While showing us the Schrodinger's cat experiment, my physics teacher defined:

$$\varphi_\text{alive} = \begin{bmatrix}1\\0\end{bmatrix},\qquad \varphi_\text{dead} = \begin{bmatrix}0\\1\end{bmatrix}, \qquad \hat{O}\varphi = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\varphi,$$ and $$\Phi = \frac{1}{\sqrt{2}}\varphi_\text{alive} + \frac{1}{\sqrt{2}}\varphi_\text{dead},$$

such that $\hat{O}\varphi_\text{alive} = \varphi_\text{alive}$ and $\hat{O}\varphi_\text{dead} = -\varphi_\text{dead}$.

He later claims that there's a 50% chance that $\hat{O}\Phi = \varphi_\text{alive}$ and a 50% chance that $\hat{O}\Phi = \varphi_\text{dead}$. But according to the definitions, $\hat{O}\Phi$ is just a matrix multiplication whose result is $\begin{bmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{bmatrix}$...

How could doing the same operation to the same vector two times give two different results, both of which are wrong?


Let me rephrase what your teacher did. They first defined an operator $\hat{O}$ given by the matrix you have. They then noticed that the operator has two eigenvectors with eigenvalues $+1$ and $-1$, given by the two vectors that you have. Notice that these eigenvectors are orthonormal.

They then interpreted the two eigenvectors as two states of a quantum system, one corresponding to "alive" and the other one to "dead". Then they supposed that you have a system which is in a specific linear combination of these eigenvectors, that is their wavefunction is the $\Phi$ you have.

They then applied the following two axioms of quantum mechanics, things which we believe are rules of nature and cannot be derived from anything else:

  • The value of the measurement of an observable (operator) is one of its eigenvalues; the system then "collapses" to the corresponding eigenvector.
  • The probability of obtaining a certain eigenvalue is given by the modulus squared of the coefficient of the orthonormal eigenvector corresponding to the eigenvalue in the expansion of the wavefunction.

Your wavefunction is already expanded in terms of orthonormal eigenvectors of $\hat{O}$. By the first axiom, a measurement of $\hat{O}$ can give either $+1$ ("alive") or $-1$ ("dead"). Before the measurement, the wavefunction is given by $\Phi$. Say the measurement yields $+1$. After the measurement, the wavefunction of the system is given by $\Phi'=\varphi_{alive}$. The wavefunction has "collapsed".

By the second axiom, the probability of getting $+1$ is given by $|1/\sqrt{2}|^2=1/2=50\%$ and is the same as the probability of getting $-1$.

In particular, the measurement of an operator is not given by applying the operator to the wavefunction. This works only if the system is in a state which is an eigenvector of the operator. In this case the system has a definite value for that particular operator (which can be for instance, position or momentum). Otherwise the system does not have a definite value for that operator, and when you measure it, you get different outcomes with different probabilities.

  • $\begingroup$ Ok, but "$\hat{O}\Phi = \varphi_\text{alive}$ half the time" is still mathematically wrong, though... Shouldn't we be saying that "after measuring the cat using $\hat{O}$, there is a $\frac{1}{2}$ chance that $\Phi'=\varphi_\text{alive}$" instead of that? $\endgroup$ – user176139 Nov 22 '17 at 1:10
  • $\begingroup$ It seems like when they write $\hat{O}\Phi$ they will later explan this to really be shorthand for $(\varphi_{alive}\cdotp\Phi)^2$, or you know for $(\varphi_{dead}\cdotp\Phi)^2$ too. $\endgroup$ – JMLCarter Nov 22 '17 at 1:23
  • $\begingroup$ .,..but $ \begin{bmatrix}\\\varphi_{alive}\cdotp\Phi\\\varphi_{dead}\cdotp\Phi\end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{bmatrix} $, so pulls out the alive and dead coefficients of $\Phi$ into a vector. The coefficients square to give probability $\endgroup$ – JMLCarter Nov 22 '17 at 1:35
  • $\begingroup$ I mean coefficients*eigenvalues not coefficients. Maybe. $\endgroup$ – JMLCarter Nov 22 '17 at 1:41
  • 1
    $\begingroup$ @user176139 Yes that's right $\endgroup$ – John Donne Nov 22 '17 at 3:13

There seems to be some confusion here. The act of observation (which collapses the superposition to an eigenstate of $\hat O$) is profoundly different from operating on the state with $\hat O$ which doesn't collapse the state but, rather, returns a different superposition of eigenstates.

For example:

$$\hat O \Phi = \frac{1}{\sqrt{2}}\phi_{alive} - \frac{1}{\sqrt{2}}\phi_{dead} \ne \Phi$$


To get the probability that your state $\Phi$ after measurement (opening the box) collapsed into the eigenstate $\varphi_{alive}$ of the $\hat{O}$ operator ($\hat{O}$ is the operator that corresponds to the observable of 'aliveness') is given by $\left|\left < \varphi_{alive} | \, \Phi\right >\right|^2$. Analogously $\left|\left < \varphi_{dead} | \, \Phi\right >\right|^2$ is the probability of collapsing the state into $\varphi_{dead}$. The fancy $\left <a|b \right >$ notation is the standard scalar product.

  • $\begingroup$ I don’t think the operator belongs in this braket, since you want to project your current state $\Phi$ on the one you want arrive in $\varphi_\text{alive}$. $\endgroup$ – Stephan Nov 22 '17 at 1:07
  • $\begingroup$ Indeed, a stupid mistake. $\endgroup$ – DrLRX Nov 22 '17 at 2:51

protected by Qmechanic Nov 22 '17 at 13:13

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