# Let's Discuss The Postulates :: Consistency of Probability Interpretation

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We will now discuss the postulates of quantum mechanics.

I want to discuss the conditions under which the probability interpretation will be consistent.

Let $$A$$ be a dynamical variable and the corresponding operator be denoted by $$\hat{A}$$.
Let the eigen values and eigen vectors be denoted by $$\alpha_k$$ and $$\ket{\alpha_k}$$ respectively.

\hat{A} \, \ket{\alpha_k} = \alpha_k \ket{\alpha_k}.

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If a system is known to be in state with value $$\alpha_1$$ of $$A$$ , what is the probability that a measurement of $$A$$ will give some other value $$\alpha_2 \ne \alpha_1$$?

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What guarantees that theory will always give the answer you expect?

Different eigenvectors should be orthogonal.

• It is given that a system is in state $$\ket{\psi}$$.

• In order to find the probabilities of different possible outcomes $$\alpha_n$$ on measurement of $$A$$, one needs to expand the vector $$\ket{\psi}$$ and write it as superposition of eigenvectors $$\hat{A}$$.

What ensures that such every vector can be written as linear combination of eigen vectors of $$\hat{A}$$?

• The eigenvectors should form a basis.
• Then only there will be a guarantee that every vector can be expanded in as a linear combination of the eigenvectors.
• The postulates give the prescription of computing the probabilities of different outcomes of measurement of $$A$$.

• That sum of all probabilities is one means $$\sum_n |c_n|^2 = |<\psi|\psi>|^2$$.
How do we see this property?

• This will be ensured in Parseval relation holds and the eigenvectors are normalized.

• The Parseval relation holds if the basis is a complete ortho normal set.
• If a system is in a state $$\ket{\psi}$$ which satisfies $$|\innerproduct{\psi}{\alpha_m}|^2=1$$, it means that outcome of measurement of $$A$$ will definitely be the value $$\alpha_m$$.

For consistency with the first part of the postulate, we must be able to prove that

the given condition, $$|\innerproduct{\psi}{\alpha_m}|^2=1$$, implies that $$\ket{\psi}$$ is an eigen vector of $$\hat{A}$$ with eigen value $$\alpha_m$$.

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• For computing probabilities, the state vector $$|\psi>$$ and all the eigenvectors should be normalized.

• This means that $<\psi|\psi> =1 \qquad \qquad |<\alpha_n|\alpha_n>|=1 \text { for all n}\.] • If the out come of measurement is to be definite and equal to $$\alpha_m$$, for some $$m$$, we must have \[|<\psi|\alpha_m>|=1.$

• Therefore we get $|<\psi|\alpha_m>|^2 = <\psi|\psi> |<\alpha_m|\alpha_m>|.$

• This then implies that \{|\psi>\} should be equal to some some constant times $$|E_m>$$. $|\psi> = K |\alpha_m>.$

WHY ? How do I see the last statement?

You have to recall the full statement of Cauchy Schwarz inequality.

The Cauchy Schwarz inequality $|<f|g>|^2 \le |f| |g|$ becomes equality if and only if $|f> = \text{const} |g>$.

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We have several requirements coming from the consistency of probability interpretation, the content of the third postulate.

All the requirements on the eigen vectors of operator $$\hat{A}$$ are met if $$\hat{A}$$  is a self adjoint operator.

4727: Diamond Point

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