right>+\left | 01 \right >+\left | 10 \right >+\left | 11 \right >} 2##; and ##\frac {\left| 00 \right>+\left |01 \right >+\left | 10 \right >} {\sqrt 3}##, because I think the decompositions are themselves.

For example, the decomposition of ##\frac {\left| 00 \right>+\left |01 \right >+\left | 10 \right >} {\sqrt 3}## should be ##\frac {\left| 00 \right>} {\sqrt 3} + \frac {\left |01 \right >} {\sqrt 3} + \frac {\left | 10 \right >} {\sqrt 3}## which is itself. The expression satisfy all the conditions required in the theorem. But I don't think I'm right.

Could you help me point out what part do I make mistakes?

ps: Nielsen's book makes me feel terrible. I have a great time learning QM with Griffth's book. But I struggle in Nielsen's book. It is more abstract and there are not many examples that I can try. I really feel depressed and wonder that QC may be not suitable for me. But I really love computer and physics since high school.

pss: What do the "Quantum Physics Workshop" section do? Should I really post this thread in that section?
 

Answers and Replies

  • #2
DrClaude
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Could you help me point out what part do I make mistakes?
I'm not very familiar with Schmidt decomposition, but I'll try and have a look.

ps: Nielsen's book makes me feel terrible. I have a great time learning QM with Griffth's book. But I struggle in Nielsen's book. It is more abstract and there are not many examples that I can try. I really feel depressed and wonder that QC may be not suitable for me. But I really love computer and physics since high school.
Maybe the problem is the book? Try Stenholm and Suominen, Quantum Approach to Informatics
https://onlinelibrary.wiley.com/doi/book/10.1002/0471739367
pss: What do the "Quantum Physics Workshop" section do? Should I really post this thread in that section?
You can't post there. See
https://www.physicsforums.com/threads/pf-spring-cleaning.970133/
 
  • #3
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  • #4
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Hi, Schmidt decomposition (at least how I learn it, maybe there are different conventions) says the following:
Given a pure state $$\left|\psi\right>$$, there exist orthogonal states $$\left\{\left|\alpha_i\right>\right\}, \left\{\left|\beta_i\right>\right\}$$ such that $$\left|\psi\right>=\sum_{i} \lambda_i \left|\alpha_i\right>\left|\beta_i\right>$$ and where $$\lambda_i\geq0, \qquad \sum_{i}\lambda_i^2=1$$.

I'll explain why this state isn't the Schmidt decomposition (at least in this sense):
In your case you have:
$$\left|\psi\right>=\frac{\left|00\right>+\left|01\right>+\left|10\right>}{\sqrt{3}},
\qquad\text{If we try }\left\{\left|\alpha_i\right>\right\}=\left\{\left|\beta_i\right>\right\}=\left\{\left|0\right>,\left|1\right>\right\}$$
Then the Schmidt decomposition must be of the form $$\left|\psi\right>=\lambda_0\left|00\right>+\lambda_1\left|11\right>$$, since the state has the term $$\left|01\right>$$ is not the Schmidt decomposition.
 
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  • #5
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Hi, Schmidt decomposition (at least how I learn it, maybe there are different conventions) says the following:
Given a pure state $$\left|\psi\right>$$, there exist orthogonal states $$\left\{\left|\alpha_i\right>\right\}, \left\{\left|\beta_i\right>\right\}$$ such that $$\left|\psi\right>=\sum_{i} \lambda_i \left|\alpha_i\right>\left|\beta_i\right>$$ and where $$\lambda_i\geq0, \qquad \sum_{i}\lambda_i=1$$.

I'll explain why this state isn't the Schmidt decomposition (at least in this sense):
In your case you have:
$$\left|\psi\right>=\frac{\left|00\right>+\left|01\right>+\left|10\right>}{\sqrt{3}},
\qquad\text{If we try }\left\{\left|\alpha_i\right>\right\}=\left\{\left|\beta_i\right>\right\}=\left\{\left|0\right>,\left|1\right>\right\}$$
Then the Schmidt decomposition must be of the form $$\left|\psi\right>=\lambda_0\left|00\right>+\lambda_1\left|11\right>$$, since the state has the term $$\left|01\right>$$ is not the Schmidt decomposition.
Ah, I see. each eigenvector can only appear once. I used them twice. I'll rethink it.

Thank you for your help, Gaussian97.
 
  • #6
373
174
I've just realised that there was a typo, of course, the state must be normalized, so $$\sum_{i}\lambda_i^2=1$$ instead of $$\sum_{i}\lambda_i=1$$
 

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