Inner product of $\langle \phi | \psi \rangle$ gives a complex value - why/meaning?

Question

Say I have the following two wave-functions:

$$ |\psi\rangle= \pmatrix{a_1+i b_1\\ a_2+ib_2 } $$

$$ |\phi\rangle= \pmatrix{c_1+i d_1\\ c_2+id_2 } $$

Since these are unit vectors of the Hilbert space, then it follows that this is true:

$$ \langle \psi | \psi \rangle=1 $$

$$ \langle \phi | \phi \rangle=1 $$

I was surprised when I tried the following:

$$ \begin{align} \langle \phi | \psi \rangle&= \pmatrix{c_1-i d_1 & c_2-id_2 } \pmatrix{a_1+i b_1\\ a_2+ib_2 }\\ &=c_1a_1+c_1ib_1-id_1a_1+d_1b_1+c_2a_2+c_2ib_2-id_2a_2+d_2b_2\\ &=(c_1a_1+d_1b_1+c_2a_2+d_2b_2)+\underbrace{i(c_1b_1-d_1a_1+c_2b_2-d_2a_2)}_{\text{imaginary part?}} \end{align} $$

Am I doing something wrong? Why is $\langle \phi | \psi \rangle$ not giving me $1$ like the other inner products? What is the physical interpretation of $\langle \phi | \psi \rangle$? Looking at the axioms of Dirac-Von-Neumman https://en.wikipedia.org/wiki/Dirac–von_Neumann_axioms it appears only inner products such as $\langle \psi | \psi \rangle$ are given a physical meaning - do products such as these $\langle \phi | \psi \rangle$ simply don't have physical meaning?

edit:

As suggested in the answers I will now take $\langle \psi | \phi \rangle$

$$ \begin{align} \langle \phi | \psi \rangle&= \pmatrix{a_1-i b_1 & a_2-ib_2 } \pmatrix{c_1+i d_1\\ c_2+id_2 }\\ &=a_1c_1+a_1id_1-ib_1c_1+b_1d_1+a_2c_2+a_2id_2-ib_2c_2+b_2d_2\\ &=a_1c_1+b_1d_1+a_2c_2+b_2d_2+a_1id_1-ib_1c_1+a_2id_2-ib_2c_2\\ &=(a_1c_1+b_1d_1+a_2c_2+b_2d_2)-i(b_1c_1-a_1d_1-a_2d_2+b_2c_2) \end{align} $$

Then, it follows that:

$$ \langle \psi | \phi \rangle\langle \psi | \phi \rangle=(a_1c_1+b_1d_1+a_2c_2+b_2d_2)^2+(b_1c_1-a_1d_1-a_2d_2+b_2c_2)^2 $$

What is then the physical interpretation of this?

Answer 1

As noted in Mike Stone's answer this is explained in any quantum mechanics textbook. The inner product is a complex inner product, in general:

$$\langle \phi|\psi\rangle\in \Bbb C \tag{1}.$$

In quantum mechanics the probability - given a system in the state $|\psi\rangle$ - of finding it after measurement in the state $|\phi\rangle$ (which will be an eigenstate of the observable in question) is given by:

$$|\langle \phi|\psi\rangle|^2\in \Bbb R. \tag{2}$$

Note that this quantity (the square magnitude of a complex number) is positive and real, exactly what you would want for something we are expecting to interpret as a probability.

Answer 2

I'm not sure why you'd expect $\langle \phi|\psi\rangle$ to be equal to $1$.

In elementary mechanics, we deal with the vector spaces $\mathbb R^2$ and $\mathbb R^3$. If you have two vectors $\vec a$ and $\vec b$, $\vec a \cdot \vec a = \vec b \cdot \vec b = 1$ means that $\vec a$ and $\vec b$ both have magnitude $1$, but if I gave you no other information it would be silly to assume that $\vec a \cdot \vec b = 1$, right? The dot product could be $1,-1$, or anything in between.

The vector space you're talking about is not $\mathbb R^2$ but rather $\mathbb C^2$. Because this vector space is complex, the inner product

$$\left<\pmatrix{a\\b}\bigg|\pmatrix{c\\d}\right> = a^*c + b^* d$$

is generally complex-valued, as you saw, though it's easy to see that the inner product of any vector with itself is both real and non-negative.

In exactly the same way as $\mathbb R^2$, the magnitude of a vector $|\psi\rangle \in \mathbb C^2$ is $\Vert \psi\Vert = \sqrt{\langle \psi|\psi\rangle}$. Just as before, the inner product of two unit vectors is typically not $1$ (which would indicate that they were in fact the same vector).

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It's hard to give a comprehensive list of physical interpretations here, because this is just the basic algebra of a Hilbert space (loosely, just a vector space with an inner product). In an abstract sense, the inner product of two vectors has the same interpretation as it does in $\mathbb R^2$ - e.g. that the inner product is a scalar product which depends on how much the two vectors overlap, and is equal to zero if they are orthogonal. In fact this is slightly backward, since the definition of orthogonality is that the inner product is equal to zero.

In quantum mechanics, if $|\psi\rangle$ represents the state the system is in and $|\phi\rangle$ represents an eigenstate of some observable $\hat A$ with eigenvalue $\lambda$, then $|\langle \phi|\psi\rangle|^2$ is essentially the probability that a measurement of $\hat A$ would yield the result $\lambda$.

There could be other physical interpretations depending on what $|\phi\rangle$ and $|\psi\rangle$ actually are.

Answer 3

The physical quantity $|\langle \psi|\phi\rangle|^2= \langle \phi|\psi\rangle \langle \psi|\phi\rangle$ is real. Its interpretation is explained in any Quantum Mechanics textbook.

Answer 4

An inner product (more specifically a positive definite inner product) defines the notions of length of a vector and projection/angle of one vector along another. While length can easily be real and positive $$\langle \psi | \psi \rangle \in \bf R^+$$ which follows from the properties of the inner product, the inner product between two different vectors inevitably has to allow imaginary values over the complex field. Suppose you had two vectors $\psi,\phi$ with $$\langle \psi|\phi\rangle \in \bf R$$ then $\phi^\prime = i\phi$ is also an allowed vector (due to complex completeness of the vector space), and hence, $$\langle \psi|\phi^\prime\rangle= i\langle \psi|\phi\rangle=imaginary$$ However, if the vectors are orthogonal, imaginary (or whatever) scaling does not change anything about that: $$0=\langle \psi|\phi^\prime\rangle= i\langle \psi|\phi\rangle=0$$