# Inner product not invariant in QM?

For simplicity, consider 2D space. Let $$\{|a\rangle, |b\rangle\}$$ and $$\{|a'\rangle, |b'\rangle\}$$ be two sets of basis kets.

Now the kets $$|a\rangle, |b\rangle$$ can be represented in its basis $$\{|a\rangle, |b\rangle\}$$ by the column vectors: $$|a\rangle = [1,0]^T, |b\rangle = [0,1]^T$$. Similarly, the kets $$|a'\rangle, |b'\rangle$$ can be represented in its basis $$\{|a'\rangle, |b'\rangle\}$$ by the column vectors: $$|a'\rangle = [1,0]^T, |b'\rangle = [0,1]^T$$.

Now the rule for finding the bra corresponding to any ket is to take the transpose-conjugate of the ket's column vector (Principles of QM by R. Shankar, Chapter 1). For e.g. $$\langle a| = [1,0]$$ and so on. The inner product is then $$\langle a|a\rangle = [1,0]*[1,0]^T=1$$. Similarly, it turns out that $$\langle b|b\rangle = 1, ~\langle a|b\rangle = 0$$. Obviously the same holds for the basis $$\{|a'\rangle, |b'\rangle\}$$ when its components are written in its own basis. This means that any arbitrarily chosen set of basis vectors is guaranteed to be orthonormal!

But this gives rise to further problems. Say $$|a\rangle = |a'\rangle + \alpha |b'\rangle, ~|b\rangle = |a'\rangle + \beta |b'\rangle$$, where $$\alpha, \beta$$ are numbers. Then $$\langle a|a\rangle = 1+|\alpha|^2$$, which for general $$\alpha$$ implies that $$|a\rangle$$ is not normalized, contrary to our earlier conclusion. In fact, for general $$\alpha, \beta$$ the basis $$\{|a\rangle, |b\rangle\}$$ turns out to be neither orthogonal nor normalized. In other words, it appears that the inner product as defined is not invariant with respect to a change of basis.

This problem is avoided in regular geometry by the intervention of a metric tensor $$\textbf{g}$$ which maps a vector to its dual (or 1-form): $$\textbf{g}(|a\rangle)\rightarrow\langle a|$$. This is not a component-level operation (like the prescription to take transpose-conjugate of a column vector in a particular basis) and actually renders the inner product invariant.

But since in QM the inner product is written as $$\langle a|b\rangle$$ without reference to any basis, it implicitly implies such an invariance. (And wouldn't an inner product without such an invariance be useless?) Any reply to clear up the confusion is much appreciated.

• "Now the rule for finding the bra corresponding to any ket is to take the transpose-conjugate of the ket's column vector..." <-- nope. This is only true in an orthonormal basis. Commented Nov 17, 2023 at 14:06
• First, a small mistake: $\langle a | a \rangle = 1 + |\alpha|^2$. Secondly, your transformation is not unitary, so it's not surprising that it doesn't preserve the inner product. In fact, as it's currently written, the transformation would be unitary iff $\alpha = 0$ and $\beta = 0$, i.e. the basis would be identical. Commented Nov 17, 2023 at 14:23
• @Prahar Assuming an orthogonal set of kets, the problem of normalization still remains. Please see my comment to Er Jio's answer.
– Deep
Commented Nov 18, 2023 at 6:37
• @mb28025 Edited, thanks. I am making a point about an arbitrary basis, whose kets are linearly independent and not necessarily orthogonal or normalized. If you limit yourself to bases obtained from each other by unitary transformations then obviously the inner product is preserved, by definition of a unitary transformation -- not an interesting case for me.
– Deep
Commented Nov 18, 2023 at 6:41
• This is the confusion I described in this answer for the specific case of $T$ being the matrix representation of the inner product. Commented Nov 18, 2023 at 19:39

The inner product is invariant, and your contradiction comes from an incorrect method of finding bra component vectors the fact that the inner product does not correspond to multiplication of components.

Now the rule for finding the bra corresponding to any ket is to take the transpose-conjugate of the ket's column vector

The rule for finding the row vector of the bra corresponding to any ket is to take the conjugate-transpose of the ket's column vector. An important distinction is that the column and row vectors corresponding to kets and bras are not equal to those kets and bras. Kets are elements of a Hilbert space $$\mathcal H$$ and bras are elements of the dual space $$\mathcal H^*$$. The corresponding column and row vectors of the components in a given basis are representations which belong to the row and column matrix spaces $$\mathbb C^{1\times n}$$ and $$\mathbb C^{n\times1}$$ (for an $$n$$ dimensional $$\mathcal H$$).

