# Derivative of a bra?

I understand that

$$\frac{\mathrm d}{\mathrm dt} \langle\psi|\psi\rangle =\left[\frac{\mathrm d}{\mathrm dt} \langle\psi|\right]|\psi\rangle + \langle\psi|\left[\frac{\mathrm d}{\mathrm dt}|\psi\rangle\right]$$

and that this can be established by a direct application of the definition of the derivative; but it is not clear to me why

$$\frac{\mathrm d}{\mathrm d t} |\psi\rangle \;=\; -i H |\psi\rangle\implies\left[\frac{\mathrm d}{\mathrm dt} \langle\psi|\right]=i\langle\psi|H^\dagger$$

given only a linear vector space and the properties of an inner product.

Should I be able to derive the above conclusion from just the properties of a linear vector space with an inner product, or is more needed?

• Assuming $H$ is a linear operator, then only the knowlege of how to take complex conjugate is needed. – Shuhao Cao Apr 19 '13 at 19:51
• @ShuhaoCao: That sounds like the start of an answer. – orome Apr 29 '13 at 18:59

Since the bra-vector is just the Hermitian conjugate of the ket-vector$\newcommand{\ddt}{\frac{\mathrm d}{\mathrm dt}}$ $$\langle\psi| = |\psi\rangle^\dagger$$ $$\ddt \langle\psi| = \ddt |\psi\rangle^\dagger =\left(\ddt |\psi\rangle\right)^\dagger= \Big(-i H |\psi\rangle\Big)^\dagger =-|\psi\rangle^\dagger H^\dagger i^\dagger = i\langle\psi|H^\dagger$$ If we assume $H$ is a linear operator defined on proper space.

For example: If $|\psi\rangle = (c_1,\ldots,c_n)^T$ is a column vector with $n$ complex entries, then $\langle\psi| = (c_1^*,\ldots,c_n^*)$ is a row vector whose entries are the complex conjugate of $c_i$, and then $H$ is an $n\times n$ complex matrix. If $|\psi\rangle$ is a function in $H^1_0$, then $\langle\psi|$ is a bounded linear functional on $H^1_0$, and $H$ is a bounded linear operator like Laplacian.

EDIT: As OP pointed out, the reason why bra-vector is the Hermitian conjugate of the ket-vector is Riesz representation theorem. In short: any Hilbert and its dual space has a 1-1 correspondence, moreover the correspondence is nice in that it preserves the norm (isometry). If $|\psi\rangle \in V$, then its dual is a bounded linear function on $V$, the space we denote it as $V'$. For any $|\psi\rangle$, we can associate a unique linear functional $l_{\psi}$, and $\langle\psi|\phi\rangle:= l_{\psi}(\phi)$, in other words, $\langle\psi|$ is this linear functional, just different notation.

A translation from physics:

If using the functional analysis notation, translating from bra-ket notation, the problem should be:

Known $\displaystyle \ddt \psi = -i H\psi$, find $\displaystyle \ddt l_{\psi} =?$ where $l_{\psi}$ is defined as: $l_{\psi}(\phi) = \langle \psi,\phi\rangle$, and $\langle \cdot,\cdot\rangle$ is the inner product on the Hilbert space $V$ involved.

For any $\phi \in V$, by the definition induced by Riesz: $$\ddt l_{\psi}(\phi) = \ddt\langle \psi,\phi\rangle$$ A possible choice for the inner product is the integration of the first input's complex conjugate times second input with respect to properly defined measure (it doesn't have to be this, can be others): $$\langle f,g\rangle = \int_X f\overline{g}$$ Then: $$\ddt\langle \psi,\phi\rangle = \langle \ddt\psi,\phi\rangle = \langle-i H\psi,\phi\rangle$$ If the Hermitian conjugate (adjoint) of the linear operator $H$ is denoted as $H^*$: $$\langle-i H\psi,\phi\rangle = \langle \psi,(iH^*)\phi\rangle = l_{\psi}(iH^*\phi)$$ Therefore: $$\ddt l_{\psi}(\phi) = l_{\psi}(iH^*\phi), \quad \forall \phi\in V.$$ This is the same as: $$\ddt\langle\psi| = i\langle\psi|H^\dagger$$ just one using functional analysis notation and the other using bra-ket notation.

• So the answer depends on this, which (though I accepted) I'm not sure I completely grasp. How from the required properties of the inner product alone (not the matrix representation) do we know that a bra is the adjoint of a ket (I'm willing to have another go at an answer to that question)? – orome Apr 29 '13 at 19:38
• @raxacoricofallapatorius Hi, I added more in my answer. The link you gave is about the Riesz representation theorem, though that guy seemed not mentioning this theorem's name directly. – Shuhao Cao Apr 29 '13 at 20:55
• Thanks, that gets me much closer to understanding. (Though I'll need to think about it: I think my questions make me seem better educated in these matters than I really am.) – orome Apr 29 '13 at 21:05
• @raxacoricofallapatorius Taking a functional analysis course would get your questions resolved completely (hopefully but it may raise more questions) :) I myself found the book "quantum field theory: a tourist guide for mathematicians" extremely useful for me to understand physics in mathematical language that I am used to immerse myself with. – Shuhao Cao Apr 29 '13 at 21:10