I am a reading a book for beginners of the quantum mechanics. In one section, the author shows the inner product of two wave functions $\langle\alpha\vert\beta\rangle$. I am wondering what's the significance of that product? I Googled that and someone call that probability amplitude, but that product could be complex, so does it tell any phyical significance?

In terms of quantum interference, shall we add the wavefunctions or the probability amplitudes before taking the square of modulus? Sorry I am just starting on learning the quantum mechanics and many concepts are pretty confusing to me.

  • $\begingroup$ This does seem a bit odd as described. Perhaps if you give the name of the text, or show the relevant equations and their context, we would be in a better position to help. I can think of a couple plausible explanations, but it's hard to be sure without context. $\endgroup$ – user10851 Dec 17 '12 at 5:17

Suppose you have some linear algebra background. The most important thing you need to know is that the inner product has the same meaning of what you have learnt in linear algebra class. The inner product $$\left\langle \phi|\psi\right\rangle =\int\phi^{*}(x)\psi(x)dx$$

has the meaning related to a projection of one vector onto another vector (for true projection, the wavefunctions needed to be normalized). It is similar to the projection of a three dimensional vector $\mathbf{\vec{v}}=a\hat{\mathbf{x}}+b\hat{\mathbf{y}}+c\hat{\mathbf{z}}$ onto another unit vector $\mathbf{\hat{x}}$ which gives you the results $\mathbf{\vec{v}}\cdot\mathbf{\hat{x}}=a$.

First, the inner product can give you the "length square" of the wavefunction: $$\left\langle \psi|\psi\right\rangle =\int\psi^{*}(x)\psi(x)dx =\int|\psi(x)|^2dx$$ similar to the $\mathbf{\vec{v}}\cdot\mathbf{\vec{v}}=a^{2}+b^{2}+c^{2}$ , so you can normalize your wavefunction by the condition $\left\langle \psi|\psi\right\rangle =1$.

Second, it allows you to show that two wavefunctions are orthogonal to each other, given by the condition that the inner product evaluated to zero $\left\langle \phi|\psi\right\rangle =0$ which is the analog to the $\mathbf{\hat{x}}\cdot\mathbf{\hat{y}}=0$.

Third, if we write the wavefunction $\psi(x)$ as a linear combination of orthonormal wavefunctions $\psi_{n}(x)$: $$\psi(x)=\sum_{n}c_{n}\psi_{n}(x)$$ similar to a general vector in linear algebra, then we will have the inner product $\left\langle \psi_{n}|\psi\right\rangle =c_{n}$. The meaning of $c_{n}$ is the probability amplitude and it is a complex number in general. So the probability $p_{n}$ of the wavefunction $\psi$ having the component $\psi_{n}$ is given by $p_{n}=|c_{n}|^{2}=|\left\langle \psi_{n}|\psi\right\rangle |^{2}$. The meaning here is very important when you learn how to preform measurement.

Lastly, you should add two wavefunctions amplitude together before you take the square, similar to adding the amplitude of two water waves. More precisely, if the new wavefunction is $\psi(x) = A[\psi_{a}(x)+\psi_{b}(x)]$, then the probability density at position $x$ is $A^2|\psi_{a}(x)+\psi_{b}(x)|^{2}$. Note that $A$ is the normalization constant given by the condition $\left\langle \psi|\psi\right\rangle=1$. It is where the quantum effect arise. Dont take the square and then add them together.

  • $\begingroup$ Thanks hwlau. In you last statement, you said I should add wavefunctions amplitude together before taking the sqaure. So does it mean something like following: |<\phi_a+\phi_b|\phi_b+\phi_b>|^2, so it gives |<\phi_a|\phi_a> + <\phi_a|\phi_b> + <\phi_b|\phi_a> + <\phi_b|\phi_b>|^2, am I right? $\endgroup$ – user1285419 Dec 18 '12 at 21:24
  • $\begingroup$ No, when you take the inner product, it is already the "length square". If you take square again, then you have "length to power 4". Also, when I say "length", it is the "length" of the whole wave function, not a particular local region (x,x+dx). You also remind me that I have missing a normalization factor when adding two wavefunction. See the edit. $\endgroup$ – unsym Dec 19 '12 at 5:15
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    $\begingroup$ Let me elaborate on one point that hwlau made. Suppose I prepare a quantum system in state $|\beta\rangle$, and then send it through a filter that picks out state $|\alpha\rangle$. For instance, the states could refer to polarization of light, and the filter could be a polarizer. What's the probability by state will make it through the filter? Answer: $P = |\langle \beta | \alpha \rangle|^2$. $\endgroup$ – emarti Dec 19 '12 at 6:24
  • $\begingroup$ To hwlau, thanks for your explain and sorry for my dummy question. I am confusing in your statement "when you take the inner product, it is already the "length squaure"", so to my understanding, $\langle\alpha|\beta\rangle$ denote the inner product, but it could be complex. For example, if $|\alpha\rangle=e^{-i\alpha}|\beta\rangle$, then $\langle\alpha|\beta\rangle=e^{i\alpha}$. So I don't understand why inner product is the length square here. Could you please explain more, thanks. $\endgroup$ – user1285419 Dec 19 '12 at 17:25
  • $\begingroup$ @user1285419 This is not dummy question, it is just a basic question. Inner product is complex in general. However, I say it is "length square" because of what you write in the first comment, this is the meaning of $<\psi|\psi>$ when the wavefunction $\psi$ in the bracket are the same. In this case, the result are always a positive real number (see Eq 2) which means length square. Also, you should note that when people talk about wavefunction, it is almost always means that it is normalized which means that $<\psi|\psi>=1$ (just like what emarti implied in his comment) $\endgroup$ – unsym Dec 19 '12 at 23:07

The dot product you mention is the probability amplitude of one of the states transforming into another. The actual probability is obtained by the norm of the amplitude. This is what happens all the time in quantum physics.

  • $\begingroup$ Also, quoting Auletta, "the expression $\langle\psi|U_t|\varphi\rangle$ is the probability amplitude that, given an initial state $|\varphi\rangle$ , measures how close it evolves unitarily to a final state $|\psi\rangle$ at time $t$" where $U_t=e^{\displaystyle -\frac{i}{\hbar}\hat{H}(t-t_0)}$. $\endgroup$ – Paracosmiste Sep 1 '14 at 13:54

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