A general 2D ket $$|\psi\rangle$$ can be expanded in terms of a (not necessarily orthonormal) basis $$B = \{|a\rangle,|b\rangle\}$$ as

$$|\psi\rangle = \psi_a|a\rangle + \psi_b|b\rangle$$ This can be represented by a column vector of its components in this basis $$|\psi\rangle_B = \begin{bmatrix} \psi_a \\ \psi_b \end{bmatrix} \in \mathbb C^{2\times1}$$

The corresponding bra $$\langle\psi|$$ can be expanded in terms of the bra basis $$B^* = \{\langle a|,\langle b|\}$$ as $$\langle \psi| = \langle a|\psi_a^* + \langle b|\psi_b^*$$ This can be represented by a row vector of its components in this basis $$\langle \psi|_{B^*} = [\psi_a^*,\psi_b^*]\in \mathbb C^{1\times2}$$

which is indeed the conjugate-transpose the column representation of $$|\psi\rangle$$.

However it is important to note that the inner product of a bra and a ket is not equivalent to matrix multiplication of their row and column representations. We can see this by expanding the inner product $$\langle\psi|\psi\rangle$$ in terms of the bases

$$\langle\psi|\psi\rangle = (\langle a|\psi_a^* + \langle b|\psi_b^*)(\psi_a|a\rangle + \psi_b|b\rangle) = \psi_a^*\psi_a\langle a|a\rangle + \psi_a^*\psi_b\langle a|b\rangle + \psi_b^*\psi_a\langle b|a\rangle +\psi_b^*\psi_b\langle b|b\rangle$$

which in general is different to $$(\langle\psi|_{B^*})(|\psi\rangle_B) =[\psi_a^*,\psi_b^*]\begin{bmatrix}\psi_a \\ \psi_b\end{bmatrix} = \psi_a^*\psi_a + \psi_b^*\psi_b$$

The difference depends on the value of the inner products, and its only in an orthonormal basis, where $$\langle a|a\rangle = 1,\langle b|b\rangle = 1, \langle a|b\rangle = 0, \langle b|a\rangle = 0$$, that the multiplication of components is equal to the inner product.

• Your answer partly clears my confusion. There are still two questions however. (1) Say the basis $\{|a\rangle, |b\rangle\}$ is orthogonal but otherwise arbitrary. We still have the conclusion $\langle a|a \rangle =1=\langle b|b \rangle$ when the components are written in its own basis. This conclusion is violated when we switch basis (see my question for details). (2) How do we even decide if $\{|a\rangle, |b\rangle\}$ are orthogonal? In geometry, to know whether two vectors $u,v$ are orthogonal we have the explicit recipe $g_{ij}u^iv^j$ where $g$ is the metric tensor. What is it in QM?
– Deep
Commented Nov 18, 2023 at 6:34
• @Deep I realized when answering this that finding the row vector for a bra by conjugate transposing the column of the ket is correct, but equating the matrix multiplication of the row and column vectors to the inner product is wrong. I've edited my answer accordingly. This directly addresses your first question: Yes the columns representing the vectors in their own basis are $|a\rangle = [1,0]^T, |b\rangle = [0,1]^T$, but the inner product is not generally given by the multiplication of the components. Second, in QM 2 vectors are orthogonal if the inner product is 0: $\langle a|b\rangle = 0$ Commented Nov 18, 2023 at 19:37

You said-"This means that any arbitrarily chosen set of basis vectors is guaranteed to be orthonormal!".... It isn't correct, you can have basis vectors which are non orthogonal for instance.. in linear algebra the criteria for being basis basis vector is neither orthogonality nor normalised( we do it to make the math easy and tidy so that we can focus more on the physics)... But regardless, you can look up coherent basis states which aren't orthogonal.

What you are doing here is taking the inner product of the respective basis ket .. suppose if you define |a'⟩ =x|a⟩ + y|b⟩, it is a different ket (linear combination of |a⟩ and |b⟩, it's inner product with itself is bound to be different.

Now if you define a |$$\psi$$⟩ with basis ket |a⟩ |b⟩ and if you give a unitary transformation to the basis , the inner product of |$$\psi$$⟩ with itself will remain same.(check the comments)