Are capacity and spectral efficiency identical for a dicrete-time digital signal? I saw this answer and saw on the wikipedia page that both spectral efficiency and capacity can be measured in bpcu i.e. 'bits/symbol'. I just want to confirm then that they are identical for a discrete-time digital signal.

This diagram is plotting $\log_2(1+SNR)$ but referring to it as 'capacity', which confirms my suspicion because this should typically be spectral efficiency before being multiplied with bandwidth to get capacity. I consistently see diagrams showing 'spectral efficiencies' and 'capacities' showing this range of values.
I do not know where the $1/2$ factor came from in the answer I linked, or why they've used variance of the discrete noise in the SNR, but it appears that capacity and spectral efficiency are the same for a digital signal.
My guess would be that 'channel use' means per sample. Imagine an OFDM with 64 subcarriers using 16QAM and pretend none of the carriers are used for pilots or guard bands. The total number of bits per symbol will be 4*64 i.e. 256, but because there are 64 samples in a symbol and hence 64 frequencies used, if you divide that by 64 you get 4. Therefore spectral efficiency = channel use in this scenario?
If 48 of the 64 subcarriers are used to transmit data and the rest are pilots and guard bands, then I'm guessing that the spectral efficiency changes to 192/64 and changes to 192/80 if you include the cyclic prefix i.e. 2.4 bpcu before taking into account error coding of 3/4 where it becomes 1.8bpcu. And modulation efficiency is 4 bpcu now? i.e. 192/48.
Is this because there is no concept of a capacity as distinct from spectral efficiency in a discrete-time signal?
If you calculate the spectral efficiency of the OFDM signal using the bps/Hz method, you get the same result i.e. $\log_2\left(1+\frac{S}{N}\right)$, but the capacity is instead $B\log_2\left(1+\frac{S}{N}\right)$ i.e. 20MHz x 4 bits = 80Mbps. You can also calculate this capacity if you know the capacity in bpcu and know the time interval of the channel use, then you do 4*sample freuquency = 4 x 20MHz = 80Mbps
The capacity is a property of a channel. A channel can transport no more than a certain amount of information per channel use. When we use a fixed modulation and coding scheme (&quot;discrete-time signal&quot;), this scheme will be able to support a certain amount of information per channel use (which for the channel to support it must be $\leq$ its capacity). It is thus strictly speaking incorrect to speak of the capacity of a certain modulation and coding scheme.
For a given modulation and coding scheme such as a QAM, you can of course compute its achievable data rate in bits per channel use (i.e., per unit bandwidth). You can also interpret this as a spectral efficiency, since it measures how efficient the scheme uses bandwidth (how many Mbit/s do I get for each MHz of bandwidth I invest?). And the results you get may be reminiscent of the capacity since under certain conditions, the schemes can get somewhat close to it.
It is still not the same. Capacity is a property of a channel, spectral efficiency is a property of a transmission scheme (modulation and coding scheme).