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- Thread starter SinaHp
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In summary, the professor wants you to do the following:Take the inner product of the vector space with itself and set it equal to 1.Keep in mind the relationship##c_i c_j \langle \psi_i | \psi_j \rangle= c_i c_j \delta_{ij}##So no cross terms should generate.Can you carry out this work and post it here?f

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- #2

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##\Psi = \sum_{i=1} c_i \psi_i## with itself and set it equal to 1.

Keep in mind the relationship

##c_i c_j \langle \psi_i | \psi_j \rangle= c_i c_j \delta_{ij}##

So no cross terms should generate.

Can you carry out this work and post it here?

- #3

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Unfortunately no, how is that exactly?

##\Psi = \sum_{i=1} c_i \psi_i## with itself and set it equal to 1.

Keep in mind the relationship

##c_i c_j \langle \psi_i | \psi_j \rangle= c_i c_j \delta_{ij}##

So no cross terms should generate.

Can you carry out this work and post it here?

- #4

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Unfortunately no, how is that exactly?

What are you referring to?

The expression with the kronecker delta is the definition of an orthonormal basis which is what is stated in the question.

When you write out the whole expression

You get terms like

##c_1 c_1 \langle \psi_1 | \psi_1 \rangle = c_1^2##

##c_1 c_2 \langle \psi_1 | psi_2 \rangle = 0##

Carry this process out with each combination and sum them. What terms are you left with?

Your professor probably wants you to do this in index notation so I’ll leave that to you. I am just giving you specific term examples so you get the idea.

While I’m here I might as well correct my other statement. Since we’re dealing with vector spaces over complex fields.

It should technically be ##\langle \psi_i | \psi_j \rangle = c_i c_j^* \delta_{ij}## but the idea is the same.

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- #6

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Yup! You got the idea. Now do it compactly with index notation for n terms instead of just 2 terms.something like this?

- #7

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Got it! Only C_{n} instead of C_{i} and C_{j}

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