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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?

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    $\begingroup$ Assuming $H$ is a linear operator, then only the knowlege of how to take complex conjugate is needed. $\endgroup$
    – Shuhao Cao
    Apr 19, 2013 at 19:51
  • $\begingroup$ @ShuhaoCao: That sounds like the start of an answer. $\endgroup$
    – orome
    Apr 29, 2013 at 18:59

2 Answers 2

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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.

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  • $\begingroup$ 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)? $\endgroup$
    – orome
    Apr 29, 2013 at 19:38
  • $\begingroup$ @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. $\endgroup$
    – Shuhao Cao
    Apr 29, 2013 at 20:55
  • $\begingroup$ 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.) $\endgroup$
    – orome
    Apr 29, 2013 at 21:05
  • $\begingroup$ @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. $\endgroup$
    – Shuhao Cao
    Apr 29, 2013 at 21:10
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Lets say that $\psi(t)$ maps from the time axis to a complex vector space, and we have an expression for $\partial_t\psi(t)$.

If there is an inner product $\langle\cdot{,}\cdot\rangle$ then we can use it to define a dual space (of covectors that map linearly from the vector space to the complex field). In particular, a time-dependent covector $I_\psi$ may be defined such that $I_\psi(v)=\langle\psi(t)\,{,}\,v\rangle$ for any vector $v$.

The time derivative of this covector shall be another linear map: $$\begin{align} (\partial_t I_\psi )(v) &= \lim_{\epsilon\rightarrow0} \frac{I_{\psi(t+\epsilon)}(v)-I_{\psi(t)}(v)}\epsilon \\&= \lim_{\epsilon\rightarrow0} \frac{\langle\psi(t+\epsilon)\,{,}\,v\rangle-\langle\psi(t)\,{,}\,v\rangle}\epsilon \\&= \lim_{\epsilon\rightarrow0} \frac{\big(\ \langle v\,{,}\,\psi(t+\epsilon)\rangle-\langle v\,{,}\,\psi(t)\rangle\ \big)^*}\epsilon \\&=\lim_{\epsilon\rightarrow0}\ \big\langle v\,{,}\,\frac{\psi(t+\epsilon)-\psi(t)}\epsilon\big\rangle^* \\&=\langle v\,{,}\,\partial_t\psi(t)\,\rangle^* \\&=\langle\ \partial_t\psi(t)\ {,}\ v\,\rangle \\&=I_{\partial_t\psi}(v) \end{align} $$

Now substituting in the Schrödinger equation, $\partial_t\psi(t)=-\frac i \hbar H(\psi)$.

$I_{\partial_t\psi}(v) =\langle-\frac i \hbar H(\psi)\ {,}\ v\,\rangle =(-\frac i \hbar)^*\ \langle H(\psi)\ {,}\ v\,\rangle =+\frac i \hbar\ \langle H(\psi)\ {,}\ v\,\rangle = \frac i \hbar\ \langle \psi\ {,}\ H(v)\rangle = (\frac i \hbar I_\psi H)(v)$

$\therefore \partial_t\,\langle \psi|=\frac i \hbar \langle\psi|H$

This result can be applied to finding the evolution of a density matrix operator $\rho=|\psi\rangle\langle\psi|$.

$$ \begin{align} \partial_t \rho &= \big(\partial_t|\psi\rangle\big)\langle\psi|+|\psi\rangle\big(\partial_t\langle\psi|\big) \\&=\big(-\frac i \hbar H\ |\psi\rangle\big)\langle\psi|+|\psi\rangle\big(\frac i \hbar \ \langle\psi|\ H\big) \\&=-\frac i \hbar\ H \rho +\frac i \hbar\ \rho H \\\therefore i\hbar\ \partial_t\rho&=[H,\rho] \end{align} $$

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