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In quantum mechanics, two fundamental properties of inner products (J.J Sakurai) Chapter 1.2, are:

  • $\langle \alpha|\beta\rangle = \langle \beta|\alpha\rangle^*$

  • $\langle \alpha|\alpha\rangle \ge 0$

But I'm not able to understand why the second postulate is required. Can't it be derived from the first postulate?

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    $\begingroup$ $\uparrow $ No. $\endgroup$
    – Qmechanic
    Commented Feb 11, 2018 at 12:12
  • $\begingroup$ Could you please elaborate a bit? @Qmechanic $\endgroup$
    – DDDAD
    Commented Feb 11, 2018 at 12:16

3 Answers 3

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A requirement is missed in your list. If $a,b \in \mathbb C$, it must also hold $$\langle \alpha| a\beta + b\beta' \rangle = a\langle \alpha| \beta \rangle + b\langle \alpha|\beta' \rangle\:.$$

Regarding your question, consider the $2\times 2$ matrix $A:= diag(-2,1)$ and define in $\mathbb C^2$ $$\langle \vec{x}| \vec{y}\rangle := {\vec{x}^*}^t A \vec{y}$$ where $\vec{z} = (a,b)^t$ with $a,b \in \mathbb C$ and $\vec{z}^* = (a^*,b^*)^t$.

With this definition the first requirement is true (together the one I added), but the second is false because, if $\vec{x}=(1,1)^t$ you find $$\langle \vec{x}| \vec{x} \rangle = -1\:.$$

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But I'm not able to understand why the second postulate is required. Can't it be derived from the first postulate?

No. What you can derive from the first postulate is that $\langle \alpha|\alpha\rangle$ must be real since $\langle \alpha|\alpha\rangle = \langle \alpha|\alpha\rangle^\ast$. There's nothing in this that says that this real number must be non-negative.

For example, consider the function $f(\alpha,\beta) = -\alpha^\ast\beta$ for $\alpha,\beta \in \mathbb C$. This satisfies the first listed postulate of an inner product over $\mathbb C$ because $f(\beta,\alpha) = -\beta^\ast\alpha = (-\alpha^\ast\beta)^\ast = f(\alpha,\beta)^\ast$. It does not the second postulate because $f(1,1) = f(i,i) = -1$

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  • $\begingroup$ but the elements of ⟨α| are conjugates of the elements of |α⟩ right? so can't we just directly say that it is greater that equal to 0 or is ⟨α|α⟩≥0 the more fundamental postulate? $\endgroup$
    – DDDAD
    Commented Feb 11, 2018 at 12:25
  • $\begingroup$ @DDDAD Think of special relativity where the (invariants) intervals computed from a scalar product can be timeline, space like or lightlike, i.e. can be $>0, <0$ or $=0$. As pointed out by ValterMoretti, the positive semidefinite property of the inner product depends on the metric used to compute scalar products. $\endgroup$ Commented Feb 11, 2018 at 13:54
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You shouldn't really say "in quantum mechanics", because the definition above (with the addition by Valter Moretti, that it is linear in the second component) is actually just the definition of a scalar product, a very general notion.

There are many possible answers why the scalar product should be positive, but the simplest is that $\langle f, f \rangle$ is supposed to give you the squared norm $\| f\|^2$ of $f$, so some sense of length, or weight, and norms or lengths or weights should be positive. In quantum mechanics, this will turn out to be a probability, and also probabilities should be non-negative real numbers.

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